On Tate’s Proof of a Theorem of Dedekind

DOI: 10.4236/ojdm.2018.83007   PDF   HTML     544 Downloads   1,108 Views   Citations
In this note we give a complete proof of a theorem of Dedekind.

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Gupta, S. (2018) On Tate’s Proof of a Theorem of Dedekind. Open Journal of Discrete Mathematics, 8, 73-78. doi: 10.4236/ojdm.2018.83007.

1. Introduction

In this note we give a complete proof of the following theorem of Dedekind. Our proof is a somewhat detailed version of the one given in Basic Algebra by Jacobson, Volume I,  and we shall keep the notations used in that proof.

Theorem 1 Let $f\left(x\right)\in ℤ\left[x\right]$ be square-free monic polynomial of degree n and p be a prime such that p does not divide the discriminant of $f\left(x\right)$ . Let $G\subset {S}_{n}$ be the Galois group of $f\left(x\right)$ over the field $ℚ$ of rational numbers. Suppose that ${f}_{p}=\stackrel{¯}{f}=f\left(\mathrm{mod}p\right)\in {ℤ}_{p}\left[x\right]$ factors as:

${f}_{p}=\stackrel{¯}{f}=\underset{i=1}{\overset{r}{\prod }}\text{ }\stackrel{¯}{{f}_{i}}$

where $\stackrel{¯}{{f}_{i}}$ are distinct monic irreducible polynomials in ${ℤ}_{p}\left[x\right]$ , degree $\left(\stackrel{¯}{{f}_{i}}\right)={d}_{i}$ , $1\le i\le r$ , and ${d}_{1}+{d}_{2}+\cdots +{d}_{r}=n$ .

Then there exists an automorphism $\sigma \in G$ which when considered as a permutation on the zeros of $f\left(x\right)$ is a product of disjoint cycles of lengths ${d}_{1},{d}_{2},\cdots ,{d}_{r}$ .

2. Preliminary Results

We shall assume that the reader is familiar with the following well-known results.

1) Let $\mathbb{F}$ be a field and $f\left(x\right)\in \mathbb{F}\left[x\right]$ be a polynomial of degree $n\ge 2$ . Then any two splitting fields of $f\left(x\right)$ are isomorphic.

2) A finitely generated Abelian group is direct sum of (finitely many) cyclic groups. (This is the fundamental theorem of finitely generated Abelian groups).

3) A system of n homogeneous equation in $m>n$ variables has a non-trivial solutions.

4) Let $\mathbb{E}/\mathbb{F}$ be an algebraic extension. Then any subring of $\mathbb{E}$ containing $\mathbb{F}$ is a subfield of $\mathbb{E}$ . Proof: Let K be a ring such that $\mathbb{F}\subset K\subset \mathbb{E}$ . Let $\alpha \in K-\mathbb{F}$ . As $\alpha$ is algebraic over $\mathbb{F}$ , $\mathbb{F}\left(\alpha \right)=\mathbb{F}\left[\alpha \right]$ . So ${\alpha }^{-1}\in \mathbb{F}\left(\alpha \right)\subset K$ .

5) (Dedekind’s Independence Theorem). Distinct characters of a monoid (a set with associative binary operation with an identity element) into a field are linearly independent. That is if ${\chi }_{1},{\chi }_{2},\cdots ,{\chi }_{n}$ are distinct characters of a monoid into a field $\mathbb{F}$ , then the only elements ${a}_{i}\in \mathbb{F}$ , $1\le i\le n$ , such that

${a}_{1}{\chi }_{1}\left(h\right)+{a}_{2}{\chi }_{2}\left(h\right)+\cdots +{a}_{n}{\chi }_{n}\left(h\right)=0$

for all $h\in H$ are ${a}_{i}=0$ , $1\le i\le n$ .

6) Let p be a prime and $GF\left({p}^{m}\right)$ be a finite field with ${p}^{m}$ elements. Then the group $\text{Aut}\left(GF\left({p}^{m}\right)\right)=〈\sigma 〉$ is cyclic of order m and the generating automorphism $\sigma$ maps $\alpha \in GF\left({p}^{m}\right)$ to ${\alpha }^{p}$ .

