chapter3.tex

\section{Chapter III\hspace{0.2em} Rings and modules}

\subsection{\textsection1. Definition of ring}
\begin{problem}[1.1]
	$\vartriangleright$ Prove that if $0 = 1$ in a ring $R$, then $R$ is a zero-ring. [\textsection1.2]
\end{problem}
\begin{solution}
	For any $x$ in the ring $R$, we have
	\[
	1\cdot x=x,\qquad 0\cdot x=0.
	\]
	Since $0 = 1$ we see that $x=0$, which implies $R$ is a ring with only one element $0$.
\end{solution}


\hypertarget{Exercise III.1.2}{}
\begin{problem}[1.2]
	$\neg$ Let $S$ be a set, and define operations on the power set $\mathscr{P}(S)$ of $S$ by setting $\forall A,B \in \mathscr{P}(S)$
	\[
	A+B :=(A \cup B) \backslash(A \cap B) \quad, \quad A \cdot B=A \cap B
	\]
	Prove that $(\mathscr{P}(S),+,\cdot)$ is a commutative ring. [2.3, 3.15]
\end{problem}
\begin{solution}
	First, we need to check that $(\mathscr{P}(S),+)$ is an abelian group:
	\begin{itemize}
		\item associativity:	
		\begin{align*}
		&\hspace{1em}(A+B)+C\\
		&=((A \cup B) \backslash(A \cap B))+C\\
	    &=((A \cup B) \cap(A^C \cup B^C))+C\\
	    &=(A\cap(A^C \cup B^C) )\cup (B\cap(A^C \cup B^C)) +C\\
	    &=(A \cap B^C) \cup(A^C \cap B)+C\\
	    &=(((A \cap B^C) \cup(A^C \cap B)) \cap C^C) \cup(((A \cap B^C) \cup(A^C \cap B))^C \cap C)\\
	    &=((A \cap B^C\cap C^C )\cup(A^C \cap B\cap C^C) ) \cup((A^C \cup B) \cap(A\cup B^C)\cap C)\\
	    &=((A \cap B^C\cap C^C )\cup(A^C \cap B\cap C^C) ) \cup((A^C\cap B^C) \cup( A\cap B) \cap C)\\
	    &=(A \cap B^C\cap C^C )\cup(A^C \cap B\cap C^C)  \cup(A^C\cap B^C\cap C) \cup( A\cap B\cap C) \\
	    &=(A \cap (B \cap C) \cup(B^C \cap C^C)) \cup((A^C \cap B \cap C^C) \cup (A^C \cap B^C \cap C))\\
	    &=(A \cap (B^C \cup C) \cap(B \cup C^C)) \cup((A^C \cap B \cap C^C) \cup (A^C \cap B^C \cap C))\\
	    &=(A \cap ((B \cap C^C) \cup(B^C \cap C))^C) \cup(A^C \cap ((B \cap C^C) \cup(B^C \cap C)))\\
	    &=A+((B \cap C^C) \cup(B^C \cap C))\\
	    &=A+(B+C);
		\end{align*}
		\item commutativity:
		\[
		A+B =(A \cup B) \backslash(A \cap B)=(B \cup A) \backslash(B \cap A)= B+A;
		\]
		\item additive identity: the additive identity is $\varnothing$ since
		\[
		A+\varnothing=(A \cup \varnothing) \backslash(A \cap \varnothing)=A; \backslash\varnothing=A
		\] 
		\item inverse: the inverse of some set $A$ is just itself since
		\[
		A+A=(A \cup A) \backslash(A \cap A)=A\backslash A=\varnothing.
		\]
	\end{itemize}
	Then we have to show that $(\mathscr{P}(S),\cdot)$ is a commutative monoid, which clearly holds with the multiplicative identity $S$. What is left to show is the distributive properties and the check is straightforward.
	\begin{align*}
	&\hspace{1em}(A+B)\cdot C\\
	&=((A \cap B^C) \cup(A^C \cap B))\cap C\\
	&=(A \cap B^C\cap C) \cup(A^C \cap B\cap C)\\
	&=(A\cap C \cap (B^C\cup C^C)) \cup((A^C\cup C^C) \cap (B\cap C))\\
	&=(A\cap C \cap (B\cap C)^C) \cup((A\cap C)^C \cap (B\cap C))\\
	&=A\cdot C+B\cdot C.
	\end{align*}
\end{solution}

\begin{problem}[1.3]
	$\neg$ Let $R$ be a ring, and let $S$ be any set. Explain how to endow the set $R^S$ of set-functions $S\to R$ of two operations $+$, $\cdot$ so as to make $R^S$ into a ring, such that $R^S$ is just a copy of $R$ if $S$ is a singleton. [2.3]
\end{problem}
\begin{solution}
	To make $(R^S,+,\cdot )$ a ring , for all $f,g\in R^S$ we define addition and multiplication as
	\begin{align*}
	f+g&:S\longrightarrow R,\quad x\longmapsto f(x)+g(x)\\
	f\cdot g&:S\longrightarrow R,\quad x\longmapsto f(x)\cdot g(x).
	\end{align*}
\end{solution}

\hypertarget{Exercise III.1.4}{}
\begin{problem}[1.4]
	$\vartriangleright$ The set of $n\times n$ matrices with entries in a ring $R$ isdenoted $\mathcal{M}_n(R)$. Prove that componentwise addition and matrix multiplication makes $\mathcal{M}_n(R)$ into a ring, for any ring $R$. The notation $\mathfrak{gl}_n(R)$ is also commonly used, especially $R=\mathbb{R}$ or $\mathbb{C}$ (although this indicates one is considering them as \emph{Lie algebras}) in parallel with the analogous notation for the corresponding groups of units, cf. \hyperlink{Exercise II.6.1}{Exercise II.6.1}. In
	fact, the parallel continues with the definition of the following sets of matrices:
	\begin{itemize}
		\item $\mathfrak{sl}_n(\mathbb{R}) = \{M \in \mathfrak{gl}_n(\mathbb{R}) | \mathrm{tr}(M) = 0\}$;
		\item $\mathfrak{sl}_n(\mathbb{C}) = \{M \in \mathfrak{gl}_n(\mathbb{C}) | \mathrm{tr}(M) = 0\}$;
		\item $\mathfrak{so}_n(\mathbb{R}) = \{M \in \mathfrak{sl}_n(\mathbb{R}) |M +M^t = 0\}$;
		\item $\mathfrak{su}(n) = \{M \in \mathfrak{sl}_n(\mathbb{C}) |M +M^\dag = 0\}$.
	\end{itemize}
	Here $\mathrm{tr}(M)$ is the trace of $M$, that is, the sum of its diagonal entries. The other notation matches the notation used in \hyperlink{Exercise II.6.1}{Exercise II.6.1}. Can we make rings of these sets, by endowing them of ordinary addition and multiplication of matrices? (These sets are all Lie algebras, cf. \hyperlink{Exercise VI.1.4}{Exercise VI.1.4}.) [\textsection1.2, 2.4, 5.9, VI.1.2, VI.1.4]
\end{problem}
\begin{solution}
	It is plain to show $\mathcal{M}_n(R)$ is a ring according to the definition. For multiplicative associativity, it follows that for all $A,B,C\in\mathcal{M}_n(R)$,
	\begin{align*}
	&\hspace{1em}((A B) C)_{\alpha, \delta}\\
	&=\sum_{i=1}^{n}(A B)_{\alpha, i} c_{i, \delta}\\
	&=\sum_{i=1}^{n}\left(\sum_{j=1}^{n} a_{\alpha, j} b_{j, i}\right) c_{i, \delta}\\
	&=\sum_{i=1}^{n} \sum_{j=1}^{n}\left(a_{\alpha, j} b_{j, i}\right) c_{i, \delta}\\
	&=\sum_{j=1}^{n} \sum_{n=1}^{n} a_{\alpha, j}\left(b_{j, i} c_{i, \delta}\right)\\
	&=\sum_{j=1}^{n} a_{\alpha, j}\left(\sum_{i=1}^{n} b_{j, i} c_{i, \delta}\right)\\
	&=\sum_{j=1}^{n} a_{\alpha, j}(B C)_{j, \delta}\\
	&=(A(B C))_{\alpha, \delta}.
	\end{align*}
	Under the ordinary addition and multiplication of matrices, $\mathfrak{sl}_n(\mathbb{R}),\mathfrak{sl}_n(\mathbb{C}),\mathfrak{so}_n(\mathbb{R}),\mathfrak{su}_n(\mathbb{C})$ are not rings. In fact, they are not closed under the multiplication.
\end{solution} 

\begin{problem}[1.5]
	Let $R$ be a ring. If $a, b$ are zero-divisors in $R$, is $a+b$ necessarily a zero-divisor?
\end{problem}
\begin{solution}
	That is not true. Let's take $\mathbb{Z}/6\mathbb{Z}$ as an counterexample. Though both $[2]_6$ and $[3]_6$ are zero-divisors, their sum $[5]_6$ is not a zero-divisor.
\end{solution}

\hypertarget{Exercise III.1.6}{}
\begin{problem}[1.6]
	$\neg$ An element $a$ of a ring $R$ is \emph{nilpotent} if $a^n = 0$ for some $n$.
	\begin{enumerate}
	\item Prove that if $a$ and $b$ are nilpotent in $R$ and $ab = ba$, then $a+b$ is also nilpotent.
	\item Is the hypothesis $ab = ba$ in the previous statement necessary for its conclusion to hold?
	\end{enumerate}
[3.12]
\end{problem}
\begin{solution}
	\begin{enumerate}
		\item Assume that $a^n=b^m=0$ and let $k=2\max\{n,m\}$. If $ab = ba$, we can get
		\[
		(a+b)^k=\sum_{p=0}^{\tfrac{k}{2}}\binom{k}{p}a^kb^{k-p}+\sum_{p=\tfrac{k}{2}+1}^{k}\binom{k}{p}a^kb^{k-p}=\sum_{p=0}^{\tfrac{k}{2}}\binom{k}{p}a^k\cdot 0+\sum_{p=\tfrac{k}{2}+1}^{k}\binom{k}{p}0\cdot b^{k-p}=0,
		\]
		which means $a+b$ is also nilpotent.
		\item The hypothesis $ab = ba$ is necessary. A counterexample can be found in the ring $\mathfrak{gl}_2(\mathbb{R})$. Let
		\[
		a=\left(
		\begin{matrix}	
		0 & 1\\	
		0 & 0	
		\end{matrix}
		\right),\quad
		b=\left(
		\begin{matrix}	
		0 & 0\\	
		1 & 0	
		\end{matrix}
		\right)
		\]
		and then we have $a^2=b^2=0$. In other words, $a$ and $b$ are nilpotent. However, by diagonalization we see that
		\[
		(a+b)^n=
		\left(
		\begin{matrix}	
		0 & 1\\	
		1 & 0	
		\end{matrix}
		\right)^n
		=\left(
		\begin{matrix}	
		-1 & 1\\	
		1 & 1	
		\end{matrix}
		\right)
		\left(
		\begin{matrix}	
		-1 & 0\\	
		0 & 1	
		\end{matrix}
		\right)^n
		\left(
		\begin{matrix}	
		-1 & 1\\	
		1 & 1	
		\end{matrix}
		\right)^{-1}\ne 
		\left(
		\begin{matrix}	
		0 & 0\\	
		0 & 0	
		\end{matrix}
		\right).
		\]
		Thus in such case, $a+b$ is no longer nilpotent.
	\end{enumerate}
\end{solution}

\begin{problem}[1.8]
	Prove that $x = \pm1$ are the only solutions to the equation $x^2 = 1$ in an integral	domain. Find a ring in which the equation $x^2 = 1$ has more than 2 solutions.
\end{problem}
\begin{solution}
	It clearly holds that $1\cdot1=1$ and $(-1)\cdot(-1)=((-1)\times(-1))1\cdot1=1$. That is to say,  $x = \pm1$ are the solutions to the equation $x^2 = 1$. Note that if there exists $x$ in an integral	domain such that $x^2=1$, then we have
	\[
	(x-1)\cdot(x+1)=x^2-1=0,
	\]
	which implies $x-1=0$ or $x+1=0$. Therefore, we can assert $x = \pm1$ are the solutions. In the ring $\mathbb{Z}/8\mathbb{Z}$, $[3]_8$ and $[5]_8$ are also the solutions to the equation $x^2 = 1$.
\end{solution}

\begin{problem}[1.10]
	Let $R$ be a ring. Prove that if $a \in R$ is a right unit, and has two or more left-inverses, then $a$ is not a left-zero-divisor, and is a right-zero-divisor.
\end{problem}
\begin{solution}
	Since $a \in R$ is a right unit, it cannot be a left-zero-divisor. Assume there exist two distinct elements $x,y\in R$ such that $xa=ya=1$ and it deduces $(y-x)a=0$. Thus we show that $a$ a right-zero-divisor.
\end{solution}

\begin{problem}[1.11]
	Construct a field with 4 elements: as mentioned in the text, the underlying
	abelian group will have to be $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$; $(0, 0)$ will be the zero element, and (1, 1) will be the multiplicative identity. The question is what $(0, 1)\cdot(0, 1)$, $(0, 1)\cdot(1, 0)$, $(1, 0)\cdot(1, 0)$ must be, in order to get a field. [\textsection1.2, \textsection V.5.1]
\end{problem}
\begin{solution}
	Define 
	\[
	(0, 1)\cdot(0, 1)=(0, 1),\quad (0, 1)\cdot(1, 0)=(0,0),\quad (1, 0)\cdot(1, 0)=(1, 0),
	\]
	and the the rest definition of multiplication will be determined uniquely according to field properties. For example, we have no alternatives but to define
	\[
	(0, 1)\cdot(1,1)=(0, 1)\cdot((0,1)+(1,0))=(0, 1)\cdot(0,1)+(0, 1)\cdot(1,0)=(0, 1)+(0,0)=(0,1).
	\]
	Then we can check $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$ forms a field by definition. 
\end{solution}

