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Abstract Algebra by Pinter, Chapter 20
Amir Taaki
- \usepackage{mathrsfs}
- \usepackage{mathtools}
- \usepackage{extpfeil}
- \DeclareMathOperator\ker{ker}
- \DeclareMathOperator\ord{ord}
- \DeclareMathOperator\gcd{gcd}
- \DeclareMathOperator\lcm{lcm}
- \DeclareMathOperator\mod{mod}
- \DeclareMathOperator\char{char}
Chapter 20 on Integral Domains
A. Characteristic of an Integral Domain Let $a$ be any nonzero element of $A$. If $n \cdot a = 0$, where $n \neq 0$, then $n$ is a multiple of the characteristic of $A$.
$$n \cdot a = 0 \implies n \cdot 1 = 0$$
but $\char(A) \cdot 1 = 0$ , so $n$ is a multiple of the characteristic of $A$ .
If $A$ has characteristic zero, $n \neq 0$, and $n \cdot a = 0$, then $a = 0$.
$$n \neq 0$$
$$n \cdot a = 0$$
$$n \cdot 1 \cdot a = 0$$
but $n \cdot 1 \neq 0$ because characteristic is zero.
$$a = 0$$
by cancellation property.
If $A$ has characteristic $3$, and $5 \cdot a = 0$, then $a = 0$.
$$3 \cdot a = 0$$
$$5 \cdot a = 3 \cdot a + 2 \cdot a = 2 \cdot a$$
$$2 \cdot a = 0 \implies a = 0$$
If there is a nonzero element $a$ in $A$ such that $256 \cdot a = 0$, then $A$ has characteristic $2$.
$$256 = 2^8$$
characteristic is prime.
$$\char(A) = 2$$
If there are distinct nonzero elements $a$ and $b$ in $A$ such that $125 \cdot a = 125 \cdot b$, then $A$ has characteristic
$$125 \cdot a = 125 \cdot b$$
$$5 \times 5 \cdot 1 \cdot (a - b) = 0$$
$$a \neq b \implies a - b \neq 0$$
$$5 \cdot 1 = 0$$
$$\char(A) = 5$$
If there are nonzero elements $a$ and in $A$ such that $(a + b)^2 = a^2 + b^2$, then $A$ has characteristic $2$.
Theorem 3: $(a + b)^p = a^p + b^p$
$$(a + b)^2 = a^2 + b^2$$
$$\char(A) = 2$$
If there are nonzero elements $a$ and $b$ in $A$ such that $10a = 0$ and $14b = 0$, then $A$ has characteristic $2$.
$$2 \times 5 \cdot 1 \cdot a = 0$$
$$2 \times 7 \cdot 1 \cdot a = 0$$
$$2 \cdot 1 \cdot (5a + 7a) = 0$$
$$\char(A) = 2$$
B. Characteristic of a Finite Integral Domain If $A$ has characteristic $q$, then $q$ is a divisor of the order of $A$.
By Lagrange's theorem, any subgroup divides the group order.
$1 + \cdots + 1 = 0$ and so forms a subgroup which divides the group order. The order of this subgroup is the characteristic of $A$ .
See this question .
If the order of $A$ is a prime number $p$, then the characteristic of $A$ must be equal to $p$.
From above the characteristic must divide the order, but the order is prime so the characteristic must equal $p$ .
If the order of $A$ is $p^m$ , where $p$ is a prime, the characteristic of $A$ must be equal to $p$.
The characteristic is prime and divides the order, hence $\char(A) = p$ .
$81 = 3 \times 3 \times 3 \times 3$ , so by above $\char(A) = 3$ .