7) If R is a commutative ring with identity and M is a maximal ideal of R then R/M is a field.

8) Let $\sigma ,\eta \in {S}_{n}$ . Then $\sigma$ and ${\eta }^{-1}\sigma \eta$ have same cyclic structure.

Let $f\left(x\right)\in ℤ\left[x\right]$ be a polynomial of degree $n\ge 1$ , and p a prime number. Then ${f}_{p}\left(x\right)\in {ℤ}_{p}\left[x\right]$ will denote the polynomial obtained by reducing the coefficients of $f\left(x\right)$ modulo p.

Theorem 2 Let $f\left(x\right)\in ℤ\left[x\right]$ be a monic polynomial of degree $n\ge 1$ and p be a prime number which does not divide the discriminant of $f\left(x\right)$ . Let $\mathbb{E}$ be a splitting field of $f\left(x\right)$ over $ℚ$ . Let ${\mathbb{E}}_{p}$ be a splitting field of ${f}_{p}\left(x\right)$ over ${ℤ}_{p}=ℤ/\left(p\right)$ . Let

$f\left(x\right)=\left(x-{r}_{1}\right)\left(x-{r}_{2}\right)\cdots \left(x-{r}_{n}\right),\text{ }{r}_{i}\in \mathbb{E}\subset ℂ,1\le i\le n$

$\begin{array}{l}R=\left\{{r}_{1},{r}_{2},\cdots ,{r}_{n}\right\},\\ {R}_{p}=\left\{\stackrel{¯}{{r}_{1}},\stackrel{¯}{{r}_{2}},\cdots ,\stackrel{¯}{{r}_{n}}\right\}\subset {\mathbb{E}}_{p}\end{array}$

where $\stackrel{¯}{{r}_{i}}$ , $1\le i\le n$ are the roots of ${f}_{p}\left(x\right)\in {ℤ}_{p}\left[x\right]$ and

$\mathbb{E}=ℚ\left({r}_{1},{r}_{2},\cdots ,{r}_{n}\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\mathbb{E}}_{p}={ℤ}_{p}\left(\stackrel{¯}{{r}_{1}},\stackrel{¯}{{r}_{2}},\cdots ,\stackrel{¯}{{r}_{n}}\right)$

Let $D=ℤ\left[{r}_{1},{r}_{2},\cdots ,{r}_{n}\right]$ be the subring generated by the roots of ${r}_{1},{r}_{2},\cdots ,{r}_{n}$ of $f\left(x\right)$ in $ℂ$ . Then

• 1) There exists a homomorphism $\psi$ of D onto ${\mathbb{E}}_{p}$ .

• 2) Any such homomorphism $\psi$ gives a bijection of the set R of the roots of $f\left(x\right)$ in $\mathbb{E}$ onto the set ${R}_{p}$ of the roots of the ${f}_{p}\left(x\right)$ in ${\mathbb{E}}_{p}$ .

• 3) If $\psi$ and ${\psi }^{\prime }$ are two such homomorphisms then there exist $\sigma \in Aut\left(\mathbb{E}/ℚ\right)=Gal\left(f\left(x\right)\right)$ , such that ${\psi }^{\prime }=\psi \cdot \sigma$ . (Note that the restriction of $\sigma$ to D is an automorphism of D).