\hypertarget{Exercise III.1.4}{}
\begin{problem}[1.12]
	Just as complex numbers may be viewed as combinations $a+bi$, where
	$a,b\in \R$, and $i$ satisfies the relation $i^2=-1$ (and commutes with $\R$), we may construct a ring $\mathbb{H}$ by considering linear combinations $a + bi + cj + dk$ where $a, b, c, d \in \R$, and $i, j, k$ commute with $\R$ and satisfy the following relations:
	\[
	i^{2}=j^{2}=k^{2}=-1 \quad, \quad i j=-j i=k \quad, \quad j k=-k j=i \quad, \quad k i=-i k=j.
	\]
	Addition in $\mathbb{H}$ is defined componentwise, while multiplication is defined by imposing	distributivity and applying the relations. For example,
	\[
	(1+i+j) \cdot(2+k)=1 \cdot 2+i \cdot 2+j \cdot 2+1 \cdot k+i \cdot k+j \cdot k=2+2 i+2 j+k-j+i=2+3 i+j+k.
	\]
	\begin{enumerate}[(i)]
		\item Verify that this prescription does indeed define a ring.
		\item Compute $(a+b i+c j+d k)(a-b i-c j-d k)$, where $a, b, c, d \in \R$.
		\item Prove that $\mathbb{H}$ is a division ring. Elements of $\mathbb{H}$ are called quaternions. Note that $Q_8 := \{\pm1,\pm i,\pm j,\pm k\}$ forms a subgroup of the group of units of $\mathbb{H}$; it is a noncommutative group of order 8, called the quaternionic group.
		\item  List all subgroups of $Q_8$, and prove that they are all normal.
		\item  Prove that $Q_8$, $D_8$ are not isomorphic.
		\item  Prove that $Q_8$ admits the presentation $\left(x, y | x^{2} y^{-2}, y^{4}, x y x^{-1} y\right)$.
	\end{enumerate}
	[\textsection II.7.1, 2.4, IV.1.12, IV.5.16, IV.5.17, V.6.19]
\end{problem}
\begin{solution}
	\begin{enumerate}[(i)]
		\item Verifying the $(\mathbb{H},+)$ is a abelian group is immediate and we just omitted it. It is easy to see the multiplicative identity is 1 and the distributive properties are guaranteed by definition. The check of the associativity of multiplication looks straightforward but tedious.
		\begin{align*}
			&\hspace{1.2em}((a_1+b_1i+c_1j+d_1k)\cdot(a_2+b_2i+c_2j+d_2k))\cdot(a_3+b_3i+c_3j+d_3k)\\
			&=[-c_3 \left(a_ 2 c_ 1+a_ 1 c_ 2+b_ 2 d_ 1-b_ 1 d_ 2\right)-b_ 3
			\left(a_ 2 b_ 1+a_ 1 b_ 2-c_ 2 d_ 1+c_ 1 d_ 2\right)\\
			&\hspace{1em}+a_ 3 \left(a_ 1a_ 2-b_ 1b_ 2-c_ 1 c_ 2-d_ 1 d_ 2\right)-d_3\left(-b_2 c_ 1+b_ 1 c_ 2+a_ 2 d_1+a_ 1 d_ 2\right)  ]\\
			&\hspace{1em}+[-c_3 \left(-b_2 c_ 1+b_ 1 c_ 2+a_ 2 d_ 1+a_ 1 d_ 2\right)+a_ 3 \left(a_ 2 b_ 1+a_ 1
			b_ 2-c_ 2 d_ 1+c_ 1 d_ 2\right)\\
			&\hspace{1em}+b_ 3 \left(a_ 1 a_ 2-b_ 1 b_ 2-c_ 1 c_ 2-d_ 1 d_ 2\right)+d_ 3\left(a_ 2 c_ 1+a_ 1 c_ 2+b_ 2 d_ 1-b_ 1 d_ 2\right) ]i\\
			&\hspace{1em}+[b_ 3 \left(-b_2 c_ 1+b_ 1
			c_ 2+a_ 2 d_ 1+a_ 1 d_ 2\right)+a_ 3 \left(a_ 2 c_ 1+a_ 1 c_ 2+b_ 2d_ 1-b_ 1 d_ 2\right)\\
			&\hspace{1em}+c_ 3 \left(a_ 1 a_ 2-b_ 1 b_ 2-c_ 1 c_ 2-d_1d_2\right)-d_ 3\left(a_ 2 b_ 1+a_ 1 b_ 2-c_ 2d_ 1+c_ 1 d_ 2\right) ]j\\
			&\hspace{1em}+[a_ 3 \left(-b_2 c_ 1+b_ 1 c_ 2+a_ 2 d_ 1+a_ 1 d_ 2\right)-b_ 3 \left(a_ 2 c_ 1+a_ 1 c_ 2+b_ 2 d_ 1-b_ 1 d_ 2\right)\\
			&\hspace{1em}+c_ 3 \left(a_2b_ 1+a_ 1 b_ 2-c_ 2d_ 1+c_ 1 d_ 2\right)+d_ 3\left(a_ 1 a_ 2-b_ 1 b_ 2-c_ 1 c_ 2-d_ 1 d_ 2\right) ]k
		\end{align*}
		\begin{align*}
		&\hspace{1.2em}(a_1+b_1i+c_1j+d_1k)\cdot((a_2+b_2i+c_2j+d_2k)\cdot(a_3+b_3i+c_3j+d_3k))\\
		&=[-d_1 \left(a_3 d_2+a_2 d_3-b_3
		c_2+b_2 c_3\right)-c_1 \left(a_3 c_2+a_2 c_3+b_3 d_2-b_2
		d_3\right)\\
		&\hspace{1em}-b_1 \left(a_3 b_2+a_2 b_3-c_3 d_2+c_2
		d_3\right)+a_1 \left(a_2 a_3-b_2 b_3-c_2 c_3-d_2
		d_3\right) ]\\
		&\hspace{1em}+[c_1 \left(a_3 d_2+a_2 d_3-b_3 c_2+b_2
		c_3\right)-d_1 \left(a_3 c_2+a_2 c_3+b_3 d_2-b_2
		d_3\right)\\
		&\hspace{1em}+a_1 \left(a_3 b_2+a_2 b_3-c_3 d_2+c_2
		d_3\right)+b_1 \left(a_2 a_3-b_2 b_3-c_2 c_3-d_2
		d_3\right) ]i\\
		&\hspace{1em}+[-b_1 \left(a_3 d_2+a_2 d_3-b_3 c_2+b_2
		c_3\right)+a_1 \left(a_3 c_2+a_2 c_3+b_3 d_2-b_2
		d_3\right)\\
		&\hspace{1em}+d_1 \left(a_3 b_2+a_2 b_3-c_3 d_2+c_2
		d_3\right)+c_1 \left(a_2 a_3-b_2 b_3-c_2 c_3-d_2
		d_3\right)]j\\
		&\hspace{1em}+[a_1 \left(a_3 d_2+a_2 d_3-b_3 c_2+b_2
		c_3\right)+b_1 \left(a_3 c_2+a_2 c_3+b_3 d_2-b_2
		d_3\right)\\
		&\hspace{1em}-c_1 \left(a_3 b_2+a_2 b_3-c_3 d_2+c_2
		d_3\right)+d_1 \left(a_2 a_3-b_2 b_3-c_2 c_3-d_2
		d_3\right) ]k
		\end{align*}
		\item Expand it by distributive properties and we get
		\begin{align*}
			&(a+b i+c j+d k)(a-b i-c j-d k)\\
			&=a^2-abi-acj-adk+abi+b^2-bck+bdj+acj+bck+c^2-cdi+adk-bdj+cdi+d^2\\
			&=a^2+b^2+c^2+d^2.
		\end{align*}
		\item Applying the results in (ii) we see that for any non-zero element $a+b i+c j+d k\in\mathbb{H}$,
		\[
		(a+b i+c j+d k)\cdot\frac{a-b i-c j-d k}{a^2+b^2+c^2+d^2}=\frac{a-b i-c j-d k}{a^2+b^2+c^2+d^2}\cdot(a+b i+c j+d k)=1,
		\]
		which implies $a+b i+c j+d k$ is a two-sided unit. Thus we show that $\mathbb{H}$ is a division ring. 
		\item $Q_8$ has 6 subgroups: $\{1\}$, $\{1,-1\}$, $\{1,-1,i,-i\}$, $\{1,-1,j,-j\}$, $\{1,-1,k,-k\}$, $Q_8$. We can just prove that they are all normal by the definition of normal subgroups.
		\item Note that $D_8=\{e,r,r^2,r^3,s_1,s_2,s_3,s_4\}$ has 7 subgroups: $\{e\}$, $\{e,r,r^2,r^3\}$, $\{e,s_1\}$, $\{e,s_2\}$, $\{e,s_3\}$, $\{e,s_4\}$, $D_8$, while $Q_8$ has 6 subgroups. Thus $Q_8$, $D_8$ are not isomorphic.
		\item 	Let $P=\left(x, y | x^{2} y^{-2}, y^{4}, x y x^{-1} y\right)$. The relation $x^{2} y^{-2}=e$ implies $x^2=y^2$ and the relation $xyx^{-1}y=e$ implies $yx=yx^{-1}x^2=x^{-1}y^{-1}x^2=x^3y^3x^2=x^3y^5=x^3y$. First, we can always replace $yx$ by $x^3y$ until we obtain a word of the form $x^iy^j$. Then applying $x^4=y^4=e$ and replace $y^2$ by $x^2$, we can transform it into the form $x^iy^j$ with $0\le i\le 3$ and $0\le j \le 1$. Thus we see $P$ has at most 8 elements. 
		
		Next we will complete our proof by means of the \hyperlink{Lemma II.1}{Lemma II.1} in the appendix. Define a mapping
		\begin{align*}
		f:\{x,y\}\longrightarrow Q_8,\quad& x\longmapsto i,\\
		& y\longmapsto j.
		\end{align*}
		Let $\varphi:F(\{x,y\})\to Q_8$ be the unique homomorphism induced by the universal property of free group. Since
		\begin{align*}
		&\varphi(x^{2} y^{-2})=i^2j^{-2}=1,\\
		&\varphi(y^{4})=j^{4}=1,\\
		&\varphi(x y x^{-1} y)=i j i^{-1} j=1,
		\end{align*}
		we see $\mathscr{R}=\{x^{2} y^{-2}, y^{4}, x y x^{-1} y\}\subset\ker\varphi$. And it is immediate to show that $Q_8$ can be generated by $\{i,j\}$. Thus according to the lemma, there exists a unique homomorphism $\psi:P\to Q_8$ such that $f=\psi\circ\pi\circ i$ and actually $\psi$ is surjective. 
		\[\xymatrix{
			P\ar@{-->}[rd]^{\exists!\psi}\\
			F(\{x,y\})\ar@{-->}[r]^{\varphi}\ar[u]^{\pi} &Q_8\\
			\{x,y\}\ar[ru]_{f}\ar[u]^{i}&    
		}\]
		Hence we get the inequality of cardinality $|P|\ge|Q_8|$. Since we have shown $|P|\le 8=|Q_8|$, there must be $|P|=|Q_8|=8$, which implies $\psi$ is indeed an isomorphism. Finally we conclude that $Q_8\cong\left(x, y | x^{2} y^{-2}, y^{4}, x y x^{-1} y\right)$ and complete our proof. 
	
	\end{enumerate}
\end{solution}


\hypertarget{Exercise III.1.14}{}
\begin{problem}[1.14]
	$\vartriangleright$ Let $R$ be a ring, and let $f(x),g(x)\in R[x]$ be nonzero polynomials. Prove that 
	\[
	\deg(f(x) + g(x))\le\max(\deg(f(x)), \deg(g(x))).
	\]
	Assuming that $R$ is an integral domain, prove that	
	\[
	\deg(f(x)\cdot g(x)) = \deg(f(x)) + \deg(g(x)).
	\] 
	[\textsection1.3]
\end{problem}
\begin{solution}
	Assume
	\[
	f(x)=\sum_{i \ge 0} a_{i} x^{i},\quad g(x)=\sum_{i \ge 0} b_{i} x^{i}, \quad a_i,b_i\in R
	\]
	and $n,m$ are respectively the
	largest integers $p,q$ for which $a_p$, $b_q$ are non-zero. In others words, we have $a_n\ne 0$, $a_i=0$ for $i>n$ and $b_m\ne 0$, $b_i=0$ for $i>m$. Since
	\[
	f(x)+g(x)=\sum_{i \ge 0} (a_{i}+b_i) x^{i}=\sum_{i =0}^{\max\{n,m\}} (a_{i}+b_i) x^{i},
	\]
	we see that
	\[
	\deg(f(x) + g(x))\le\max\{n,m\}=\max(\deg(f(x)), \deg(g(x))).
	\]
	Now Suppose that $R$ is an integral domain. Noticing $a_n\ne 0$ and $b_m\ne 0$ implies $a_nb_m\ne 0$, we can see
	\[
	f(x) \cdot g(x) =\sum_{k \geq 0} \sum_{i+j=k} a_{i} b_{j} x^{i+j}=\sum_{k = 0}^{n+m} \sum_{i+j=k} a_{i} b_{j} x^{i+j}
	\]
	has a degree of $n+m$. That is,
	\[
	\deg(f(x)\cdot g(x)) = \deg(f(x)) + \deg(g(x)).
	\]
\end{solution}

\begin{problem}[1.15]
	$\vartriangleright$ Prove that $R[x]$ is an integral domain if and only if $R$ is an integral domain.	[\textsection1.3]
\end{problem}
\begin{solution}
	Assume $R$ is an integral domain. \hyperlink{Exercise III.1.14}{Exercise III.1.14} tells us if $f(x)$, $g(x)\in R[x]$ are nonzero polynomials, we have 
	\[
	\deg(f(x)\cdot g(x)) = \deg(f(x)) + \deg(g(x)),
	\]
	which implies $f(x)\cdot g(x)$ is also nonzero polynomial. Thus we show $R[x]$ is a integral domain. 
	
	Conversely, assume $R[x]$ is an integral domain. Note that given any $a,b\in R$, they also belong to $R[x]$. Hence we obtain
	\[
	a\ne0,b\ne0\implies ab\ne0,
	\]
    which means $R$ is an integral domain.
\end{solution}

\begin{problem}[1.16]
	Let $R$ be a ring, and consider the ring of power series $R[[x]]$ (cf. \textsection1.3).
	\begin{enumerate}
		\item Prove that a power series $a_0+a_1x+a_2x^2+\cdots$ is a unit in $R[[x]]$ if and only if	$a_0$ is a unit in $R$. What is the inverse of $1-x$ in $R[[x]]$?
		\item Prove that $R[[x]]$ is an integral domain if and only if $R$ is.
	\end{enumerate}
\end{problem}
\begin{solution}
	\begin{enumerate}
		\item If $a_0$ is a unit in $R$ then we can assume there exists $b_0\in R$ such that $a_0b_0=1$. Let 
		\[
		f(x)=\sum_{n \ge 0} a_{n} x^{n},\quad g(x)=\sum_{n \ge 0} b_{n} x^{n}, 
		\]
		where
		\[
		b_n = -b_0 \sum_{i=1}^n a_i b_{n-i},\quad n\ge1.
		\]
		Noticing that 
		\[
		a_0b_n= -a_0b_0 \sum_{i=1}^n a_i b_{n-i}=- \sum_{i=1}^n a_i b_{n-i},\quad n\ge1,
		\]
		we have
		\begin{align*}
		f(x)g(x)&=\sum_{n \ge 0}\sum_{i=0}^na_{n-i}b_{i}x^n\\
		&=1+\sum_{n \ge 1}\sum_{i=0}^na_{i}b_{n-i}x^n\\
		&=1+\sum_{n \ge 1}\left(a_0b_n+\sum_{i=1}^na_{i}b_{n-i}\right)x^n\\
		&=1+\sum_{n \ge 1}\left(a_0b_n-a_0b_n\right)x^n\\
		&=1.
		\end{align*}
		Hence we show $f(x)=a_0+a_1x+a_2x^2+\cdots$ is a unit.
		
		For the other direction, supposing $f(x)=a_0+a_1x+a_2x^2+\cdots$ is a unit, then there exists $g(x)=b_0+b_1x+b_2x^2+\cdots$ such that 
		\[
		f(x)g(x)=a_0b_0+\sum_{n \ge 1}\sum_{i=0}^na_{i}b_{n-i}x^n=1.
		\]
		By comparing the both sides of the equality we can find $a_0b_0=1$, which implies $a_0$ is a unit in $R$.
		
