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To show closure under addition is trivial for positive powers since
$$x 2^y + v 2^w = x 2^{(y - w)} 2^w + v 2^w = (x 2^{(y - w)} + v) 2^w$$
Now for the negative case, assume $y > w$, hence $y - w$ is positive and the formulation still holds.
Q4
The sum and product of continuous functions are continuous.
Q5
The sum and product on any interval $[0, 1]$ also remains continuous, and hence also includes $\mathcal{C}$
Q6
Addition and negatives remain in $\mathcal{M}_2(\mathbb{R})$ as does multiplication
$$
\begin{pmatrix}
0 & 0\
0 & x
\end{pmatrix}
\begin{pmatrix}
0 & 0\
0 & y
\end{pmatrix}
\begin{pmatrix}
0 & 0\
0 & xy
\end{pmatrix}
$$
B. Examples of Ideals
Q1
Identify which of the following are ideals of $\mathbb{Z} \times \mathbb{Z}$
${(n, n) : n \in \mathbb{Z}}$
\begin{align*}
(n, n) + (m, m) &= (m + n, m + n) \in I \
-(n, n) &= (-n, -n) \in I \
(n, n) \cdot (a, b) &= (na, nb) \notin I
\end{align*}
Not an ideal.
${(5n, 0) : n \in \mathbb{Z}}$
\begin{align*}
(5m, 0) + (5n, 0) &= (5(m + n), 0) \in I \
-(5n, 0) &= (5(-n), 0) \in I \
(5n, 0) \cdot (a, b) = (5(na), 0) \in I
\end{align*}
Is an ideal.
${(n, m) : n + m \text{ is even }}$
\begin{align*}
(n_1, m_1) + (n_2, m_2) &= (n_1 + n_2, m_1 + m_2) \in I \
-(n, m) \in I \
(n, m) \cdot (a, b) &= (na, mb)
\end{align*}
$na$ is even and $mb$ is even, so $na + mb$ is even so $(na, mb) \in I$.
Is an ideal.
${(2n, 3m) : n, m \in \mathbb{Z}}$
\begin{align*}
(2n_1, 3m_1) + (2n_2, 3m_2) &= (2(n_1 + n_2), 3(m_1 + m_2)) \in I \
-(2n, 3m) &= (2(-n), 3(-m)) \in I \
(2n, 3m) \cdot (a, b) &= (2na, 3mb) \in I
\end{align*}
Is an ideal
Q2
List all the ideals of $\mathbb{Z}_{12}$
$\mathbb{Z}_{12} = \langle 1 \rangle$ and is cyclic. All subgroups are also cyclic.
Let $m \in \bar{m} = $, then $ = {mj : j \in \mathbb{Z}_{12}}$
Let $x \in $ and $y \in \mathbb{Z}{12}$, since $x \in $, then $x = mj$ for some $j \in \mathbb{Z}{12}$, thus $xy = mjy$, thus $$ is an ideal of $\mathbb{Z}_{12}$.
Ideals are $<0>, <1>, <2>, <3>, <4>, <6>$
The product of a continuous and non-continuous function are non-continuous, hence $\mathcal{C}(\mathbb{R})$ is not an ideal of $\mathcal{F}(\mathbb{R})$
Q6
a
Assume he means multiplication here.
$$f(x) \cdot g(x) = 0 \quad \forall x \in \mathbb{Q}$$
Thus $f \cdot g \in I$
b
Likewise $f(0) g(0) = 0 g(0) = 0$, so $f \cdot g \in I$
Q7
Ideals of $P_3$ such that $AB = A \cap B \in I$. See also 17D5
Let $x \in B$ and since $B$ is a ring then $0 \in B$, thus $0 - x = -x \in B$.
So $B$ is closed wrt negatives and hence addition since $x - (-y) = x + y \in B$
Q2
As per part 1
Q3
A ring is a group under addition. Hence order of a subring divides ring by Lagrange.
Q4
$A$ has no zero divisors, hence neither does $B$$\implies$$B$ is an integral domain.
Q5
$B$ is a subring of field $F$. Let $b \in B, b \neq 0$, then $b^{-1} \in F$ (because $F$ is a field and contains inverses). Every field is an integral domain, hence so is $B$.
Q6
$F$ is a commutative ring with inverses and unity.
Since $B$ is a subring, it also is commutative.
Since $B$ also contains inverses and is closed wrt multiplication, it must contain $1_F$.
$$h: P_C \rightarrow P_C$$$$h(A) = A \cap D$$$$D \subseteq C$$
\begin{align*}
h(A + B) &= h((A - B) \cup (B - A)) \
&= [(A - B) \cup (B - A)] \cap D \
&= [(A - B) \cap D] \cup [(B - A) \cap D] \
&= [A\cap D - B \cap D] \cup [B \cap D - A \cap D] \
&= h(A) + h(B) \
h(AB) &= h(A \cap B) = A \cap B \cap D \
&= (A \cap D) \cap (B \cap D) \
&= h(A) h(B)
\end{align*}
$$K = { A \in P_C : A \cap D = \varnothing }$$
Range is every subset of $D$.