Proof 1) One has that:

$\mathbb{E}=ℚ\left({r}_{1},{r}_{2},\cdots ,{r}_{n}\right)=ℚ\left[{r}_{1},{r}_{2},\cdots ,{r}_{n}\right]$

We claim that $D=ℤ\left[{r}_{1},{r}_{2},\cdots ,{r}_{n}\right]$ is a finitely generated (additive) Abelian group. Since each ${r}_{i}$ is a root of the monic polynomial $f\left(x\right)\in ℤ\left[x\right]$ of degree n any positive power of ${r}_{i},1\le i\le n$ can be expressed as an integral linear combination of $1,{r}_{i},{r}_{i}^{2},\cdots {r}_{i}^{n-1}$ . It follows that

$D=\underset{0\le {e}_{i}\le n-1}{\sum }ℤ{r}_{1}^{{e}_{1}}{r}_{2}^{{e}_{2}}\cdots {r}_{n}^{{e}_{n}}.$

Therefore D is a finitely generated (additive) Abelian group generated by at most nn elements. By the Fundamental Theorem for Finitely Generated Abelian Groups D is a direct sum of finitely many cyclic groups. Since $D\subset ℂ$ , none of these cyclic groups is finite. So D is a direct sum of finitely many infinite cyclic groups. Let $\left\{{u}_{1},{u}_{2},\cdots ,{u}_{N}\right\}$ be a set consisting of an independent generating system of D. We have

$D=ℤ{u}_{1}\oplus ℤ{u}_{2}\oplus \cdots \oplus ℤ{u}_{N},\text{ }N\le {n}^{n}.$

We claim that $\left\{{u}_{1},{u}_{2},\cdots ,{u}_{N}\right\}$ is a basis of $\mathbb{E}/ℚ$ . Obviously $\left\{{u}_{1},{u}_{2},\cdots ,{u}_{N}\right\}$ is linearly independent over $ℚ$ . Let $ℚD={\sum }_{1\le i\le N}ℚ{u}_{i}$ . Then $ℚD$ is a ring and $ℚ\subset ℚD\subset \mathbb{E}$ therefore $ℚD$ is a field. Since ${r}_{i}\in D$ for $1\le i\le n$ , by (4) $ℚD=\mathbb{E}$ and $\left\{{u}_{1},{u}_{2},\cdots ,{u}_{N}\right\}$ is a basis of $E/ℚ$ . As $D=ℤ{u}_{1}\oplus ℤ{u}_{2}\oplus \cdots \oplus ℤ{u}_{N}$ ,

$pD=ℤ\left(p{u}_{1}\right)\oplus ℤ\left(p{u}_{2}\right)\oplus \cdots \oplus ℤ\left(p{u}_{N}\right)$

is an ideal of D and

$D/pD=\left\{\stackrel{¯}{{a}_{1}{u}_{1}+{a}_{2}{u}_{2}+\cdots +{a}_{N}{u}_{N}}:0\le {a}_{i}\le p-1\right\}.$

Therefore the $D/pD$ is finite of order ${p}^{N}$ . Let M be a maximal ideal of D containing pD. That is $pD\subset M\subset D$ and $D/M$ is a finite field of characteristic p and so it has a subfield isomorphic to ${ℤ}_{p}=ℤ/pℤ$ which we will identify as ${ℤ}_{p}$ in what follows. As

$D/M\approx \frac{D/pD}{M/pD}$

the order of $D/M$ is ${p}^{m}$ , $1\le m\le N$ . Consider the canonical epimorphism

$\nu :D\to D/M$

whose kernel is M and $pℤ\subset M$ . Therefore $\nu \left(ℤ\right)={ℤ}_{p}$ . We note that as $D=ℤ\left[{r}_{1},{r}_{2},\cdots ,{r}_{n}\right]$ we have for $1\le i\le n$

$\nu \left({r}_{i}\right)={r}_{i}+M=\stackrel{¯}{{r}_{i}},\text{ }\nu \left(D\right)={ℤ}_{p}\left[\stackrel{¯}{{r}_{1}},\stackrel{¯}{{r}_{2}},\cdots ,\stackrel{¯}{{r}_{n}}\right]$

As $\nu$ is an epimorphism we have

$\nu \left(D\right)=D/M={ℤ}_{p}\left[\stackrel{¯}{{r}_{1}},\stackrel{¯}{{r}_{2}},\cdots ,\stackrel{¯}{{r}_{n}}\right]$

is a splitting field of ${f}_{p}\left(x\right)$ over ${ℤ}_{p}$ . As both $D/M$ and ${\mathbb{E}}_{p}$ are splitting fields of ${f}_{p}\left(x\right)$ over ${ℤ}_{p}$ they are isomorphic. Let

$\varphi :D/M\to {ℤ}_{p}$

be such an isomorphism. Then $\psi =\varphi \cdot \nu$ is a homomorphism of D onto ${\mathbb{E}}_{p}$ .