		We can check that the inverse of $1-x$ in $R[[x]]$ is $1+x+x^2+\cdots$ since
		\[
		(1-x)\sum_{i \ge 0}x^i=\sum_{i \ge 0}x^i-\sum_{i \ge 0}x^{i+1}=1.
		\]
		\item Suppose $R$ is an integral domain. If $f(x)$, $g(x)\in R[x]$ are nonzero polynomials, we can assume that
		\[
		f(x)=\sum_{i \ge 0} a_{i} x^{i},\quad g(x)=\sum_{i \ge 0} b_{i} x^{i}, \quad a_i,b_i\in R
		\]
		and that $n,m$ are respectively the smallest integers $p,q$ for which $a_p$, $b_q$ are non-zero. In others words, we have $a_n\ne 0$, $a_i=0$ for $i<n$ and $b_m\ne 0$, $b_i=0$ for $i<m$.  Noticing $a_n\ne 0$ and $b_m\ne 0$ implies $a_nb_m\ne 0$, we can see
		\[
		f(x) \cdot g(x) =\sum_{k \geq 0} \sum_{i+j=k} a_{i} b_{j} x^{i+j}= a_{n} b_{m} x^{n+m}+\sum_{k\ge n+m+1}\sum_{i+j=k} a_{i} b_{j} x^{i+j}\ne 0.
		\]
		Thus we show $R[[x]]$ is an integral domain.
		
			
		Conversely, assume that $R[[x]]$ is an integral domain. Note that given any $a,b\in R$, they also belong to $R[[x]]$. Hence we obtain 
		\[
		a\ne0,b\ne0\implies ab\ne0,
		\]
		which means that $R$ is also an integral domain.
	\end{enumerate}
\end{solution}


\subsection{\textsection2. The category $\mathsf{Ring}$}

\begin{problem}[2.1]
	Prove that if there is a homomorphism from a zero-ring to a ring $R$, then $R$ is a zero-ring [\textsection2.1]
\end{problem}
\begin{solution}
	Suppose that $\varphi$ is a homomorphism from a zero-ring $O$ to a ring $R$. Since $\varphi(0_O)=0_R$, $\varphi(1_O)=1_R$, $0_O=1_O$, we have $0_R=1_R$, which implies that $R$ is a zero-ring.
\end{solution}

\hypertarget{Exercise III.2.4}{}
\begin{problem}[2.4]
	Define functions $\mathbb{H} \rightarrow \mathfrak{g l}_{4}(\mathbb{R})$ and $\mathbb{H} \rightarrow \mathfrak{g l}_{2}(\mathbb{C})$ (cf. \hyperlink{Exercise III.1.4}{Exercise III.1.4} and \hyperlink{Exercise III.1.4}{1.12}) by
	\begin{align*}
	a+b i+c j+d k& \longmapsto\left(\begin{array}{cccc}{a} & {b} & {c} & {d} \\ {-b} & {a} & {-d} & {c} \\ {-c} & {d} & {a} & {-b} \\ {-d} & {-c} & {b} & {a}\end{array}\right)\\
	a+b i+c j+d k &\longmapsto\left(\begin{array}{cc}{a+b i} & {c+d i} \\ {-c+d i} & {a-b i}\end{array}\right)
	\end{align*}
	for all $a, b, c, d \in\R$. Prove that both functions are injective ring homomorphisms.
	Thus, quaternions may be viewed as real or complex matrices.
\end{problem}
\begin{solution}
	Let $f$ be the function $\mathbb{H} \rightarrow \mathfrak{g l}_{4}(\mathbb{R})$ described above. For simplicity, we omit trivial check and only verify $f$ preserves multiplication
	\begin{align*}
	&f((a_1+b_1i+c_1j+d_1k)\cdot(a_2+b_2i+c_2j+d_2k))\\
	&=f((a_1 a_2-b_1 b_2-c_1 c_2-d_1 d_2)+(a_2
	b_1+a_1 b_2-c_2 d_1+c_1 d_2)i\\
	&\hspace{1em}+(a_2 c_1+a_1 c_2+b_2 d_1-b_1
	d_2)j+(a_2 d_1+a_1 d_2-b_2 c_1+b_1 c_2)k)\\
	&=\left(\begin{array}{cccc}{a_1} & {b_1} & {c_1} & {d_1} \\ {-b_1} & {a_1} & {-d_1} & {c_1} \\ {-c_1} & {d_1} & {a_1} & {-b_1} \\ {-d_1} & {-c_1} & {b_1} & {a_1}\end{array}\right)
	\left(\begin{array}{cccc}{a_2} & {b_2} & {c_2} & {d_2} \\ {-b_2} & {a_2} & {-d_2} & {c_2} \\ {-c_2} & {d_2} & {a_2} & {-b_2} \\ {-d_2} & {-c_2} & {b_2} & {a_2}\end{array}\right)\\
	&=f(a_1+b_1i+c_1j+d_1k)f(a_2+b_2i+c_2j+d_2k)
	\end{align*}
\end{solution}

\hypertarget{Exercise III.2.5}{}
\begin{problem}[2.5]
	The norm of a quaternion $w=a+b i+c j+d k$, with $a, b, c, d \in \mathbb{R}$, is the real number $N(w)=a^{2}+b^{2}+c^{2}+d^{2}$.
	Prove that the function from the multiplicative group $\mathbb{H}^{*}$ of nonzero quaternions to the multiplicative group $\mathbb{R}^{+}$ of positive real numbers, defined by assigning to each nonzero quaternion its norm, is a homomorphism. Prove that the kernel of this homomorphism is isomorphic to $\mathrm{SU}_{2}(\C)$ (cf. \hyperlink{Exercise II.6.3}{Exercise II.6.3}). $[4.10, \mathrm{IV} .5 .17$ $\mathrm{V} .6 .19]$
\end{problem}
\begin{solution}
	According to \hyperlink{Exercise III.2.4}{Exercise III.2.4}, $w\in\mathbb{H}^{*}$ can be viewed as a matrix $i(w)\in\mathfrak{g l}_{2}(\mathbb{C})$ where $i:\mathbb{H}\to\mathfrak{g l}_{2}(\mathbb{C})$ is a monomorphism in $\mathsf{Ring}$. Then the function $N:\mathbb{H}^{*}\to \mathbb{R}^{+}$ can be just viewed as the determinant mapping $\det:i(\mathbb{H}^{*})\subset\mathfrak{g l}_{2}(\mathbb{C})\to\mathbb{R}^{+}$. More precisely, it means $N=\det\circ \;i$. We can check that
	\[
	N(w_1w_2)=\det(i(w_1w_2))=\det(i(w_1)i(w_2))=\det(i(w_1))\det(i(w_2))=N(w_1)N(w_2)
	\]
	and
	\[
	w\in\ker N\iff N(w)=\det(i(w))=1\iff i(w)\in\mathrm{SU}_{2}(\C).
	\]
	Therefore, $N$ is a homomorphism and $\ker N$ isomorphic to $\mathrm{SU}_{2}(\C)$.
\end{solution}


\begin{problem}[2.6]
	Verify the ‘extension property’ of polynomial rings, stated in Example 2.3.
	[\textsection2.2]
\end{problem}
\begin{solution}
	Define the following ring homomorphisms 
	\begin{align*}
	\alpha:\ &R\longrightarrow S,\quad r\longmapsto \alpha(r)\\
	\epsilon:\ &R\longrightarrow R[x],\quad r\longmapsto r,
	\end{align*}
	and functions
	\begin{align*}
	j:\ &\{s\}\longrightarrow R[x],\quad s\longmapsto x,\\
	i:\ &\{s\}\longrightarrow S,\quad s\longmapsto s.	
	\end{align*}
	Assume that $s \in S$ is an element commuting with $\alpha(r)$ for all $r \in R$, we are to show that there exists a unique ring homomorphism $\overline{\alpha}: R[x]\to S$ such that the following diagram commutes.
	\[\xymatrix{
		R\ar@{->}[d]_{\epsilon} \ar@{->}[rd]^{\alpha}\\
		R[x]\ar@{-->}[r]^{\exists!\overline{\alpha}} & S\\
		\{s\}\ar@{^{(}->}[ru]_{i}\ar@{|->}[u]^{j}  
	}\]
	\textbf{Uniqueness}. If $\overline{\alpha}$ exists, then the postulated commutativity of the diagram means that for all $f(x)=\sum_{n \ge 0} a_{n}\in R[x]$, there must be
	\[
	\overline{\alpha}\left(f(x)\right)=\overline{\alpha}\left(\sum_{n \ge 0} a_{n} x^{n}\right)=\sum_{n \ge 0} \overline{\alpha}\left( a_n\right)\overline{\alpha}\left(x\right)^{n}=\sum_{n \ge 0} \alpha\left( a_n\right)s^{n}.
	\]
	That is, $\overline{\alpha}$ is unique. 
	
	\noindent\textbf{Existence}. The only choice is to define
	\[
	\overline{\alpha}:\ R[x]\longrightarrow S,\quad \sum_{n \ge 0} a_{n} x^{n}\longmapsto \sum_{n \ge 0} \alpha\left( a_n\right)s^{n}
	\] 
	and to check whether it is a ring homomorphism.
	\begin{enumerate}
		\item Preserving addition:
		\begin{align*}
		\overline{\alpha}\left(\sum_{n \ge 0} a_{n} x^{n}+\sum_{n \ge 0} b_{n} x^{n}\right)&=\overline{\alpha}\left(\sum_{n \ge 0}(a_{n}+b_{n})x^{n}\right)\\
		&=\sum_{n \ge 0}\alpha\left(a_{n}+b_{n}\right)s^{n}\\
		&=\sum_{n \ge 0}\alpha\left(a_{n}\right)s^{n}+\sum_{n \ge 0}\alpha\left(b_{n}\right)s^{n}\\
		&=\overline{\alpha}\left(\sum_{n \ge 0} a_{n} x^{n}\right)+\overline{\alpha}\left(\sum_{n \ge 0} b_{n} x^{n}\right).
		\end{align*}
		\item Preserving multiplication:
		\begin{align*}
		\overline{\alpha}\left(\sum_{n \ge 0} a_{n} x^{n}\sum_{n \ge 0} b_{n} x^{n}\right)&=\overline{\alpha}\left(\sum_{n \ge 0} \sum_{i+j=n} a_{i} b_{j} x^{n}\right)\\
		&=\sum_{n \ge 0}\alpha\left(\sum_{i+j=n} a_{i} b_{j}\right)s^{n}\\
		&=\sum_{n \ge 0}\sum_{i+j=n}\alpha\left(a_{i} \right)s^{i}\alpha\left( b_{j}\right)s^{j}\\
		&=\left(\sum_{n \ge 0}\alpha\left(a_{n}\right)s^{n}\right)\left(\sum_{n \ge 0}\alpha\left(b_{n}\right)s^{n}\right)\\
		&=\overline{\alpha}\left(\sum_{n \ge 0} a_{n} x^{n}\right)\overline{\alpha}\left(\sum_{n \ge 0} b_{n} x^{n}\right).
		\end{align*}
		\item Preserving identity element:
		\begin{align*}
		\overline{\alpha}(1_R)=\alpha(1_R)=1_S.
		\end{align*}
	\end{enumerate}
	Integrating the two parts we finally conclude there exists a unique ring homomorphism $\overline{\alpha}$ such that the diagram commutes.
	
\end{solution}

\begin{problem}[2.7]
	Let $R=\Z/2\Z$, and let $f(x)=x^2-x$; note $f(x)\ne 0$. What is the polynomial function $R\to R$ determined by $f(x)$? [\textsection2.2, \textsection V.4.2, \textsection V.5.1]
\end{problem}
\begin{solution}
	It determines a function $f:\Z/2\Z\to\Z/2\Z$ sends all elements to identity, that is, $f([0]_2)=[0]_2$, $f([1]_2)=[0]_2$.
\end{solution}

\begin{problem}[2.8]
	Prove that every subring of a field is an integral domain.
\end{problem}
\begin{solution}
	Suppose $\varphi:R\hookrightarrow K$ is a inclusion homomorphism. If $a\ne0$, we have
	\[
	ab=ac\implies \varphi(a)\varphi(b)=\varphi(a)\varphi(c)\implies \varphi(b)=\varphi(c)\implies b=c.
	\]
	Due to the community of field it also holds that $ba=ca$. Thus we show $R$ is an integral domain.
\end{solution}

\hypertarget{Exercise III.2.9}{}
\begin{problem}[2.9]
$\neg$ The \emph{center} of a ring $R$ consists of the elements a such that $ar = ra$ for all $r\in R$. Prove that the center is a subring of $R$. Prove that the center of a division ring is a field. [2.11, IV.2.17, VII.5.14,VII.5.16]
\end{problem}
\begin{solution}
Denote the center of $R$ by $Z(R)$. We can check that
\begin{enumerate}
	\item for all $x,y\in Z(R)$, for all $r\in R$, $$(x-y)r=xr-yr=rx-ry=r(x-y)\implies x-y\in Z(R);$$ 
	\item for all $r\in R$,
	\[
	1r=r1\implies 1\in Z(R);
	\]
	\item for all $x,y\in Z(R)$, for all $r\in R$,
	\[
	(xy)r=xry=r(xy)\implies xy\in Z(R).
	\]
\end{enumerate}
Thus we show that $Z(R)$ is a subring of $R$. If $R$ is a division ring, then $Z(R)$ is a also a division ring. Note that for all $x,y\in Z(R)$, $xy=yx$, we see that $Z(R)$ is a commutative division ring, namely field.
\end{solution}


\hypertarget{Exercise III.2.10}{}
\begin{problem}[2.10]
$\neg$ The \emph{centralizer} of an element $a$ of a ring $R$ consists of the elements $r \in R$ such that $ar = ra$. Prove that the centralizer of $a$ is a subring of $R$, for every $a\in R$. Prove that the center of $R$ is the intersection of all its centralizers. Prove that every centralizer in a division ring is a division ring. [2.11, IV.2.17, VII.5.16]
\end{problem}
\begin{solution}
Denote the centralizer of an element $a$ of $R$ by $Z_a(R)$. That is,
\[
Z_a(R)=\{r\in R\mid ar=ra\}.
\] 
We can check that
\begin{enumerate}
	\item for all $x,y\in Z_a(R)$, $$(x-y)a=xa-ya=ax-ay=a(x-y)\implies x-y\in Z_a(R);$$ 
	\item 
	\[
	1a=a1\implies 1\in Z_a(R);
	\]
	\item for all $x,y\in Z_a(R)$,
	\[
	(xy)a=xay=a(xy)\implies xy\in Z_a(R).
	\]
\end{enumerate}
Thus we show that $Z_a(R)$ is a subring of $R$. 

By definition we have $Z(R)\subseteq Z_a(R)$ for all $a\in R$, which implies $Z(R)\subseteq \bigcap_{a\in R} Z_a(R)$. Assume $s\in\bigcap_{a\in R} Z_a(R)$, then we see $sa=as$ for all $a\in R$, which means $s\in Z(R)$ and accordingly $\bigcap_{a\in R} Z_a(R)\subseteq Z(R)$. Thus we deduce that $Z(R)=\bigcap_{a\in R} Z_a(R)$.