Prove $f(A) = { f(x) : x \in A }$ is a subring of $B$.
Since $f$ is a homomorphism, ring operations are obeyed in the homomorphism. For negatives we note that $f(0_A) = 0_B = 1_B + (-1_B)$ and every negative is expressible as $(-1_B) \cdot a$ where $a \in B$.
Q2
Prove the kernel of $f$ is an ideal of $A$.
$$K = { x \in A: f(x) = 0_B }$$
From $f$ being a homomorphism, we conclude $K$ is a subring of $A$.
To show it's an ideal, for any $a \in A$ and $x \in K$, then $f(ax) = 0 = f(x)$. So $K$ absorbs the product $ax$.
Thus the kernel of a homomorphism is an ideal of the input ring.
Q3
Prove $f(0) = 0$, and for every $a \in A, f(-a) = -f(a)$.
$\implies x = 0 \qquad \text{[since $f$ is injective]}$
It follow $K = { 0 }$
Thus $f$ is injective $\implies K = { 0 }$
Now suppose $K = { 0 }$. Then
$f(x) = f(y)$
$\implies f(x) - f(y) = 0$
$\implies f(x - y) = 0$
$\implies x - y \in K$
$\implies x - y = 0 \qquad \text{[since $K = { 0 }$]}$
$\implies x = y$
Hence $f$ is injective.
Thus $K = { 0 } \implies f$ is injective.
Hence $f$ is injective $\iff K = { 0 }$
Q5
If $B$ is an integral domain, then either $f(1) = 1$ or $f(1) = 0$. If $f(1) = 0$ then $f(x) = 0$ for every $x \in A$. If $f(1) = 1$, the image of every invertible element of $A$ is an invertible element of $B$.
Integral domain has the cancellation property such that $ab = ac \implies b = c$.
If $f(1) = 0$ then $\forall a \in A$, $f(a) = f(1 \cdot a) = f(1) f(a) = 0$
If $f(1) = 1$ and $\exists x, y \in A$ such that $xy = 1$
$$f(xy) = f(x)f(y) \text{ where } f(y) = (f(x))^{-1}$$
Q6
Any homomorphic image of a commutative ring is a commutative ring. Any homomorphic image of a field is a field.
Let $a, b \in A$, then $f(a)f(b) = f(b)f(a)$ because $f(ab) = f(ba)$.
If $A$ is a field, then $\forall x \in A$, $\exists x^{-1} \in A$. So by the last exercise, $f(x^{-1}) = (f(x))^{-1}$ and so the inverse of $f(x)$ is a member of $B$.
$$ (f(x))^{-1} \in B$$
Q7
If the domain $A$ of the homomorphism $f$ is a field, and if the range of $f$ has more than one element, then $f$ is injective.
Since $A$ is a field, the kernel of $A$ is either ${0}$ or $A$ itself.
But the range of $f$ is more than one element, so the kernel of $A$ cannot be $A$ and must be ${0}$.
Since the kernel of $f$ is ${0}$, then $f$ is injective.
G. Examples of Isomorphisms
Q1
$$a \oplus b = a + b + 1$$
$$a \otimes b = ab + a + b$$
Addition
\begin{align*}
f(a + b) &= a + b - 1 \
f(a) \oplus f(b) &= (a - 1) + (b - 1) - 1 \
&= a + b - 1
\end{align*}
Multiplication
\begin{align*}
f(ab) &= ab - 1 \
f(a) \otimes f(b) &= (a - 1)(b - 1) + (a - 1) + (b - 1) \
&= ab - b - a + 1 + a + b - 1 - 1 \
&= ab - 1
\end{align*}
Q2
$$\mathcal{J} = {
\begin{pmatrix}
a & b\\
-b & a
\end{pmatrix}
: a, b \in \mathbb{R} }$$
$$f: \mathbb{C} \rightarrow \mathcal{J}$$
$$f(a + bi) =
\begin{pmatrix}
a & b\\
-b & a
\end{pmatrix}$$
$$a + bi = c + di \implies f(a + bi) = f(c + di)$$
Addition
\begin{align*}
f((a + bi) + (c + di)) &=
\begin{pmatrix}
a + c & b + d\
-(b + d) & a + c
\end{pmatrix} \
f(a + bi) + f(c + di) &=
\begin{pmatrix}
a + c & b + d\
-(b + d) & a + c
\end{pmatrix} \
\end{align*}
Multiplication
\begin{align*}
f((a + bi)(c + di)) &= f((ac - bd) + (ad + bc)i) \
&=
\begin{pmatrix}
a + c & b + d\
-(b + d) & a + c
\end{pmatrix} \
f(a + bi)f(c + di) &=
\begin{pmatrix}
a & b\
-b & a
\end{pmatrix}
\begin{pmatrix}
c & d\
-d & c
\end{pmatrix} \
&=
\begin{pmatrix}
a + c & b + d\
-(b + d) & a + c
\end{pmatrix} \
\end{align*}
Q3
$$A = {(x, x) : x \in \mathbb{Z}}$$$\forall x, y \in \mathbb{Z}, (x, x) \in A, (y, y) \in A, (x + y, x + y) \in A \text{ and } (xy, xy) \in A$
Thus $A$ is a subring of $\mathbb{Z} \times \mathbb{Z}$
The homomorphism $f: \mathbb{Z} \rightarrow A$ by $f(x) = (x, x)$ is isomorphic because it is one to one
$$f(x) = f(y) \implies x = y$$
and onto
$$\forall (x, x) \in A, \exists x \in \mathbb{Z} \text{ such that } f(x) = (x, x)$$
Thus
$${(x, x): x \in \mathbb{Z}} \cong \mathbb{Z}$$
Q4
Addition and negatives trivially remain inside the set.