2) Let $\psi :D\to {\mathbb{E}}_{p}$ be a homomorphism. So $\psi \left(1\right)=1$ . As $ℤ\subset D$ , and ${\mathbb{E}}_{p}$ has characteristic p, $\psi \left(p\right)=0$ , so $\psi \left(ℤ\right)={ℤ}_{p}\subset {\mathbb{E}}_{p}$ . $\psi$ can be extended to a homomorphism of the polynomial rings $D\left[x\right]\to {\mathbb{E}}_{p}\left[x\right]$ . Under this mapping $f\left(x\right)\to {f}_{p}\left(x\right)$ . As

$f\left(x\right)=\left(x-{r}_{1}\right)\left(x-{r}_{2}\right)\cdots \left(x-{r}_{n}\right)$

$\psi \left(f\left(x\right)\right)={f}_{p}\left(x\right)=\left(x-\psi \left({r}_{1}\right)\right)\left(x-\psi \left({r}_{2}\right)\right)\cdots \left(x-\psi \left({r}_{n}\right)\right),$

$\psi \left({r}_{i}\right),1\le i\le n$ are the roots of the ${f}_{p}\left(x\right)$ in ${\mathbb{E}}_{p}$ and therefore the restriction of $\psi$ to R

${\psi }_{|R}:\left\{{r}_{1},{r}_{2},\cdots ,{r}_{n}\right\}\to \left\{\stackrel{¯}{{r}_{1}},\stackrel{¯}{{r}_{2}},\cdots ,\stackrel{¯}{{r}_{n}}\right\}$

is a bijection of the set R of roots of $f\left(x\right)$ in $\mathbb{E}$ to the set ${R}_{p}$ of the roots of ${f}_{p}\left(x\right)$ in ${\mathbb{E}}_{p}$ .

3) We have seen that given a homomorphism $\psi :D\to {\mathbb{E}}_{p}$ , and $\sigma \in Gal\left(f\right)=Aut\left(\mathbb{E}/ℚ\right)$ , ${\psi }^{\prime }=\psi \cdot \sigma$ is also a homomorphism from D to ${\mathbb{E}}_{p}$ . We note that the restriction of $\sigma \in Aut\left(\mathbb{E}/ℚ\right)$ to $D=ℤ\left[{r}_{1},{r}_{2},\cdots ,{r}_{n}\right]$ is also an automorphism of the ring D. Since $\left[\mathbb{E}:ℚ\right]=N$ , the group $Aut\left(\mathbb{E}/ℚ\right)$ has order N. Let

$Aut\left(\mathbb{E}/ℚ\right)=\left\{{\sigma }_{1},{\sigma }_{2},\cdots ,{\sigma }_{N}\right\}$

So given a non-trivial homomorphism $\psi :D\to {\mathbb{E}}_{p}$ , we get N distinct homomorphisms ${\psi }_{j}=\psi \cdot {\sigma }_{j}$ , $1\le j\le N$ , from D to ${\mathbb{E}}_{p}$ . We claim that these are all the homomorphisms from D to ${\mathbb{E}}_{p}$ . Suppose that there is a homomorphism from D to ${\mathbb{E}}_{p}$ which is different from ${\psi }_{j}$ , $1\le j\le N$ . Let us denote it by ${\psi }_{N+1}$ . By Dedekind Independence Theorem the set $\left\{{\psi }_{1},{\psi }_{2},\cdots ,{\psi }_{N},{\psi }_{N+1}\right\}$ of $N+1$ homomorphisms from D to ${\mathbb{E}}_{p}$ is linearly independent over the field ${\mathbb{E}}_{p}$ .