If $R$ is a division ring and $r\in Z_a(R)$, we can assume that there exists $a\in R$ such as $ar=ra$, which means that
\[
r^{-1}(ar)r^{-1}=r^{-1}(ra)r^{-1}\implies r^{-1}a=ar^{-1}.
\]
According to the definition of $Z_a(R)$, we see $r^{-1}\in Z_a(R)$. Thus we show that $Z_a(R)$ is a division ring.
\end{solution}

\hypertarget{Exercise III.2.11}{}
\begin{problem}[2.11]
$\neg$ Let $R$ be a division ring consisting of $p^{2}$ elements, where $p$ is a prime. Prove that $R$ is commutative, as follows:
\begin{itemize}
	\item If $R$ is not commutative, then its center $C$ (\hyperlink{Exercise III.2.9}{Exercise III.2.9}) is a proper subring of $R .$ Prove that $C$ would then consist of $p$ elements.
	\item Let $r \in R, r \notin C .$ Prove that the centralizer of $r$ (\hyperlink{Exercise III.2.10}{Exercise III.2.10}) contains both $r$ and $C$.
	\item Deduce that the centralizer of $r$ is the whole of $R$.
	\item Derive a contradiction, and conclude that $R$ had to be commutative (hence, a field).
\end{itemize}
This is a particular case of Wedderburn's theorem: every finite division ring is a field. [IV.2.17, VII.5.16]
\end{problem}
\begin{solution}
	If $R$ is not commutative, then its center $Z(R)$ is a proper subring of $R$, which means $|Z(R)|<p^2$. By considering $Z(R)$ as a subgroup of the underlying abelian group $R$, we can deduce that $|Z(R)|$ divides $p^2$ according to the Lagrange theorem. Thus we see that $Z(R)$ consist of $p$ elements. Given any $r \in R-Z(R)$, in \hyperlink{Exercise III.2.10}{Exercise III.2.10} we have shown that $Z_r(R)$ is a subring of $R$ and $Z(R)\in Z_r(R)$. By the definition of $Z_r(R)$, it is clear that $r\in Z_r(R)$. Hence we have $Z(R)\cup \{r\}\subseteq Z_r(R)$ and $|Z_r(R)|>p$. Again by Lagrange theorem we have $|Z_r(R)|$ divides $p^2$, which forces $|Z_r(R)|=p^2$. Thus we show that $Z_r(R)=R$. Note that $Z_a(R)=R$ for all $a\in Z(R)$. We have $Z_a(R)=R$ for all $a\in R$. In \hyperlink{Exercise III.2.10}{Exercise III.2.10}, we have derived that $\bigcap_{a\in R} Z_a(R)\subseteq Z(R)$, which implies $R\subseteq Z(R)$. Thus we have $Z(R)=R$, which contradicts with the previous deduction that $Z(R)$ is a proper subring of $R$.  Therefore, we can conclude that $R$ is commutative.
\end{solution}



\begin{problem}[2.15]
	For $m>1,$ the abelian groups $(\mathbb{Z},+)$ and $(m \mathbb{Z},+)$ are manifestly isomorphic: the function $\varphi: \mathbb{Z} \rightarrow m \mathbb{Z}, n \mapsto mn$ is a group isomorphism. Use this isomorphism to transfer the structure of `ring without identity' $(m \mathbb{Z},+, \cdot)$ back onto $\mathbb{Z}:$ give an explicit formula for the `multiplication' $\bullet$ this defines on $\mathbb{Z}$ (that is, such that $\varphi(a \bullet b)=\varphi(a) \cdot \varphi(b))$. Explain why structures induced by different positive integers $m$ are non-isomorphic as `rings without 1'.
	
	(This shows that there are many different ways to give a structure of ring without identity to the \emph{group} $(\mathbb{Z},+)$. Compare this observation with Exercise 2.16.) [\textsection 2.1]
\end{problem}
\begin{solution}

\end{solution}


\subsection{\textsection3. Ideals and quotient rings}

\begin{problem}[3.1]
Prove that the image of a ring homomorphism $\varphi: R \to S$ is a subring of $S$. What can you say about $\varphi$, if its image is an ideal of $S$? What can you say about $\varphi$, if its kernel is a subring of $R$?
\end{problem}
\begin{solution}
We can see that $\im \varphi$ is a subring of $S$ from the canonical decomposition
\[\xymatrix{
	R\ar@/^2pc/@{->}[rrr]^{\varphi}\ar@{->>}[r]^{}&R/\ker \varphi \ar@{->}[r]^{\hspace{1em}\sim\hspace{4pt}}_{\hspace{1em}\tilde{\varphi}\hspace{4pt}}&\im \varphi\hspace{2pt}\ar@{^{(}->}[r]^{}&S
}\] 
If $\im \varphi$ is an ideal, then $s\in S,1\in\im\varphi \implies s\in \im\varphi$. So $\im \varphi=S$ and $\varphi$ is an epimorphism. Since $\ker\varphi$ is a ideal, if it is also a subring, we have $\ker \varphi=R$.
\end{solution}

\hypertarget{Exercise III.3.2}{}
\begin{problem}[3.2]
Let $\varphi: R\to S$ be a ring homomorphism, and let $J$ be an ideal of $S$. Prove
that $I = \varphi^{-1}(J)$ is an ideal of $R$. [\textsection3.1]
\end{problem}
\begin{solution}
In $\Ab$ we see $\varphi^{-1}(J)$ is a subgroup of $R$. For all $r\in R$, $a\in \varphi^{-1}(J)$, we have
\[
\varphi(ra)=\varphi(r)\varphi(a)\in J\implies ra\in\varphi^{-1}(J).
\]
Similarly we can obtain $ar\in\varphi^{-1}(J)$. Therefore, we conclude that $I = \varphi^{-1}(J)$ is an ideal of $R$.
\end{solution}

\hypertarget{Exercise III.3.3}{}
\begin{problem}[3.3]
$\neg$ Let $\varphi : R \to S$ be a ring homomorphism, and let $J$ be an ideal of $R$.
\begin{itemize}
	\item Show that $\varphi(J)$ need not be an ideal of $S$.
	\item Assume that $\varphi$ is surjective; then prove that $\varphi(J)$ is an ideal of $S$.
	\item Assume that $\varphi$ is surjective, and let $I = \ker\varphi$; thus we may identify $S$ with $R/I$.
	Let $\overline{J}= \varphi(J)$, an ideal of $R/I$ by the previous point. Prove that
	\[
	\frac{R / I}{\overline{J}} \cong \frac{R}{I+J}
	\]
\end{itemize}
(Of course this is just a rehash of Proposition 3.11.) [4.11]
\end{problem}
\begin{solution}
\begin{itemize}
	\item Let $\varphi:\Z\hookrightarrow \Q$ and $J=\Z$. It is clear that $\varphi(J)=\Z$ is not an ideal of $\Q$.
	\item Assume that $\varphi$ is surjective. In $\Ab$ we see $\varphi(J)$ is a subgroup of $S$. For all $a'=\varphi(a)\in\varphi(J)$, $r'=\varphi(r)\in S$,
	\[
	ra\in J\implies r'a'=\varphi(r)\varphi(a)=\varphi(ra)\in\varphi(J).
	\]
	Similarly we can obtain $a'r'\in\varphi(J)$. Therefore, we conclude that $\varphi(J)$ is an ideal of $S$.
	\item Assume that $\varphi$ is surjective. The universal property yields a unique homomorphism
	\begin{align*}
	\psi:R/I&\longrightarrow R/(I+J),\\
	r+I &\longmapsto r+I+J.
	\end{align*}
	Since
	\begin{align*}
	\ker \psi&=\{r+I\in R/I\mid r\in I+J\}\\
	&=\{a+b+I\in R/I\mid a\in I,b\in J\}\\
	&=\{b+I\in R/I\mid b\in J\}\\
	&=\{\varphi(b)\in S\mid b\in J\}\\
	&= \varphi(J)=\overline{J}
	\end{align*}
	and $\psi$ is surjective,
	\[
	\frac{R / I}{\overline{J}}=\frac{R / I}{\ker \psi} \cong \frac{R}{I+J}.
	\]
\end{itemize}
\end{solution}

\hypertarget{Exercise III.3.7}{}
\begin{problem}[3.7]
Let $R$ be a ring, and let $a\in R$. Prove that $Ra$ is a left-ideal of $R$, and $aR$ is a right-ideal of $R$. Prove that $a$ is a left-, resp. right-unit if and only if $R = aR$, resp. $R = Ra$.
\end{problem}
\begin{solution}
For all $r\in R$, $r(Ra)\subseteq Ra$, $(aR)r\subseteq aR$. Therefore, $Ra$ is a left-ideal of $R$, and $aR$ is a right-ideal of $R$. Since $aR\subseteq R$, $R\subseteq aR$ actually amounts to $R = aR$. 
\[
a\text{ is a left-unit}\iff\exists b\in R, ab=1\implies \forall r\in R,r=abr\in aR\implies R\subseteq aR
\]
\[
R\subseteq aR\implies\forall r\in R,\exists r'\in R,r=ar'\implies \exists r'\in R, ar'=1\iff a\text{ is a left-unit}
\]
Therefore, $a$ is a left-unit if and only if $R = aR$. Similarly we can prove $a$ is a right-unit if and only if $R = Ra$.
\end{solution}

\begin{problem}[3.8]
Prove that a ring $R$ is a division ring if and only if its only left-ideals and right-ideals are $\{0\}$ and $R$.

In particular, a commutative ring $R$ is a field if and only if the only ideals of $R$ are $\{0\}$ and $R$. [3.9, \textsection4.3]
\end{problem}
\begin{solution}
Assume the only left-ideals and right-ideals that ring $R$ have are $\{0\}$ and $R$. If $a\ne0$, we have $Ra=aR=R$. As a result of \hyperlink{Exercise III.3.7}{Exercise III.3.7}, it implies that $a$ is two-side unit and that accordingly $R$ is a division ring.

Now assume that $R$ is a division ring. Suppose $I$ is a nonzero left-ideal of $R$ and that $a\in I$ is not 0. Note that the condition of division ring guarantees there exists $b\in R$ such that $ba=1$. Since for all $r\in R$, $r=(rb)a\in I$, there must be $I=R$. Supposing that $I'$ is a nonzero right-ideal of $R$ and that $a'\in I'$ is not 0, in a similar way we can deduce $I'=R$. Therefore, we conclude that the only left-ideals of $R$ and right-ideals of $R$ are $\{0\}$ and $R$.
\end{solution}

\begin{problem}[3.11]
Let $R$ be a ring containing $\C$ as a subring. Prove that there are no ring homomorphisms $R \to \R$.
\end{problem}
\begin{solution}
	Suppose $f:R \to \R$ is a homomorphism. On the one hand, we have
	\[
	f(1)=f(1*1)=f(1)^2\ge0.
	\]
	On the other hand, we can calculate $f(1)$ by
	\[
	f(1)=f(-i*i)=-f(i)^2\le0,
	\]
	which forces $f(1)$ to be 0. Thus we see $f$ sends some nonzero element in $R$ to 0 in $\R$, which is a contradiction.
\end{solution}

\hypertarget{Exercise III.3.12}{}
\begin{problem}[3.12]
Let $R$ be a commutative ring. Prove that the set of nilpotent elements of $R$ is an ideal of $R$. (Cf. \hyperlink{Exercise III.1.6}{Exercise III.1.6}. This ideal is called the nilradical of $R$.)

Find a non-commutative ring in which the set of nilpotent elements is not an ideal. [3.13, 4.18, V.3.13, \textsection VII.2.3]
\end{problem}
\begin{solution}
Suppose $N$ is the set of nilpotent elements of $R$. In \hyperlink{Exercise III.1.6}{Exercise III.1.6} we have shown that if $R$ is commutative, then $a+b\in N$ for all $a,b\in N$. Since for all $r\in R$, $a\in N$,
\[
a^n=0\implies r^na^n=a^nr^n=0\implies ra,ar\in N,
\]
we prove that $N$ is an ideal of $R$. A counterexample for non-commutative ring can be found in the ring $\mathfrak{gl}_2(\mathbb{R})$, as is shown in \hyperlink{Exercise III.1.6}{Exercise III.1.6}.
\end{solution}

\hypertarget{Exercise III.3.13}{}
\begin{problem}[3.13]
$\neg$ Let $R$ be a commutative ring, and let $N$ be its nilradical (cf. \hyperlink{Exercise III.3.12}{Exercise III.3.12}). Prove that $R/N$ contains no nonzero nilpotent elements. (Such a ring is said to be reduced.) [4.6, VII.2.8]
\end{problem}
\begin{solution}
Suppose there exists a nilpotent element $r+N\in R/N$ and $n>0$ such that
\[
r^n+N=N\iff r^n\in N.
\]
Then we have $r^{nm}=0$ for some $m>0$, which implies $r\in N$. Therefore, the only nilpotent element in $R/N$ is $N$. 
\end{solution}

\begin{problem}[3.14]
$\neg$ Prove that the characteristic of an integral domain is either 0 or a prime
integer. Do you know any ring of characteristic 1?
\end{problem}
\begin{solution}
Suppose the characteristic of the integral domain $R$ is $pq$ where $p,q$ are positive prime integers. Then we have $p1_R\ne 0$ and $q1_R\ne0$, since the order of $1_R$ is $pq$. However, we can deduce
\[
(p1_R)(q1_R)=pq1_R=0_R,
\]
which contradicts the assumption that $R$ is an integral domain. 

If the characteristic of the integral domain $R$ is 1, then the inclusion homomorphism $i:\Z\to R$ will send all integers to $0_R$, which means $0_R=1_R$ and $R$ is actually a zero ring instead of an integral domain. Thus the characteristic of an integral domain is either be 0 or a prime integer.
\end{solution}


\hypertarget{Exercise III.3.15}{}
\begin{problem}[3.15]
	$\neg$ A ring $R$ is boolean if $a^{2}=a$ for all $a \in R$. Prove that $\mathscr{P}(S)$ is boolean, for every set $S$ (cf. \hyperlink{Exercise III.1.2}{Exercise III.1.2}). Prove that every boolean ring is commutative, and has characteristic 2. Prove that if an integral domain $R$ is boolean, then $R \cong \mathbb{Z} / 2 \mathbb{Z}$. $[4.23, \mathrm{~V}.6.3]$
\end{problem}
\begin{solution}
Since $A\cap A=A$ for all $A\in \mathscr{P}(S)$, $\mathscr{P}(S)$ is boolean.

Assume that $R$ is a boolean ring. For any $x\in R$,
\[
x+x=(x+x)^2=x^2+x^2+x^2+x^2=x+x+x+x\implies x+x=0_R\implies x=-x.
\]
For any $a,b\in R$,
\[
a+b=(a+b)^2=a^2+ab+ba+b^2=a+ab+ba+b\implies ab=-ba=ba.
\]
Thus we show every boolean ring is commutative. Since $1_R+1_R=0_R$, boolean ring has characteristic 2.