$$
\begin{pmatrix}
0 & 0\
0 & x
\end{pmatrix}
\begin{pmatrix}
0 & 0\
0 & y
\end{pmatrix}
\begin{pmatrix}
0 & 0\
0 & xy
\end{pmatrix}
$$
Hence the set is a subring.
$$A = {
\begin{pmatrix}
0 & 0\
0 & x
\end{pmatrix}
: x \in \mathbb{R}}$$
Define $f: \mathbb{R} \rightarrow A$ by $f(x) =
\begin{pmatrix}
0 & 0\
0 & x
\end{pmatrix}$, then $f$ is an homomorphism from $\mathbb{R}$ to $A$.
Hence $A \cong \mathbb{R}$
Q5
$$f : k \mathbb{Z} \rightarrow l \mathbb{Z}$$
$$f(k) = ln \text{ for some } n \neq 0$$
$x \in A$ and $a \in \rad J$ then $(xa)^n = x^n a^n \in J$, so $xa \in \rad J$.
$a, b \in \rad J$, then $(a + b)^{m + n} \in J$ and so $a + b \in \rad J$.
Q4
For any $a \in A$, ${ x \in A : ax = 0 }$ is an ideal (called the annihilator of $a$).
Furthermore, ${ x \in A : ax = 0 \text{ for every } a \in A }$ is an ideal (called the annihilating ideal of $A$). If $A$ is a ring with unity, its annihilating ideal is equal to ${ 0 }$.
Let $b \in A$, then $bx \in Ann(a)$ because $ax = 0$ so $b(ax) = bxa = 0$.
Let $x, y \in Ann(a)$ then $a(x + y) = 0$ so $x + y \in Ann(a)$.
$$I = { x \in A : ax = 0 \text{ for every } a \in A }$$
If $A$ is a ring with unity then $a = 1 \implies x = 0$ so $I = { 0 }$.
Q5
Show that ${0}$ and $A$ are ideals of $A$. (They are trivial ideals; every other ideal of $A$ is a proper ideal.) A proper $J$ of $A$ is called maximal if it is not strictly contained in any strictly larger proper ideal: that is if $J \subseteq K$, where $K$ is an ideal containing some element not in $J$, then necessarily $K = A$.
Show the following is an example of a maximal ideal: in $\mathcal{F}(\mathbb{R})$, the ideal $J = { f : f(0) = 0 }$.
Continuous function with a nonzero value is invertible.
$h(x) - g(x) = -g(0) \in K$ but $g(0) \neq 0$ so $-1/g(0) \in A$.
But since $K$ is an ideal, that is
$$g(0) \cdot 1 / g(0) \in K$$
but this equals $1$, and $1 \in K$ so $K = A$ and is maximal.
I. Further Properties of Homomorphisms
Q1
If $f: A \rightarrow B$ is a homomorphism from $A$ onto $B$ with kernel $K$, and $J$ is an ideal of $A$ such that $K \subseteq J$, then $f(J)$ is an ideal of B$.
$f$ is onto $\exists x: f(x) = y$ so it's an ideal. Closed under addition and negatives and absorbs products.
If $f : A \rightarrow B$ is a homomorphism from $A$ onto $B$, and $B$ is a field, then the kernel of $f$ is a maximal ideal.
The kernel $K$ is a subset of the ideal for $A$. As shown above $f(J)$ is an ideal of $B$, which by D6 can only be ${0}$ or $B$ itself. Since the homomorphism is onto, then $f(A)$ maps to $B$, but $A$ is a trivial ideal of $A$. Thus $K$, the kernel of $f$ is the proper ideal for $A$ which maps to ${0}$ in $B$.
Q3
There are no nontrivial homomorphisms from $\mathbb{Z}$ to $\mathbb{Z}$.
$a$ is not a divisor of zero $\implies \forall x \in A, ax \neq 0$, thus ring $A$ has cancellation property
$$\pi_a(x) = \pi_a(y) = ax = ay \implies x = y$$
Q3
If $a$ is invertible then $\forall y \in A$, $y = a(a^{-1}y)$ so $x = a^{-1}y$, $f(x) = y$, thus $\pi_a$ is surjective.
$\forall \pi_a \in \mathcal{A}, \exists a \in A : \pi_a = \phi(a)$ by definition.
If $a$ has no divisors of zero, then to show injective property, note that
$$ax = bx \implies a = b$$$$\pi_a = pi_b \implies \pi_a(x) = ax = \pi_b(x) = bx \implies a = b$$
From the cancellation property since it has no divisors of zero.