Consider the following system of N homogeneous equations in $N+1$ variables $\left\{{x}_{1},{x}_{2},\cdots ,{x}_{N},{x}_{N+1}\right\}$ ,

$\underset{i=1}{\overset{N+1}{\sum }}\text{ }{x}_{i}{\psi }_{i}\left({u}_{j}\right)=0,\text{ }1\le j\le N.$

Since there are more variables than the equations this system of equations has a non-trivial solution. Let this non-trivial solution be ${x}_{i}={a}_{i}\in {\mathbb{E}}_{p}$ , $1\le i\le N+1$ . So we have

$\underset{i=1}{\overset{i=N+1}{\sum }}{a}_{i}{\psi }_{i}\left({u}_{j}\right)=0,\text{ }1\le j\le N.$

Let $y\in D=ℤ{u}_{1}\oplus ℤ{u}_{2}\oplus \cdots \oplus ℤ{u}_{N}$ . So $y={n}_{1}{u}_{1}+{n}_{2}{u}_{2}+\cdots +{n}_{N}{u}_{N}$ , ${n}_{k}\in ℤ$ , $1\le k\le N$ . Then for $1\le i\le N+1$ we have

${\psi }_{i}\left(y\right)=\stackrel{¯}{{n}_{1}}{\psi }_{i}\left({u}_{1}\right)+\stackrel{¯}{{n}_{2}}{\psi }_{i}\left({u}_{2}\right)+\cdots +\stackrel{¯}{{n}_{N}}{\psi }_{i}\left({u}_{N}\right)=\underset{j=1}{\overset{j=N}{\sum }}\stackrel{¯}{{n}_{j}}{\psi }_{i}\left(uj\right)$

where $\stackrel{¯}{{n}_{j}}={n}_{j}+\left(p\right)$ . We shall show that

$\underset{i=1}{\overset{i=N+1}{\sum }}{a}_{i}{\psi }_{i}\left(y\right)=0,$

which will contradict the linear independence of $\left\{{\psi }_{1},{\psi }_{2},\cdots ,{\psi }_{N},{\psi }_{N+1}\right\}$ over ${\mathbb{E}}_{p}$ .

$\begin{array}{l}\underset{i=1}{\overset{i=N+1}{\sum }}{a}_{i}{\psi }_{i}\left(y\right)\\ =\underset{i=1}{\overset{i=N+1}{\sum }}{a}_{i}\underset{j=1}{\overset{j=N}{\sum }}\stackrel{¯}{{n}_{j}}{\psi }_{i}\left({u}_{j}\right)\\ =\underset{i=1}{\overset{i=N+1}{\sum }}{a}_{i}\left(\stackrel{¯}{{n}_{1}}{\psi }_{i}\left({u}_{1}\right)+\stackrel{¯}{{n}_{2}}{\psi }_{i}\left({u}_{2}\right)+\cdots +\stackrel{¯}{{n}_{N}}{\psi }_{i}\left({u}_{N}\right)\right)\\ =\stackrel{¯}{{n}_{1}}\underset{i=1}{\overset{i=N+1}{\sum }}{a}_{i}{\psi }_{i}\left({u}_{1}\right)+\stackrel{¯}{{n}_{2}}\underset{i=1}{\overset{i=N+1}{\sum }}{a}_{i}{\psi }_{i}\left({u}_{2}\right)+\cdots +\stackrel{¯}{{n}_{N}}\underset{i=1}{\overset{i=N+1}{\sum }}{a}_{i}{\psi }_{i}\left({u}_{N}\right)\\ =0.\end{array}$