If an integral domain $R$ is boolean, we can define 
\begin{align*}
	\psi:\mathbb{Z} / 2 \mathbb{Z}&\longrightarrow R ,\\
	[n]_2 &\longmapsto n\cdot1_R.
\end{align*}
$\psi$ is well-defined since for all $n,k\in\mathbb{Z}$, $n\cdot1_R=(n+2k)\cdot1_R$. Since $\psi([0]_2)\ne\psi([1]_2)$, $\psi$ is injective. For any $a\in R$, we have $a(a-1_R)=0_R$. Since $R$ is an integral domain, there must be $a=0_R$ or $a-1_R=0$, which implies $\psi$ is surjective. Therefore, we show $\psi$ is an isomorphism and $R \cong \mathbb{Z} / 2 \mathbb{Z}$.
\end{solution}

\begin{problem}[3.17]
Let $I, J$ be ideals of a ring $R$. State and prove a precise result relating the ideals $(I + J)/I$ of $R/I$ and $J/(I \cap J)$ of $R/(I \cap J)$. [\textsection3.3]
\end{problem}
\begin{solution}
As abelian groups, the second isomorphism theorem ensures $(I + J)/I\cong J/(I \cap J)$.
\end{solution}



\subsection{\textsection4. Ideals and quotients: remarks and examples. Prime and maximal ideals}
\hypertarget{Exercise III.4.2}{}
\begin{problem}[4.2]
Prove that the homomorphic image of a Noetherian ring is Noetherian. That is, prove that if $\varphi: R \to S$ is a surjective ring homomorphism, and $R$ is Noetherian, then $S$ is Noetherian. [\textsection6.4]
\end{problem}
\begin{solution}
	According to \hyperlink{Exercise III.3.2}{Exercise III.3.2}, given any ideal $J$ of $S$, we see $\varphi^{-1}(J)$ is an ideal of $R$. Since $R$ is a Noetherian ring, we have $\varphi^{-1}(J)=(a_1,a_2,\cdots,a_n)$. Since $\varphi$ is surjective, there must be
	$$
	J=\varphi(\varphi^{-1}(J))=(\varphi(a_1),\varphi(a_2),\cdots,\varphi(a_n)),
	$$
	which means $J$ is finitely generated. Thus we conclude $S$ is Noetherian.
\end{solution}

\begin{problem}[4.3]
	Prove that the ideal $(2, x)$ of $\Z[x]$ is not principal.
\end{problem}
\begin{solution}
	Suppose $(f)=(2,x)$. Since it is easy to see $f\ne 0$ and $f\ne 1$, there must be 
	$$
	2=gf\implies f=2.
	$$ 
	However, it is impossible to find some $h\in \Z[x]$ such that
	\[
	2+x=hf=2h,
	\]
	which leads to a contradiction. Thus we show that the ideal $(2, x)$ of $\Z[x]$ is not principal.
\end{solution}

\hypertarget{Exercise III.4.5}{}
\begin{problem}[4.5]
Let $I, J$ be ideals in a commutative ring $R$, such that $I+J = (1)$. Prove that $IJ = I \cap J$.[\textsection4.1]
\end{problem}
\begin{solution}
For any $k\in IJ$, we can assume that $k=ab$, ($a\in I$, $b\in J$). Note that $k\in aJ=J$ and $k\in Ib=I$. It deduces that $k\in I\cap J$. Thus we show $IJ\subseteq I\cap J$.

\noindent Suppose $l\in I\cap J$. If $1=a+b$ ($a\in I$, $b\in J$), Then we have $l=1*l=(a+b)l=al+lb\in IJ$, which implies that $I\cap J\subseteq IJ$. Therefore, we show $IJ = I \cap J$.
\end{solution}

\begin{problem}[4.6]
	Let $I, J$ be ideals in a commutative ring $R$. Assume that $R/(IJ)$ is reduced (that is, it has no nonzero nilpotent elements; cf. \hyperlink{Exercise III.3.13}{Exercise III.3.13}). Prove that $IJ = I \cap J$.
\end{problem}
\begin{solution}
	The notation $(IJ)$ suggests $R$ is commutative. As is shown in \hyperlink{Exercise III.4.5}{Exercise III.4.5}, it holds that $IJ\subseteq I\cap J$. Thus we are left to show $I\cap J\subseteq IJ$. Suppose $l\in I\cap J$. The condition that $R/(IJ)$ is reduced tells that $\forall r\in R$, 
	\[
	r^n\in IJ\implies r\in IJ.
	\]
	Noticing $l\in I$ and $l\in J$, it is clear that $l^2\in IJ$ which implies $l\in IJ$. There we show $I\cap J\subseteq IJ$ and complete the proof. 
\end{solution}

\begin{problem}[4.7]
	$\vartriangleright$ Let $R = k$ be a field. Prove that every nonzero (principal) ideal in $k[x]$ is generated by a unique \emph{monic} polynomial. [\textsection4.2, \textsection VI.7.2]
\end{problem}
\begin{solution}
    Suppose $I$ is an nonzero ideal in $k[x]$ and the least degree of nonzero polynomials in $I$ is $d$. Since $k$ is a field, we can find a monic polynomial $f(x)=k_0x^d+k_1x^{d+1}+\cdots+x^{d+n}$ in $I$. Given any $g(x)\in I$, there exist unique polynomials $q(x), r(x)\in k[x]$ such that $g(x) = f(x)q(x) + r(x)$ and $\deg r(x) < \deg f(x)=d$. Since $r(x)=g(x)-f(x)q(x)\in I$ and the least degree of nonzero polynomials in $I$ is $d$, there must be $r(x)=0$. Thus we show that $I$ is generated by a monic polynomial $f(x)$. Suppose $I=(f(x))$ can be also generated by a monic polynomial $\bar{f}(x)$. Then we have $\bar{f}(x)=cf(x)$ for some $c\ne0$. Since the two monic polynomials $\bar{f}(x),f(x)$ have the same degree, they are forced to be equal. Therefore, we conclude that every nonzero ideal in $k[x]$ is generated by a unique monic polynomial.
\end{solution}

\begin{problem}[4.8]
	$\vartriangleright$ Let $R$ be a ring, and $f(x)\in R[x]$ a monic polynomial. Prove that $f(x)$ is not a (left-, or right-) zero-divisor. [\textsection4.2, 4.9]
\end{problem}
\begin{solution}
	Suppose $f(x)=x^{d}+a_{d-1} x^{d-1}+\cdots+a_{1} x+a_{0}$ is a monic polynomial in $R[x]$ and $f(x)g(x)=0$ for some $g(x)=b_sx^{s}+b_{s-1} x^{s-1}+\cdots+b_{1} x+b_{0}\in R[x]$. Since the term of the degree of $d+s$ of $f(x)g(x)$ is $b_sx^{d+s}$, there must be $b_s=0$. Then the term of the degree of $d+s-1$ of $f(x)g(x)$ is $b_{s-1}x^{d+s-z}$, which implies $b_{s-1}=0$. Repeating this process we can show that $b_s=b_{s-1}=\cdots=b_0=0$, that is, $g(x)=0$. Thus we see $f(x)$ is not a left-zero-divisor. In a similar way we can show that $f(x)$ is not a right-zero-divisor. 
\end{solution}

\hypertarget{Exercise III.4.10}{}
\begin{problem}[4.10]
	$\neg$ Let $d$ be an integer that is not the square of an integer, and consider the subset of $\mathbb{C}$ defined by 
	\[
	\mathbb{Q}(\sqrt{d}):=\{a+b \sqrt{d} | a, b \in \mathbb{Q}\}
	\]
	\begin{itemize}
		\item Prove that $\mathbb{Q}(\sqrt{d})$ is a subring of $\mathbb{C}$.
		\item Define a function $N: \mathbb{Q}(\sqrt{d}) \rightarrow \mathbb{Q}$ by $N(a+b \sqrt{d}):=a^{2}-b^{2} d .$ Prove that
		\[
		N(z w)=N(z) N(w), \text { and that } N(z) \neq 0 \text { if } z \in \mathbb{Q}(\sqrt{d}), z \neq 0
		\]
		The function $N$ is a `norm'; it is very useful in the study of $\mathbb{Q}(\sqrt{d})$ and of its subrings. (Cf. also \hyperlink{Exercise III.2.5}{Exercise III.2.5}.)
		\item Prove that $\mathbb{Q}(\sqrt{d})$ is a field, and in fact the smallest subfield of $\mathbb{C}$ containing both $\mathbb{Q} \text { and } \sqrt{d} . \text { (Use } N .)$
		\item Prove that $\mathbb{Q}(\sqrt{d})\cong \mathbb{Q}[t] /\left(t^{2}-d\right)$ . (Cf. Example 4.8.)\\
		$[\mathrm{V} .1 .17, \mathrm{V} .2 .18, \mathrm{V} .6 .13, \mathrm{VII} .1 .12]$
	\end{itemize}	
\end{problem}
\begin{solution}
	\begin{itemize}
		\item We only show the check on multiplication
		\[
		(a_1+b_1 \sqrt{d})(a_2+b_2 \sqrt{d})=(a_1a_2+b_1b_2d)+(a_1b_2+a_2b_1)\sqrt{d}\in\Q(\sqrt{d}).
		\]
		\item It is immediate to check $N(z w)=N(z) N(w)$.	Let $z \in \mathbb{Q}(\sqrt{d})$ and $z=a+b\sqrt{d} \neq 0$. Suppose $N(z)=a^2-b^2d=0$. If $b=0$, we have $a=0$, which contradicts with $a+b\sqrt{d} \neq 0$. Otherwise we have $b\ne0$ and $d=(a/b)^2$. Thus we get a contradiction again.
		\item We have known $\mathbb{Q}(\sqrt{d})$ is a commutative ring. For any $z=a+b\sqrt{d}\in\mathbb{Q}(\sqrt{d})$ such that $z\ne0$,
		\[
		N(z)=(a+b \sqrt{d})(a-b \sqrt{d})=a^{2}-b^{2}d\ne0.
		\]
		Therefore
		\[
		\left(a+b \sqrt{d}\right)\left(\frac{a}{N(z)}-\frac{b}{N(z)} \sqrt{d}\right)=1
		\]
		and $\mathbb{Q}(\sqrt{d})$ is a field.
		\item The mapping
		\begin{align*}
		\overline{\varphi}:\mathbb{Q}[t] /\left(t^{2}-d\right)& \longrightarrow \Q(\sqrt{d}),\\
		a+bt+(t^{2}-d)& \longmapsto a+b\sqrt{d}.
		\end{align*}
		is well-defined since if $(a_1+b_1t)-(a_2+b_2t)=g(t)(t^2-d)$, then
		\begin{align*}
		\overline{\varphi}(a_1+b_1t+(t^{2}-d))-\overline{\varphi}(a_2+b_2t+(t^{2}-d))&=\left(a_1+b_1\sqrt{d}\right)-\left(a_2+b_2\sqrt{d}\right)\\
		&=g\left(\sqrt{d}\right)\left(\left(\sqrt{d}\right)^2-d\right)\\
		&=0.
		\end{align*}
		It is clear that $\overline{\varphi}$ preserves addition. Then we can check $\overline{\varphi}$ preserve multiplication:
		\begin{align*}
		&\hspace{1em}\overline{\varphi}\left((a_1+b_1t+(t^{2}-d))(a_2+b_2t+(t^{2}-d))\right)\\
		&=\overline{\varphi}\left((a_1a_2+(a_1b_2+a_2b_1)t+b_1b_2t^2+(t^{2}-d)\right)\\
		&=\overline{\varphi}\left(((a_1a_2+b_1b_2d)+(a_1b_2+a_2b_1)t+b_1b_2(t^2-d)+(t^{2}-d)\right)\\
		&=(a_1a_2+b_1b_2d)+(a_1b_2+a_2b_1)\sqrt{d}\\
		&=(a_1+b_1\sqrt{d})(a_2+b_2\sqrt{d})\\
		&=\overline{\varphi}\left(a_1+b_1t+(t^{2}-d)\right)\overline{\varphi}\left(a_2+b_2t+(t^{2}-d)\right).
		\end{align*}
		Thus we see $\overline{\varphi}$ is a ring homomorphism. Note
		\[
		a+bt+(t^{2}-d)\in\ker \overline{\varphi}\iff a+b\sqrt{d}=0\iff a=b=0.
		\]
		It implies that $\ker \overline{\varphi}=\{0+(t^2-d)\}$ and $\overline{\varphi}$ is injective. It is clear that $\overline{\varphi}$ is surjective. Therefore, $\overline{\varphi}$ is an isomorphism and $\mathbb{Q}(\sqrt{d})\cong \mathbb{Q}[t] /\left(t^{2}-d\right)$.
	\end{itemize}
\end{solution}

\begin{problem}[4.11]
	Let $R$ be a commutative ring, $a \in R,$ and $f_{1}(x), \ldots, f_{r}(x) \in R[x]$.
	\begin{itemize}
		\item Prove the equality of ideals
		\[
		\left(f_{1}(x), \ldots, f_{r}(x), x-a\right)=\left(f_{1}(a), \ldots, f_{r}(a), x-a\right)
		\]
		\item Prove the useful substitution trick
		\[
		\frac{R[x]}{\left(f_{1}(x), \ldots, f_{r}(x), x-a\right)} \cong \frac{R}{\left(f_{1}(a), \ldots, f_{r}(a)\right)}
		\]
		(Hint: \hyperlink{Exercise III.3.3}{Exercise III.3.3}.)
	\end{itemize}	
\end{problem}
\begin{solution}
	\begin{itemize}
		\item According to the polynomial remainder theorem, we have 
		\[
		f_i(x)=(x-a)q(x)+f_i(a),
		\]
		which suffices to show that $\left(f_{1}(x), \ldots, f_{r}(x), x-a\right)=\left(f_{1}(a), \ldots, f_{r}(a), x-a\right)$.
		\item Define
		\begin{align*}
		\varphi:R[x]& \longrightarrow R,\\
		f(x)& \longmapsto f(a).
		\end{align*}
		We can check that $\varphi$ is a surjective ring homomorphism and $\ker\varphi=(x-a)$. According to \hyperlink{Exercise III.3.3}{Exercise III.3.3}, we have
		\[
		\frac{R[x]}{\left(f_{1}(x), \cdots, f_{r}(x), x-a\right)}\cong\frac{R[x]}{\left(f_{1}(a), \cdots, f_{r}(a), x-a\right)}\cong \frac{R[x] / (x-a)}{\overline{\left(f_{1}(a), \cdots, f_{r}(a)\right)}},
		\]
		where
		\[
		\overline{\left(f_{1}(a), \cdots, f_{r}(a)\right)}=\left(f_{1}(a)+(x-a), \cdots, f_{r}(a)+(x-a)\right).
		\]
		The ring isomorphism 
		\begin{align*}
		\psi:R[x]/(x-a)& \longrightarrow R,\\
		f(x)+(x-a)& \longmapsto f(a)
		\end{align*}
		gives the following isomorphism 
		\[
		\frac{R[x] / (x-a)}{\overline{\left(f_{1}(a), \cdots, f_{r}(a)\right)}}\cong \frac{R}{\left(f_{1}(a), \ldots, f_{r}(a)\right)},
		\]
		which completes the proof.
	\end{itemize}
\end{solution}