If $A$, with addition alone, is a cyclic group, the order of $A$ is a prime number.
$$A = \langle 1, + \rangle$$
$$\ord(1) = |A|$$
$$\ord(1) \cdot 1 = 0 = |A| \cdot 1$$
but
$$\char(1) \cdot 1 = 0$$
Hence $\char(1)| |A|$
But $\ord(1)$ is the smallest $n$ such that $\ord(1)\cdot 1 = 0$ so $\ord(1) | \char(1)$ by Lagrange, and $\ord(1) = |A|$
Thus $|A|$ is prime since $|A| = \char(1)$ which is also prime.
See this question
Prove every nonzero element of $A$ is either a divisor of zero or invertible.
$$A = { 0, 1, a_1, a_2, \dots, a_n }$$
$$|A| = n + 2$$
$$a_i 0, a_i 1, a_i a_2, \dots, a_i a_n$$
The size of this subgroup divides $|A|$ .
If its size is less than $|A|$ , then there exists $a_i x = 0$ where $x \neq 0$ . So $a_i$ is a divisor of zero.
Otherwise if its size equals $|A|$ , then only $a_i 0 = 0$ and so $a_i x = 1$ meaning $a_i$ is invertible.
Prove: If $a \neq 0$ is not a divisor of zero, then some positive power of $a$ is equal to $1$.
$a \neq 0$ is not a divisor of zero $\implies \ord(a) = |A| \implies A = \langle a \rangle$
But $1 \in A$ , so for some $x$ , $a^x = 1$
If $a$ is invertible, then $a^{-1}$ is equal to a positive power of $a$.
If $a$ is invertible, $ax = 0 \implies (a^{-1} \cdot a) x = 0 \implies x = 0$ .
Therefore $a$ is not a divisor of zero.
$$\ord(a) = |A|$$
$$A = \langle a \rangle$$
So $a^{-1} = a^k$
D. Field of Quotients of an Integral Domain \begin{align*}
[a, b] = [r, s] &\implies as = br \
[c, d] = [t, u] &\implies cu = dt
\end{align*}
\begin{align*}
[a, b] + [c, d] &= [ad + bc, bd] \
[r, s] + [t, u] &= [ru + st, su]
\end{align*}
$$[ad + bc, bd] = [ru + st, su] \implies (ad + bc)su = bd(ru + st)$$
$$adsu + bcsu = bdru + bdst$$
$$as \cdot du + cu \cdot bs = br \cdot du + dt \cdot bs$$
Since $as = br$ and $cu = dt$
$$br \cdot du + dt \cdot bs = br \cdot du + dt \cdot bs$$
So
$$[a, b] + [c,d] = [r, s] + [t, u]$$
\begin{align*}
[a, b] \cdot [c,d] &= [ac, bd] \
[r, s] \cdot [t, u] &= [ru, su]
\end{align*}
$$[ac, bd] = [rt, su] \implies acsu = bdrt$$
$$as \cdot cu = br \cdot dt$$
But $as = br$ and $cu = dt$
Thus
$$[a,b] \cdot [c, d] = [r, s] \cdot [t, u]$$
$$(u, v) \sim (a, b) \text{ and } (u, v) \sim (c, d) \implies (a, b) \sim (c, d)$$
\begin{align*}
(u, v) \sim (a, b) &\implies av = bu \
(u, v) \sim (c, d) &\implies cv = du
\end{align*}
$$v = c^{-1}du$$
$$av = ac^{-1}du = bu$$
$$ad = bc$$
$$(a, b) \sim (c, d)$$
\begin{align*}
[a, b] + ([c, d] + [e, f]) &= [a, b] + [cf + de, df] \
&= [adf + bcf + bde, bdf] \
([a, b] + [c, d]) + [e, f] &= [ad + bc, bd] + [e, f] \
&= [adf + bcf + bde, bdf]
\end{align*}
$$[a, b] + [c, d] = [c, d] + [a, b]$$
\begin{align*}
[a, b] \cdot ([c, d] \cdot [e, f]) &= [a, b] \cdot [ce, df] \
&= [ace, bdf] \
([a, b] \cdot [c, d]) \cdot [e, f] &= [ac, bd] \cdot [e, f] \
&= [ace, bdf]
\end{align*}
$$[a, b] \cdot [c, d] = [c, d] \cdot [a, b]$$
\begin{align*}
[a, b] \cdot ([c, d] + [e, f]) &= [a, b] \cdot [cf + de, df] \
&= [acf + bde, bdf] \
[a, b] \cdot [c, d] + [a, b] \cdot [e, f] &= [ac, bd] + [ae, bf] \
&= [acbf + bdae, bdbf]
\end{align*}
Both are equivalent so distributive.