3. Proof of the Main Theorem

Since the field ${\mathbb{E}}_{p}$ has order ${p}^{m}$ , the group $Aut\left({\mathbb{E}}_{p}\right)$ has order m and $\pi :{\mathbb{E}}_{p}\to {\mathbb{E}}_{p}$ where $\pi \left(a\right)={a}^{p}$ for all $a\in {\mathbb{E}}_{p}$ , is the generating automorphism of $Aut\left({\mathbb{E}}_{p}\right)$ . So if $\psi :D\to {\mathbb{E}}_{p}$ is any homomorphism then so is $\pi \cdot \psi$ . Since $\psi$ and $\pi \cdot \psi$ are two homomorphisms from D to ${\mathbb{E}}_{p}$ there exist $\sigma \in Aut\left(\mathbb{E}/ℚ\right)$ such that $\pi \cdot \psi =\psi \cdot \sigma$ or ${\psi }^{-1}\cdot \pi \cdot \psi =\sigma$ . This proves that the action on $\sigma$ on $\left\{{r}_{1},{r}_{2},\cdots ,{r}_{n}\right\}$ is similar to the action of $\pi$ on $\left\{\stackrel{¯}{{r}_{1}},\stackrel{¯}{{r}_{2}},\cdots ,\stackrel{¯}{{r}_{n}}\right\}$ . Note: In the following diagram the mapping

$D\stackrel{\sigma }{\to }D$

is the restriction of $\sigma \in Aut\left(\mathbb{E}/ℚ\right)$ to D and we are only concerned with the effect of the mappings $\sigma$ , $\psi$ and $\pi$ on $\left\{{r}_{1},{r}_{2},\cdots ,{r}_{n}\right\}$ and $\left\{\stackrel{¯}{{r}_{1}},\stackrel{¯}{{r}_{2}},\cdots ,\stackrel{¯}{{r}_{n}}\right\}$ . Clearly

$\begin{array}{l}\left\{{r}_{1},{r}_{2},\cdots ,{r}_{n}\right\}\stackrel{\sigma }{\to }\left\{{r}_{1},{r}_{2},\cdots ,{r}_{n}\right\}\\ \left\{\stackrel{¯}{{r}_{1}},\stackrel{¯}{{r}_{2}},\cdots ,\stackrel{¯}{{r}_{n}}\right\}\stackrel{\pi }{\to }\left\{\stackrel{¯}{{r}_{1}},\stackrel{¯}{{r}_{2}},\cdots ,\stackrel{¯}{{r}_{n}}\right\}\\ \left\{{r}_{1},{r}_{2},\cdots ,{r}_{n}\right\}\stackrel{\psi }{\to }\left\{\stackrel{¯}{{r}_{1}},\stackrel{¯}{{r}_{2}},\cdots ,\stackrel{¯}{{r}_{n}}\right\}\end{array}$ As ${\psi }^{-1}\cdot \pi \cdot \psi =\sigma$ and $\psi \cdot \sigma \cdot {\psi }^{-1}=\pi$ the effect of $\sigma$ on $\left\{{r}_{1},{r}_{2},\cdots ,{r}_{n}\right\}$ is similar to the effect of $\pi$ on $\left\{\stackrel{¯}{{r}_{1}},\stackrel{¯}{{r}_{2}},\cdots ,\stackrel{¯}{{r}_{n}}\right\}$ . This is further illustrated by the following:

$\sigma \left({r}_{i}\right)={r}_{j}⇒\pi \left(\stackrel{¯}{{r}_{i}}\right)=\stackrel{¯}{{r}_{j}}$

$\stackrel{¯}{{r}_{i}}\stackrel{{\psi }^{-1}}{\to }{r}_{i}\stackrel{\sigma }{\to }{r}_{j}\stackrel{\psi }{\to }\stackrel{¯}{{r}_{j}}$

$\pi \left(\stackrel{¯}{{r}_{i}}\right)=\stackrel{¯}{{r}_{j}}⇒\sigma \left({r}_{i}\right)={r}_{j}$

${r}_{i}\stackrel{\psi }{\to }\stackrel{¯}{{r}_{i}}\stackrel{\pi }{\to }\stackrel{¯}{{r}_{j}}\stackrel{{\psi }^{-1}}{\to }{r}_{j}$

Conflicts of Interest

The authors declare no conflicts of interest.

   Jacobson, N. (2014) Basic Algebra. 2nd Edition, Dover Publications, Inc., Mine-ola, New York.

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