\hypertarget{Exercise III.4.12}{}
\begin{problem}[4.12]
	$\vartriangleright$ Let $R$ be a commutative ring, and $a_{1}, \cdots, a_{n}$ elements of $R .$ Prove that
	\[
	\frac{R\left[x_{1}, \ldots, x_{n}\right]}{\left(x_{1}-a_{1}, \ldots, x_{n}-a_{n}\right)} \cong R
	\]
	$[\mathrm{VII} .2 .2]$	
\end{problem}
\begin{solution}
	\[
	R\cong \frac{R\left[x_{1}\right]}{\left(x_{1}-a_{1}\right)}\cong \frac{R\left[x_{1},x_{2}\right]}{\left(x_{1}-a_{1},x_2-a_2\right)}
	\]
	The mapping
	\begin{align*}
	\overline{\varphi}:\frac{R\left[x_{1}, \ldots, x_{n}\right]}{\left(x_{1}-a_{1}, \ldots, x_{n}-a_{n}\right)}& \longrightarrow \frac{R\left[x_{1}, \ldots, x_{n-1}\right]}{\left(x_{1}-a_{1}, \ldots, x_{n-1}-a_{n-1}\right)},\\
	f(x_1,\cdots,x_n)+\left(x_{1}-a_{1}, \ldots, x_{n}-a_{n}\right)& \longmapsto f(x_1,\cdots,x_{n-1},a_n)+\left(x_{1}-a_{1}, \ldots, x_{n-1}-a_{n-1}\right)
	\end{align*}
	is well-defined since if $f_1\left(x_1,\cdots,x_n\right)-f_2\left(x_1,\cdots,x_n\right)=\sum\limits_{i=1}^n g_i(x_1,\cdots,x_n)(x_i-a_i)$, then
	\begin{align*}
	\overline{\varphi}\left(\overline{f_1(x_1,\cdots,x_n)}\right)-\overline{\varphi}\left(\overline{f_2(x_1,\cdots,x_n)}\right)&=f_1(x_1,\cdots,x_{n-1},a_n)-f_2(x_1,\cdots,x_{n-1},a_n)\\
	&=\sum_{i=1}^{n-1} g_i(x_1,\cdots,x_{n-1},a_n)(x_i-a_i)+g_n(t)(a_n-a_n)\\
	&=\sum_{i=1}^{n-1} g_i(x_1,\cdots,x_{n-1},a_n)(x_i-a_i).
	\end{align*}
	It is clear that $\overline{\varphi}$ preserves addition and multiplication. Thus we see $\overline{\varphi}$ is a ring homomorphism. Note
	\begin{align*}
	&\hspace{3em}f(x_1,\cdots,x_n)\in\ker \overline{\varphi}\\
	&\iff f(x_1,\cdots,x_{n-1},a_n)=\sum_{i=1}^{n-1} g_i(x_1,\cdots,x_{n-1},a_n)(x_i-a_i)\\
	&\iff f(x_1,\cdots,x_{n-1},x_n)=\sum_{i=1}^{n-1} g_i(x_1,\cdots,x_{n-1},a_n)(x_i-a_i)+g_n(x_1,\cdots,x_{n-1},a_n)(x_n-a_n)\\
	&\iff f(x_1,\cdots,x_n)\in\left(x_{1}-a_{1}, \ldots, x_{n}-a_{n}\right),
	\end{align*}
	where the last but one line can be deduced by the polynomial remainder theorem if we fix $x_1,\cdots,x_{n-1}$ and regard $x_n$ as a variable. It implies that $\ker \overline{\varphi}=\{0+\left(x_{1}-a_{1}, \ldots, x_{n}-a_{n}\right)\}$ and $\overline{\varphi}$ is injective. It is clear that $\overline{\varphi}$ is surjective. Therefore, $\overline{\varphi}$ is an isomorphism and 
	\[
	R\cong \frac{R\left[x_{1}\right]}{\left(x_{1}-a_{1}\right)}\cong \frac{R\left[x_{1},x_{2}\right]}{\left(x_{1}-a_{1},x_2-a_2\right)}\cong\cdots \cong \frac{R\left[x_{1}, \ldots, x_{n}\right]}{\left(x_{1}-a_{1}, \ldots, x_{n}-a_{n}\right)}.
	\]
\end{solution}


\begin{problem}[4.13]
$\triangleright$ Let $R$ be an integral domain. For all $k=1, \ldots, n$ prove that $\left(x_{1}, \ldots, x_{k}\right)$ is prime in $R\left[x_{1}, \ldots, x_{n}\right] .[\S 4.3]$
\end{problem}
\begin{solution}
	Note
	\[
	\frac{R\left[x_{1}, \cdots, x_{n}\right]}{\left(x_{1}, \cdots, x_{k}\right)}\cong \frac{R\left[x_{k+1},\cdots,x_n\right]\left[x_{1}, \cdots, x_{k}\right]}{\left(x_{1}, \cdots, x_{k}\right)}\cong R\left[x_{k+1},\cdots,x_n\right].
	\]
	Since $R$ is an integral domain, we can deduece $R\left[x_{k+1},\cdots,x_n\right]$ is an integral domain. Therefore, $\left(x_{1}, \cdots, x_{k}\right)$ is prime in $R\left[x_{1}, \cdots, x_{n}\right]$.
\end{solution}



\begin{problem}[4.17]
	$\neg$ (If you know a little topology...) Let $K$ be a compact topological space, and let $R$ be the ring of continuous real-valued functions on $K,$ with addition and multiplication defined pointwise.
	\begin{enumerate}[(i)]
		\item For $p \in K,$ let $M_{p}=\{f \in R | f(p)=0\} .$ Prove that $M_{p}$ is a maximal ideal in $R$
		\item Prove that if $f_{1}, \ldots, f_{r} \in R$ have no common zeros, then $\left(f_{1}, \ldots, f_{r}\right)=(1)$ (Hint: consider $f_{1}^{2}+\cdots+f_{r}^{2}$ )
		\item Prove that every maximal ideal $M$ in $R$ is of the form $M_{p}$ for some $p \in K .$ (Hint:
		you will use the compactness of $K$ and (ii).) 
	\end{enumerate}
	If further $K$ is Hausdorff (and, as Bourbaki would have it, compact spaces are Hausdorff), then Urysohn's lemma shows that for any two points $p \neq q$ in $K$ there exists a function $f \in R$ such that $f(p)=0$ and $f(q)=1 .$ If this is the case, conclude that $p \mapsto M_{p}$ defines a bijection from $K$ to the set of maximal ideals of $R .$ (The set of maximal ideals of a commutative ring $R$ is called the \emph{maximal spectrum} of $R$; it is contained in the (prime) spectrum Spec $R$ defined in $\$ 4.3 .$ Relating commutative rings and 'geometric' entities such as topological spaces is the business of \emph{algebraic geometry}.)
	
	The compactness hypothesis is necessary: cf. Exercise V.3.10. [V.3.10]	
\end{problem}
\begin{solution}
	\begin{enumerate}[(i)]
		\item Suppose all functions in $R$ that have same value in a neighborhood of $p$ are identified. It is easy to check that $M_p$ is an ideal and $R/M_p$ is commutative. Given any $f\in R-M_p$, we have $f(p)\ne0$ and
		\[
		(f+M_p)\left(\frac{1}{f}+M_p\right)=1+M_p
		\]
		Therefore, $R/M_p$ is a field and  $M_{p}$ is a maximal ideal in $R$.
		\item If $f_{1}, \ldots, f_{r} \in R$ have no common zeros, $\left(f_{1}, \ldots, f_{r}\right)=(1)$ follows from
		\[
		\sum_{i=1}^r\frac{f_i}{f_1^2+\cdots+f_r^2}f_i=1.
		\]
		\item Suppose $M$ is a maximal ideal in $R$ but there is no such $p\in K$ that $M=M_p$. Thus for any $a\in K$, there exists $f_a\in M$ such that $f_a(a)\ne0$. Since $f_a$ is continous, there exists a open neighborhood of $a$ denoted by $U_a$ such that 
		\[
		\forall x\in U_a, f(x)\ne0.
		\]
		Note that $\{U_a\}_{a\in K}$ is a open cover of $K$. According to the compactness of $K$, we can suppose $\{U_{p_1},U_{p_2},\cdots,U_{p_n}\}$ covers $K$. Since given any $x\in K$, there exists $i\in\{1,2,\cdots,n\}$ such that $x\in U_{p_i}$ and $f_{p_i}(x)\ne 0$, we can deduce that $f_{p_1},f_{p_2}, \cdots, f_{p_n} \in R$ have no common zeros. The conclusion of (ii) tells us $\left(f_{p_1},f_{p_2}, \cdots, f_{p_n}\right)=(1)$, which implies $M=(1)$. However, this contradicts with the assumption that $M$ is a maximal ideal. Therefore, we prove that every maximal ideal $M$ in $R$ is of the form $M_{p}$ for some $p \in K$.
	\end{enumerate}
If further $K$ is Hausdorff, for any two points $p \neq q$ in $K$, there exists a function $f \in R$ such that $f(p)=0$ and $f(q)=1$, which implies
\[
f\in M_p, \ f\notin M_q\implies M_p\ne M_q.
\] 
Thus we see the mapping 
\begin{align*}
	\varphi:K& \longrightarrow \text{maximal spectrum of } R,\\
	p& \longmapsto M_{p}
\end{align*}
 is injective. In (iii) we have shown that $\varphi$ is surjective. Therefore, $\varphi$ is a bijection.
\end{solution}


\begin{problem}[4.23]
A ring $R$ has Krull dimension 0 if every prime ideal in $R$ is maximal. Prove that fields and boolean rings (\hyperlink{Exercise III.3.15}{Exercise III.3.15}) have Krull dimension $0 .$
\end{problem}
\begin{solution}
Assume that $R$ is a field. If $I$ be a prime ideal in $R$, then $R/I$ is an integral domain. Given any $r+I\in R/I$, we have
\[
(r+I)(r^{-1}+I)=(r^{-1}+I)(r+I)=1_R+I,
\]
which means $R/I$ is a field. Thus $I$ is maximal and $R$ has Krull dimension $0$.

Assume that $R$ is a boolean ring. If $I$ be a prime ideal in $R$, then $R/I$ is an integral domain. Given any $r+I\in R/I$, we have
\[
(r+I)(r+I)=r^2+I=r+I\implies (r+I)\left((r+I)-(1_R+I)\right)=0_R+I,
\]
which implies
\[
r+I=0_R+I\text{  or  }r+I=1_R+I.
\]
Thus we see $R/I\cong \mathbb{Z}/2\mathbb{Z}$ and $R/I$ is a field.
\end{solution}


\subsection{\textsection5. Modules over a ring}
\begin{problem}[5.1]
$\vartriangleright$ Let $R$ be a ring. The \emph{opposite} ring $R^{\circ}$ is obtained from $R$ by reversing the multiplication: that is, the product $a \bullet b$ in $R^{\circ}$ is defined to be $b a \in R .$ Prove that the identity map $R \rightarrow R^{\circ}$ is an isomorphism if and only if $R$ is commutative. Prove that $\mathcal{M}_{n}(\mathbb{R})$ is isomorphic to its opposite (not via the identity map!). Explain how to turn right-$R$-modules into left-$R$-modules and conversely, if $R \cong R^{\circ}$. [\textsection5.1,VIII.5.19 ]
\end{problem}
\begin{solution}
	Let $i$ denote the identity map $R \rightarrow R^{\circ}$. If $R$ is commutative, we have
	\[
	i(ab)=ab=b\bullet a=i(b)\bullet i(a)=i(a)\bullet i(b).
	\]
	Given that $i(a+b)=a+b$ and identity map is a bijection, we see that $i$ is an isomorphism.
	
	If $i$ is an isomorphism, we have
	\[
	ab=b\bullet a=i^{-1}(b\bullet a)=i^{-1}(b)i^{-1}(a)=ba,
	\]
	which implies that $R$ is commutative.
	
	Suppose $A,B\in\mathcal{M}_{n}(\mathbb{R})$. We can show that the transpose of matrix $\cdot^T:A\mapsto A^T$ 
	is an isomorphism by checking
	\[
	(AB)^T=B^TA^T=A^T \bullet B^T.
	\]
	Let $M$ be a right-$R$-module with right multiplication $\odot$. If $R \cong R^{\circ}$ and $f:R \rightarrow R^{\circ}$ is an isomorphism, then
	\[
	f(ab)=f(a)\bullet f(b)=f(b)f(a)
	\]
	Define left multiplication $\odot_L$ as
	\[
	r\odot_L m:=m\odot f(r),\qquad\forall r\in R,m\in M.
	\]
	We can check that
	\begin{align*}
		&1\odot_L m=m \odot f(1)=1,\\
		&(rs)\odot_L m=m\odot f(rs)=m\odot (f(s)f(r))=(m\odot f(s))\odot f(r)\\
		&\hspace{4.5em}=(s\odot_L m)\odot f(r)=r\odot_L(s\odot_L m),\\
		&r\odot_L(m_1+m_2)=(m_1+m_2)\odot f(r)=m_1\odot f(r)+m_2\odot f(r)=r\odot_L m_1+r\odot_L m_2.
	\end{align*}
	Therefore, we show that $M$ is a left-$R$-module with right multiplication $\odot_L$.
	
	If $M$ is a left-$R$-module with left multiplication $*$ and $f:R \rightarrow R^{\circ}$ is an isomorphism, then we can show that $M$ is a right-$R$-module with right multiplication $*_R$ defined as
	\[
	m*_R r := f^{-1}(r),\qquad\forall r\in R,m\in M.
	\]
\end{solution}

\begin{problem}[5.3]
$\vartriangleright$ Let $M$ be a module over a ring $R .$ Prove that $0 \cdot m=0$ and that $(-1) \cdot m=$ $-m,$ for all $m \in M$. [\textsection 5.2]
\end{problem}
\begin{solution}
	\begin{align*}
	&0\cdot m=(0+0)\cdot m=0\cdot m+0\cdot m\implies0\cdot m=0,\\
	&0=(1-1)\cdot m=1\cdot m+(-1) \cdot m\implies(-1) \cdot m=-m.
	\end{align*}
\end{solution}