$$\phi(a) = [a, 1]$$
\begin{align*}
\phi(ab) &= [ab, 1] \
&= \phi(a) \phi(b) = [a, 1] \cdot [b, 1] = [ab, 1]
\end{align*}
\begin{align*}
\phi(a + b) &= [a + b, 1] \
&= \phi(a) + \phi(b) = [a, 1] + [b, 1] \
&= [a + b, 1]
\end{align*}
E. Further Properties of the Characteristic of an Integral Domain $$p \cdot a = 0$$
$$n = p\cdot m + r$$
\begin{align*}
n \cdot a &= pm \cdot a + r \cdot a \\
&= m(p \cdot a) + r \cdot a \\
&= r \cdot a
\end{align*}
but $n \cdot a = 0$ and $r \neq 0$ because $p$ does not divide $n$
$$r \cdot a = 0 \implies a = 0$$
Characteristic is prime and since $a \neq 0$ then the characteristic must be $p$ .
$p^m$ is a multiple of $p$ . So the characteristic is $p$ .
$$f(ab) = a^p b^p = f(a)f(b)$$
$$f(a + b) = (a + b)^p = a^p + b^p = f(a) + f(b)$$
Order is prime and group is cyclic because
$$A \cong \mathbb{Z}_p$$
For any $a \in A : p \neq 0$
$$A = \langle a \rangle$$
\begin{align*}
(a + b)^{p^2} &= [(a + b)^p]^p = [a^p + b^p]^p \
&= a^{p^2} + b^{p^2}
\end{align*}
Assume true for $n = k$
\begin{align*}
(a + b)^{p^k} &= a^{p^k} + b^{p^k} \
(a + b)^{p^{k + 1}} &= [(a + b)^{p^k}]^p \
&= [a^{p^k} + b^{p^k}]^p \
&= a^{p^{k + 1}} + b^{p^{k + 1}}
\end{align*}
\begin{align*}
(a_1 + a_2 + \cdots + a_r)^{p^n} &= [(a_1 + a_2 + \cdots) + a_r]^{p^n} \
&= (a_1 + a_2 + \cdots)^{p^n} + a_r^{p^n} \
&= (a_1 + a_2 + \cdots)^{p^n} + a_{r-1}^{p^n} + a_r^{p^n} \
&= a_1^{p^n} + a_2^{p^n} + \cdots + a_r^{p^n}
\end{align*}
$1 \in A$ and $1 \in B$
$n \cdot 1 = 0$ is true both in $A$ and $B$
$$\implies \char(A) = \char(B)$$
A finite field has order prime $p$ and so is isomorphic to the cyclic group (see E5).
That is
$$A = \langle 1, + \rangle$$
Since $A$ is finite order, $\char(A) \neq 0$
$$f(a) = a^p$$
To show injective(onto):
$$f(x) = f(y) \implies x = y$$
From 18F7, the domain of $f$ is a field and so $f$ is injective.
This can also be shown by
$$f(a) = a^p = f(b) = b^p$$
$$\implies a^p - b^p = 0$$
$$\implies (a - b)^p = 0$$
$$\implies a = b$$
$f$ is injective. $A$ has $p$ elements, so the image of $f$ has at least $p$ elements. But the image of $f$ is contained in $A$ , so it has at most $p$ elements.
Therefore $f$ is surjective.