\hypertarget{Exercise III.5.4}{}
\begin{problem}[5.4]
$\neg$ Let $R$ be a ring. A nonzero $R$-module $M$ is \emph{simple} (or \emph{irreducible}) if its only submodules are $\{0\}$ and $M$. Let $M$, $N$ be simple modules, and let $\varphi : M \to N$ be a homomorphism of $R$-modules. Prove that either $\varphi= 0$, or $\varphi$ is an isomorphism. (This rather innocent statement is known as Schur's lemma.) [5.10, 6.16, VI.1.16]
\end{problem}
\begin{solution}
For convenience, we talk about the identity of modules up to isomorphism. Since the nonzero $R$-module $M$ is simple, $\ker\varphi $ is either $\{0\}$ or $M$. Thus $\im \varphi= M/ \ker\varphi$ is either $\{0\}$ or $M$. Note that $\im \varphi\subset N$ is either $\{0\}$ or $N$. If $\im \varphi=\{0\}$, then we have $\varphi=0$. If $\im \varphi=M$, then we have $\im \varphi=M=N$. Therefore we show that either $\varphi= 0$, or $\varphi$ is an isomorphism. 
\end{solution}

\begin{problem}[5.5]
Let $R$ be a commutative ring, viewed as an $R$-module over itself, and let $M$ be an $R$-module. Prove that $\mathrm{Hom}_{R-\mathrm{Mod}}(R, M) \cong M$ as $R$-modules.
\end{problem}
\begin{solution}
	Define
	\begin{align*}
	\varphi:\mathrm{Hom}_{R-\mathrm{Mod}}(R, M)& \longrightarrow M,\\
	f& \longmapsto f(1)
	\end{align*}
	Since
	\begin{align*}
	\varphi(f+g)&=(f+g)(1)=f(1)+g(1)=\varphi(f)+\varphi(g),\\
	\varphi(rf)&=(rf)(1)=rf(1)=r\varphi(f),
	\end{align*}
	we see $\varphi$ is a homomorphism.
	If $\varphi(f_1)=\varphi(f_2)$, we have $f_1(1)=f_2(1)$. Multiply both sides by any $r\in R$ and we get
	\[
	rf_1(1)=rf_2(1)\implies f_1(r)=f_2(r),
	\]
	which means $f_1=f_2$. Thus we show $\varphi$ is injective. Given any $m\in M$, let
	\begin{align*}
	h_m:R& \longrightarrow M,\\
	r& \longmapsto rm
	\end{align*}
	Since $\varphi(h_m)=h_m(1)=m$, we show that $\varphi$ is surjective. Therefore, we conclude that $\varphi$ is an isomorphism and $\mathrm{Hom}_{R-\mathrm{Mod}}(R, M) \cong M$ as $R$-modules.
\end{solution}

\begin{problem}[5.6]
Let $G$ be an abelian group. Prove that if $G$ has a structure of $\mathbb{Q}$-vector space, then it has only one such structure. (Hint: First prove that every nonidentity element of $G$ has necessarily infinite order. Alternative hint: The unique ring homomorphism $\mathbb{Z} \rightarrow \mathbb{Q}$ is an epimorphism.)
\end{problem}
\begin{solution}
Assume that $G$ has two structures of $\mathbb{Q}$-vector space with scalar multiplication operations $\cdot$ and $*$ respectively. Note that $1\cdot g=1*g=g$ for all $g\in G$. With the conventional notation $\sum_{i=1}^{n}g=ng$, we have for all $g\in G$,
\[
\sum_{i=1}^{n}1\cdot g=\sum_{i=1}^{n}1* g=ng\implies \left(\sum_{i=1}^{n}1\right)\cdot g=\left(\sum_{i=1}^{n}1\right)* g=ng\implies n\cdot g=n*g=ng.
\]
Since
\begin{align*}
\sum_{i=1}^{m}\frac{1}{m}\cdot h&=\left(\sum_{i=1}^{m}\frac{1}{m}\right)\cdot h=h,\qquad\forall h\in G,\\
\sum_{i=1}^{m}\frac{1}{m}* h&=\left(\sum_{i=1}^{m}\frac{1}{m}\right)* h=h,\qquad\forall h\in G,
\end{align*}
it holds that for all $h\in G$,
\[
\sum_{i=1}^{m}\frac{1}{m}\cdot h=\sum_{i=1}^{m}\frac{1}{m}* h\implies\sum_{i=1}^{m}\left(\frac{1}{m}\cdot h-\frac{1}{m}* h\right)=0\implies m\cdot\left(\frac{1}{m}\cdot h-\frac{1}{m}* h\right)=0.
\]
According to the property of vector space, we have $m=0$ or $\frac{1}{m}\cdot h-\frac{1}{m}* h=0$. However, $m$ is a positive integer, which forces $\frac{1}{m}\cdot h=\frac{1}{m}* h$. Thus we can deduce that for all $h\in G$, $n,m\in \mathbb{Z}_+$,
\[
n\cdot \left(\frac{1}{m}\cdot h\right)=n*\left(\frac{1}{m}* h\right)\implies \frac{n}{m}\cdot h=\frac{n}{m}*h.
\]
In other words, for all $h\in G$, $q\in \mathbb{Q}_+$, we have $q\cdot h=q* h$. Note that $(-q)\cdot h=(-q)* h$ and $0\cdot h=0* h$, finally we obtain that for all $h\in G$, $q\in \mathbb{Q}$, 
\[
q\cdot h=q* h.
\]
Therefore, the two scalar multiplication operations $\cdot$ and $*$ coincide, which completes the proof.
\end{solution}


\begin{problem}[5.7]
Let $K$ be a field, and let $k \subseteq K$ be a subfield of $K$. Show that $K$ is a vector space over $k$ (and in fact a $k$-algebra) in a natural way. In this situation, we say that $K$ is an extension of $k.$
\end{problem}
\begin{solution}
	Define the scalar multiplication $\cdot$ as
	\[
	a\cdot x:=ax, \qquad\forall a\in k,x\in K.
	\]
	Then we can check that for all $a,b\in k$, $x,y\in K$,
	\begin{align*}
	&1\cdot x=x,\\
	&(ab)\cdot x=(ab) x=a(bx)=a\cdot(b\cdot x),\\
	&(a+b)\cdot x=(a+b)x=ax+bx=a\cdot x+b\cdot x,\\
	&a\cdot(x+y)=a(x+y)=ax+ay=a\cdot x+a\cdot y,\\
	&(a\cdot x)(b\cdot y)=(ax)(by)=(ab)(xy)=(ab)\cdot(xy).
	\end{align*}
	Therefore, $K$ is a $k$-vector space and a $k$-algebra as well.
\end{solution}

\begin{problem}[5.8]
	What is the initial object of the category $R$-$\mathsf{Alg}$?
\end{problem}
\begin{solution}
    The ring $R$ can be seen as a $R$-algebra if it is endowed with a scalar multiplication $\cdot$ in a natural way, that is 
	\[
	r\cdot x:=rx, \qquad\forall r\in R,x\in R.
	\]
	Given any $R$-algebra $A$, define the following map
	\begin{align*}
	f:R& \longrightarrow A,\\
	r& \longmapsto r\cdot 1_A.
	\end{align*}
	We can check that
	\begin{align*}
	f(r_1+r_2)&=(r_1+r_2)\cdot 1_A=r_1\cdot 1_A+r_2\cdot 1_A=f(r_1)+f(r_2),\\
	f(r_1r_2)&=(r_1r_2)\cdot 1_A=r_1\cdot(r_2\cdot 1_A)=r_1\cdot (1_A(r_2\cdot 1_A))=(r_1\cdot 1_A)(r_2\cdot 1_A)=f(r_1)f(r_2),\\
	f(r_1\cdot r_2)&=f(r_1r_2)=r_1\cdot(r_2\cdot 1_A)=r_1\cdot f(r_2).
	\end{align*}
	Hence $f$ is a morphism in the category $R$-$\mathsf{Alg}$.
	
	Suppose $g:R\to A$ is a morphism in $R$-$\mathsf{Alg}$. Then for all $r_1,r_2\in R$,
	\[
	g(r_1r_2)=g(r_1\cdot r_2)\implies g(r_1)g(r_2)=r_1\cdot g(r_2)=r_1\cdot(1_Ag(r_2))=(r_1\cdot1_A)g(r_2).
	\]
	Take $r_2=1_R$ and then for all $r_1\in R$,
	\[
	g(r_1)g(1_R)=(r_1\cdot1_A)g(1_R)\implies g(r_1)=r_1\cdot1_A.
	\]
	Thus we have $g=f$. Therefore, we show that for any $R$-algebra $A$, there exists a unique morphism $f:R\to A$ in $R$-$\mathsf{Alg}$. In other words, $R$ is the initial object of the category $R$-$\mathsf{Alg}$.
\end{solution}


\begin{problem}[5.9]
	$\neg$ Let $R$ be a commutative ring, and let $M$ be an $R$-module. Prove that the operation of composition on the $R$-module $\mathrm{End}_{R-\mathsf{Mod}}(M)$ makes the latter an $R$-algebra in a natural way.
	
	Prove that $\mathcal{M}_{n}(R)$ (cf. \hyperlink{Exercise III.1.4}{Exercise III.1.4} ) is an $R$-algebra, in a natural way. $[\mathrm{VI} .1 .12,$ $\mathrm{VI} .2 .3]$
\end{problem}
\begin{solution}
	In textbook we have show that $\mathrm{End}_{R-\mathsf{Mod}}(M)$ is an $R$-module with natural addition and scalar multiplication.
	We can check that for all $f,g,h\in\mathrm{End}_{R-\mathsf{Mod}}(M)$, $r,s\in R$, $x\in M$,
	\begin{align*}
	(id_M\circ f)(x)&=id_M(f(x))=f(x),\\
	((f+g)\circ h)(x)&=(f+g)(h(x))=f(h(x))+g(h(x))=(f\circ h)(x)+(g\circ h)(x)\\
	&=(f\circ h+g\circ h)(x),\\
	((r\cdot f)\circ(sg))(x)&=(r\cdot f)((s\cdot g)(x))=r\cdot f(s\cdot g(x))=r\cdot (s\cdot f(g(x)))\\
	&=(rs) \cdot(f(g(x))=(rs) \cdot((f\circ g)(x))=((rs) \cdot(f\circ g))(x).
	\end{align*}
	Thus we prove that the operation of composition $\circ$ on the $R$-module $\mathrm{End}_{R-\mathsf{Mod}}(M)$ makes $\mathrm{End}_{R-\mathsf{Mod}}(M)$ an $R$-algebra.
	
	In \hyperlink{Exercise III.1.4}{Exercise III.1.4} we have shown that $\mathcal{M}_{n}(R)$ is a ring. Let the scalar multiplication $\cdot$ be componentwise multiplication, namely
	\[
	r\cdot(a_{ij})_{n\times n}:=(ra_{ij})_{n\times n},\qquad\forall r\in R,(a_{ij})_{n\times n}\in \mathcal{M}_{n}(R).
	\]
	We can check that
	\begin{align*}
	1_R\cdot(a_{ij})_{n\times n}&=(1_Ra_{ij})_{n\times n}=(a_{ij})_{n\times n},\\
	(r+s)\cdot(a_{ij})_{n\times n}&=((r+s)a_{ij})_{n\times n}=(ra_{ij})_{n\times n}+(sa_{ij})_{n\times n}\\
	&=r\cdot(a_{ij})_{n\times n}+s\cdot(a_{ij})_{n\times n},\\
	r\cdot\left(\left(a_{ij}\right)_{n\times n}+\left(b_{ij}\right)_{n\times n}\right)&=r\cdot\left(a_{ij}+b_{ij}\right)_{n\times n}=\left(r\left(a_{ij}+rb_{ij}\right)\right)_{n\times n}\\
	&=\left(ra_{ij}\right)_{n\times n}+\left(rb_{ij}\right)_{n\times n}=r\cdot\left(a_{ij}\right)_{n\times n}+r\cdot\left(b_{ij}\right)_{n\times n},\\
	(rs)\cdot\left(a_{ij}\right)_{n\times n}&=	\left((rs)a_{ij}\right)_{n\times n}=r\cdot\left(sa_{ij}\right)_{n\times n}=r\cdot \left(s\cdot\left(a_{ij}\right)_{n\times n}\right),\\
	\left(r\cdot \left(a_{ij}\right)_{n\times n}\right)\left(s\cdot \left(b_{ij}\right)_{n\times n}\right)&=(ra_{ij})_{n\times n}(sb_{ij})_{n\times n}=\left(\sum_{k=1}^n\left(ra_{ik}\right)\left(sb_{kj}\right)\right)_{n\times n}\\
	&=\left((rs)\sum_{k=1}^n a_{ik}b_{kj}\right)_{n\times n}=(rs)\cdot\left(\sum_{k=1}^n a_{ik}b_{kj}\right)_{n\times n}\\
	&=(rs)\cdot\left(\left(a_{ij}\right)_{n\times n}\left(b_{ij}\right)_{n\times n}\right).
	\end{align*}
	Therefore, $\mathcal{M}_{n}(R)$ is an $R$-algebra.
\end{solution}


\begin{problem}[5.11]
	$\vartriangleright$ Let $R$ be a commutative ring, and let $M$ be an $R$-module. Prove that there is a bijection between the set of $R[x]$-module structures on $M$ (extending the given $R$-module structure) and $\mathrm{End}_{R-\mathsf{Mod}}(M)$. $[\$ \mathrm{VI} .7 .1]$
\end{problem}
\begin{solution}	
	According to \hyperlink{Exercise III.5.9}{Exercise III.5.9}, $\mathrm{End}_{R-\mathsf{Mod}}(M)$ has an $R$-algebra structure, which can induce an $R[x]$-module structure on $\mathrm{End}_{R-\mathsf{Mod}}(M)$. That is, for all $f(x)=r_0+r_1x+\cdots+ r_nx^n\in R[x],\varphi\in \mathrm{End}_{R-\mathsf{Mod}}(M)$,
	\[
	f(x)\cdot \varphi:=f(\varphi)=r_0+r_1\varphi+\cdots+ r_n\varphi^n.
	\] 
	Given any $\varphi\in\mathrm{End}_{R-\mathsf{Mod}}(M)$, define the following $R[x]$-module structures on $M$ with scalar multiplication $\cdot_\varphi$,
	\[
	f(x)\cdot_\varphi m:=(f(\varphi))(m), \qquad\forall f(x)\in R[x],m\in M.
	\]
	What we need is to show that the map $\varphi\mapsto\cdot_\varphi$ is a bijection.
	

	If $\cdot_\varphi=\cdot_\eta$, we have $(f(\varphi))(m)=(f(\eta))(m)$. Take $f(x)=x$ and then we have $\varphi(m)=\eta(m)$ for all $m\in M$, which implies $\varphi=\eta$. Hence the map $\varphi\mapsto\cdot_\varphi$ is injective.
	
	Suppose $\bullet$ is a scalar multiplication which makes $M$ an $R[x]$-module.
	\begin{itemize}
		\item $f(x)\bullet (m+n)=f(x)\bullet m+f(x)\bullet n$
		\item $(f(x)+g(x))\bullet m=f(x)\bullet m+g(x)\bullet m$
		\item $(f(x) g(x))\bullet m=f(x)\bullet(g(x)\bullet m)$
		\item $1\bullet m=m$
	\end{itemize}
	
	
\end{solution}

\subsection{\textsection6. Products, coproducts, etc. in $R$-$\Mod$}
\begin{problem}[6.3]
Let $R$ be a ring, $M$ an $R$-module, and $p : M \to M$ an $R$-module homomorphism such that $p^2 = p$. (Such a map is called a projection.) Prove that $M\cong\ker p \oplus\im p$.
\end{problem}
\begin{solution}
For any $x\in\ker p \cap \im p$, we can assume $x=py$ and deduce that $0=px=p^2y=py=x$. Thus we have $\ker p \cap \im p=\{0\}$. Furthermore, for any $m\in M$,
\[
	p(m - p(m)) = p(m) - p^2(m) = p(m) - p(m) = 0\implies m - p(m)\in \ker p.
\] 
This implies $m = \left(m - p(m)\right) + p(m)\in \ker p \oplus \im p$, thereby establishing the isomorphism $M \cong \ker p \oplus \im p$.
\end{solution}


\hypertarget{Exercise III.6.16}{}
\begin{problem}[6.16]
	$\vartriangleright$ Let $R$ be a ring. A (left-) $R$-module $M$ is \emph{cyclic} if $M=\langle m\rangle$ for some $m \in M .$ Prove that simple modules (cf. \hyperlink{Exercise III.5.4}{Exercise III.5.4}) are cyclic. Prove that an $R$-module $M$ is cyclic if and only if $M \cong R / I$ for some (left-)ideal $I$. Prove that every quotient of a cyclic module is cyclic. $[6.17$, \textsection $\mathrm{VI}.4.1]$
\end{problem}
\begin{solution}
	Let $M$ be a simple module and $m\in M$. We see $\langle m\rangle$ is a submodule of $M$. Since the submodule of simple modules can only be $\{0\}$ or itself, there must be $\langle m\rangle=M$, which implies $M$ is cyclic.
	
	If $M \cong R / I$ for some (left-)ideal $I$, we can suppose $f:R / I\to M$ is an isomorphism. Since for any $m\in M$, there exists $r+I\in R/I$ such that
	\[
	m=f(r+I)=rf(1_R+I),
	\]
	we have $M=\langle f(1_R+I)\rangle$. Thus we show $M$ is cyclic.
	
	If $M$ is cyclic, we can suppose $M=\langle m\rangle$ where $m\in M$. Define
	\begin{align*}
		\varphi:R^1 &\longrightarrow M ,\\
		r&\longmapsto rm.
	\end{align*}
	Since any element in $M$ is in the form of $rm$ where $r\in R$ and $m\in M$, $\varphi$ must be surjective. Note that $\varphi$ preseves addition and scalar mutiplication, which implies $\varphi$ is a ring homomorphism. Thus we have $R/\ker\varphi\cong M$, where $\ker\varphi$ is an ideal of $R$.
	
	Suppose $M=\langle m_0\rangle$ is a cyclic module and $M/N$ is a quotient of $M$. Given any element $m+N \in M/N$, we have
	\[
	m+N=rm_0+N=r(m_0+N),\quad r\in R,
	\]
	which implies $M/N=\langle m_0+N\rangle$. Therefore, we show that every quotient of a cyclic module is cyclic.

\end{solution}

\hypertarget{Exercise III.6.17}{}
\begin{problem}[6.17]
	$\neg$ Let $M$ be a cyclic $R$-module, so that $M \cong R / I$ for a (left-)ideal $I$ (\hyperlink{Exercise III.6.16}{Exercise III.6.16}), and let $N$ be another $R$-module.
	\begin{itemize}
		\item Prove that $\operatorname{Hom}_{R-\operatorname{Mod}}(M, N) \cong\{n \in N \mid(\forall a \in I), a n=0\}$
		\item For $a, b \in \mathbb{Z},$ prove that $\operatorname{Hom}_{R-\operatorname{Mod}}(\mathbb{Z} / a \mathbb{Z}, \mathbb{Z} / b \mathbb{Z}) \cong \mathbb{Z} / \operatorname{gcd}(a, b) \mathbb{Z}$.
	\end{itemize}
	[7.7]
\end{problem}
\begin{solution}
	\begin{itemize}
		\item First of all it is easy to check 
		\[
		\{n \in N \mid(\forall a \in I), a n=0\}
		\]
		is a module. Let $M=\langle m_0\rangle$ and let $\psi: R / I\to M,r+I\mapsto rm_0$ be an isomorphism. Define
\begin{align*}
	\varphi:\operatorname{Hom}_{R-\operatorname{Mod}}(M, N) &\longrightarrow \{n \in N \mid(\forall a \in I), a n=0\} ,\\
	f&\longmapsto f(m_0)=f(\psi(1_R+I)),
\end{align*}
where $f(\psi(1_R+I))\in\{n \in N \mid(\forall a \in I), a n=0\}$ holds because 
\[
\forall a \in I,\quad af(\psi(1_R+I))=f(a\psi(1_R+I))=f(\psi(a(1_R+I)))=f(\psi(0_R+I))=0_N.
\]
It is straightforward to check $\varphi$ is a homomorphism. Note
\[
f(rm_0)=rf(m_0)=rf(\psi(1_R+I)),\quad \forall r\in R.
\]
We have 
\[
\varphi(f)=\varphi(g)\implies f(\psi(1_R+I))=g(\psi(1_R+I))\implies f=g,
\]
which implies $\varphi$ is a monomorphism. 

For any $n_0\in\{n \in N \mid(\forall a \in I), a n=0\}$, let
\begin{align*}
	f_0:M &\longrightarrow N ,\\
	rm_0&\longmapsto rn_0.
\end{align*}
It is clear to see $f_0\in \operatorname{Hom}_{R-\operatorname{Mod}}(M, N)$ and
\[
\varphi(f_0)=f_0(m_0)=n_0.
\]
Thus $\varphi$ is an epimorphism.
Therefore, we show that $\varphi$ is an isomorphism and 
\[
\operatorname{Hom}_{R-\operatorname{Mod}}(M, N) \cong\{n \in N \mid(\forall a \in I), a n=0\}.
\]
\item What is left to show is 
\[
\{[n]_b \in \mathbb{Z} / b \mathbb{Z} :(\forall k \in  a \mathbb{Z}), k [n]_b=0\}\cong\mathbb{Z} / \operatorname{gcd}(a, b) \mathbb{Z}.
\]
Note
\[
\forall k \in  a \mathbb{Z}, k [n]_b=0\iff \forall r\in\mathbb{Z},b\mid ran \iff b\mid an \iff  b/\gcd(a, b)\mid n.
\]
We have
\[
\{[n]_b \in \mathbb{Z} / b \mathbb{Z} :(\forall k \in  a \mathbb{Z}), k [n]_b=0\}=\{[n]_b \in \mathbb{Z} / b \mathbb{Z} : b/\gcd(a, b)\mid n\}.
\]
Define
\begin{align*}
	h:\{[n]_b \in \mathbb{Z} / b \mathbb{Z} : b/\gcd(a, b)\mid n\} &\longrightarrow\mathbb{Z}/\operatorname{gcd}(a, b) \mathbb{Z} ,\\
	[kb/\gcd(a, b)]_b&\longmapsto [k]_{\operatorname{gcd}(a, b)}.
\end{align*}
Since $h$ is the restriction of the projection $\pi:\mathbb{Z} / b \mathbb{Z}\to \mathbb{Z}/\operatorname{gcd}(a, b) \mathbb{Z}$, $h$ is a homomorphism. Noticing $h$ is surjective and 
\[
\left|\{[n]_b \in \mathbb{Z} / b \mathbb{Z} : b/\gcd(a, b)\mid n\}\right|=\left|\mathbb{Z}/\operatorname{gcd}(a, b) \mathbb{Z}\right|=\operatorname{gcd}(a, b),
\]
we can conclude
\[
\operatorname{Hom}_{R-\operatorname{Mod}}(\mathbb{Z} / a \mathbb{Z}, \mathbb{Z} / b \mathbb{Z}) \cong\{[n]_b \in \mathbb{Z} / b \mathbb{Z} : b/\gcd(a, b)\mid n\} \cong \mathbb{Z} / \operatorname{gcd}(a, b) \mathbb{Z}.
\]
\end{itemize}
\end{solution}


\subsection{\textsection7. Complexes and homology}
\begin{problem}[7.1]
Assume that the complex
\[
\cdots \longrightarrow 0 \longrightarrow M \longrightarrow 0 \longrightarrow \cdots
\]
is exact. Prove that $M\cong0$. [\textsection7.3]
\end{problem}
\begin{solution}
Assume that $f:0\to M$ and $g:M\to 0$. Since the the complex is exact, we have $$\{0\}=\im f=\ker g=M.$$
\end{solution}



\begin{problem}[7.2]
Assume that the complex
\[
\cdots \longrightarrow 0 \longrightarrow M \longrightarrow M^{\prime} \longrightarrow 0 \longrightarrow \cdots
\]
is exact. Prove that $M \cong M^{\prime}$.
\end{problem}
\begin{solution}
	Suppose that  that $f:M\to M'$. According to the exactness of the complex, we have $f$ is injective and surjective. Thus we show that $M \cong M^{\prime}$.
\end{solution}


\begin{problem}[7.3]
Assume that the complex
\[
\cdots \longrightarrow 0 \longrightarrow L \longrightarrow M \stackrel{\varphi}{\longrightarrow} M^{\prime} \longrightarrow N \longrightarrow 0 \longrightarrow \cdots
\]
is exact. Show that, up to natural identifications, $L=\operatorname{ker} \varphi$ and $N=\operatorname{coker} \varphi$.

\end{problem}
\begin{solution}
Suppose that $f:L\to M$ and $g:M'\to N$. According to the exactness of the complex, we have $f$ is injective, $g$ is surjective. Thus we have
$$
\ker\varphi=\im f\cong L
$$
and
$$
\operatorname{coker} \varphi= M'/\im \varphi=M'/\ker g\cong N.
$$
\end{solution}

\begin{problem}[7.4]
Construct short exact sequences of $\mathbb{Z}$-modules
\[
0 \longrightarrow \mathbb{Z}^{\oplus \mathbb{N}} \longrightarrow \mathbb{Z}^{\oplus \mathbb{N}} \longrightarrow \mathbb{Z} \longrightarrow 0
\]
and
\[
0 \longrightarrow \mathbb{Z}^{\oplus \mathbb{N}} \longrightarrow \mathbb{Z}^{\oplus \mathbb{N}} \longrightarrow \mathbb{Z}^{\oplus \mathbb{N}} \longrightarrow 0
\]
(Hint: David Hilbert's Grand Hotel.)
\end{problem}

\begin{solution}
Assume
\[
0 \longrightarrow \mathbb{Z}^{\oplus \mathbb{N}} \operatornamewithlimits{\longrightarrow}^{f_1} \mathbb{Z}^{\oplus \mathbb{N}} \operatornamewithlimits{\longrightarrow}^{g_1} \mathbb{Z} \longrightarrow 0
\]	
Define a monomorphism
$$
f_1((x_1,x_2,x_3,\cdots))=(0,x_1,x_2,\cdots)
$$
and an epimorphism
\[
g_1((x_1,x_2,x_3,\cdots))=x_1.
\]
It is clear to see $g_1\circ f_1=0$, which implies the exactness of the complex.

\noindent Given the complex
\[
0 \longrightarrow \mathbb{Z}^{\oplus \mathbb{N}} \operatornamewithlimits{\longrightarrow}^{f_2} \mathbb{Z}^{\oplus \mathbb{N}} \operatornamewithlimits{\longrightarrow}^{g_2} \mathbb{Z}^{\oplus \mathbb{N}}\longrightarrow 0,
\]
where
$$
f_2((x_1,x_2,x_3,x_4,\cdots))=(x_1,0,x_2,0,\cdots)
$$
and 
\[
g_2((x_1,x_2,x_3,x_4,\cdots))=(x_2,x_4,x_6,x_8,\cdots),
\]
we can check that $f_2$ is a monomorphism, $g_2$ is an epimorphism and $g_2\circ f_2=0$. Therefore, we have constructed a short exact sequence.
\end{solution}

\begin{problem}[7.5]
$\vartriangleright$ Assume that the complex
\[
\cdots \longrightarrow L \longrightarrow M\longrightarrow N \rightarrow\cdots
\]
is exact, and that $L$ and $N$ are Noetherian. Prove that $M$ is Noetherian. [\textsection7.1]
\end{problem}
\begin{solution}
Suppose that $f:L\to M$ and $g:M\to N$. According to the exactness of the complex, we have $\ker g=\im f$. Since $N$ is Noetherian and $\im g$ is a submodule of $N$, $\im g$ is Noetherian. Note that $\im g\cong M/\ker g=M/\im f$, we see $M/\im f$ is Noetherian. Since $L$ is Noetherian and $\im f\cong L/\ker f$, we have $\im f$ is Noetherian. Since $M/\im f$ and $\im f$ are both Noetherian, we can conclude $M$ is Noetherian.
\end{solution}


\begin{problem}[7.7]
$\vartriangleright$ Let
\[
0 \longrightarrow M \longrightarrow N \longrightarrow P \longrightarrow0
\]
be a short exact sequence of $R$-modules, and let $L$ be an $R$-module.
\begin{enumerate}[(i)]
\item Prove that there is an exact sequence
	\[
	0 \longrightarrow \mathrm{Hom}_{R\text{-}\mathrm{Mod}}(P, L) \longrightarrow \mathrm{Hom}_{R\text{-}\mathrm{Mod}}(N, L) \longrightarrow \mathrm{Hom}_{R\text{-}\mathrm{Mod}}(M, L)
	\]
\item Redo \hyperlink{Exercise III.6.17}{Exercise III.6.17}. (Use the exact sequence $0 \rightarrow I \rightarrow R \rightarrow R / I \rightarrow 0 .)$
\item Construct an example showing that the rightmost homomorphism in (i) need not be onto.
\item Show that if if the original sequence splits, then the rightmost homomorphism in (i) is onto.
\end{enumerate}
$[7.9, \mathrm{VIII} .3 .14, \$ \mathrm{VIII.} 5.1]$

\end{problem}
\begin{solution}
	\begin{enumerate}[(i)]
		\item
Suppose that 
\[
0 \longrightarrow M \operatornamewithlimits{\longrightarrow}^{f} N \operatornamewithlimits{\longrightarrow}^{g} P \longrightarrow0.
\]
Define
\begin{align*}
	g^*:\mathrm{Hom}_{R\text{-}\mathrm{Mod}}(P, L) &\longrightarrow \mathrm{Hom}_{R\text{-}\mathrm{Mod}}(N, L) \\
	\varphi&\longmapsto \varphi\circ g
\end{align*}
and 
\begin{align*}
f^*:\mathrm{Hom}_{R\text{-}\mathrm{Mod}}(N, L) &\longrightarrow \mathrm{Hom}_{R\text{-}\mathrm{Mod}}(M, L) \\
\psi&\longmapsto \psi\circ f
\end{align*}
Since $g$ is surjective,
$$
g^*(\varphi_1)=g^*(\varphi_2)\implies \varphi_1\circ g=\varphi_2\circ g\implies\varphi_1=\varphi_2,
$$
which means $g^*$ is injective.
Since $\im f=\ker g$, for all $\psi\in\mathrm{Hom}_{R\text{-}\mathrm{Mod}}(N, L)$, we have
\begin{align*}
&\psi\in\ker f^*\iff \psi\circ f=0\iff\ker g=\im f\subset \ker\psi\\
\iff&\exists\varphi\in\mathrm{Hom}_{R\text{-}\mathrm{Mod}}(P, L),\psi=\varphi \circ g\iff\psi\in\im g^*.
\end{align*}
The deduction 
$$
\ker g\subset \ker\psi\implies\exists\varphi\in\mathrm{Hom}_{R\text{-}\mathrm{Mod}}(P, L),\psi=\varphi \circ g
$$
holds because $N/\ker g\cong \im g=P$ and the following diagram commute

\[\xymatrix@C=40pt@R=40pt{
	N\ar@{->>}[r]^{g}\ar[dr]^{\psi}\ar@{->>}[d]_{\pi}   &  P\ar@{-->}[d]^{\exists \varphi:=\bar{\psi}\circ j}\\
	N/\ker g\ar[r]_{\bar{\psi}}&L
}\]
where $j$ is an isomorphism from $P$ to $N/\ker g$. Thus we show $\ker f^*=\im g^*$ and 
\[
0 \longrightarrow \mathrm{Hom}_{R\text{-}\mathrm{Mod}}(P, L) \operatornamewithlimits{\longrightarrow}^{g^*} \mathrm{Hom}_{R\text{-}\mathrm{Mod}}(N, L) \operatornamewithlimits{\longrightarrow}^{f^*} \mathrm{Hom}_{R\text{-}\mathrm{Mod}}(M, L)
\]
is an exact sequence.
\item
\[
0 \longrightarrow \mathrm{Hom}_{R\text{-}\mathrm{Mod}}(I, N) \longrightarrow \mathrm{Hom}_{R\text{-}\mathrm{Mod}}(R, N) \longrightarrow \mathrm{Hom}_{R\text{-}\mathrm{Mod}}(R/I, N)
\]
\end{enumerate}
\end{solution}