Shiv Gupta
Department of Mathematics, West Chester University, West Chester, PA, USA.
DOI: 10.4236/ojdm.2018.83007   PDF    HTML   XML   1,213 Downloads   3,497 Views   Citations

Abstract

In this note we give a complete proof of a theorem of Dedekind.

Keywords

Galois Group of a Polynomial

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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, [1] and we shall keep the notations used in that proof.

Theorem 1 Let $f\left(x\right)\in \mathbb{Z}\left[x\right]$ 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 $\mathbb{Q}$ of rational numbers. Suppose that $f_p=\bar{f}=f\left(\mathrm{mod}p\right)\in {\mathbb{Z}}_p\left[x\right]$ factors as:

$f_p=\bar{f}={\displaystyle \underset{i=1}{\overset{r}{\prod }}}\overline{f_i}$

where $\overline{f_i}$ are distinct monic irreducible polynomials in ${\mathbb{Z}}_p\left[x\right]$ , degree $\left(\overline{f_i}\right)=d_i$ , $1\le i\le r$ , 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 $d_1\mathrm{,}{d}_2\mathrm{,}\cdots \mathrm{,}{d}_r$ .

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 $\mathbb{E}\mathrm{<mo>\; /\; </mo>}\mathbb{F}$ 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 ${\chi }_1\mathrm{,}{\chi }_2\mathrm{,}\cdots \mathrm{,}{\chi }_n$ 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 $\text{Aut}\left(GF\left(p^m\right)\right)=\langle \sigma \rangle$ 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 $\sigma \mathrm{,}\eta \in S_n$ . Then $\sigma$ and ${\eta }^{-1}\sigma \eta$ have same cyclic structure.

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

Theorem 2 Let $f\left(x\right)\in \mathbb{Z}\left[x\right]$ 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 $\mathbb{Q}$ . Let ${\mathbb{E}}_p$ be a splitting field of $f_p\left(x\right)$ over ${\mathbb{Z}}_p=\mathbb{Z}/\left(p\right)$ . Let

$f\left(x\right)=\left(x-r_1\right)\left(x-r_2\right)\cdots \left(x-r_n\right)\mathrm{,}{r}_i\in \mathbb{E}\subset ℂ\mathrm{,1}\le i\le n$

$\begin{array}{l}R=\left\{r_1,r_2,\cdots ,r_n\right\},\\ R_p=\left\{\overline{r_1},\overline{r_2},\cdots ,\overline{r_n}\right\}\subset {\mathbb{E}}_p\end{array}$

where $\overline{r_i}$ , $1\le i\le n$ are the roots of $f_p\left(x\right)\in {\mathbb{Z}}_p\left[x\right]$ and

$\mathbb{E}=\mathbb{Q}\left(r_1,r_2,\cdots ,r_n\right),{\mathbb{E}}_p={\mathbb{Z}}_p\left(\overline{r_1},\overline{r_2},\cdots ,\overline{r_n}\right)$

Let $D=\mathbb{Z}\left[r_1,r_2,\cdots ,r_n\right]$ be the subring generated by the roots of $r_1\mathrm{,}{r}_2\mathrm{,}\cdots \mathrm{,}{r}_n$ 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 $\sigma \in Aut\left(\mathbb{E}\mathrm{<mo>\; /\; </mo>}\mathbb{Q}\right)=Gal\left(f\left(x\right)\right)$ , 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:

$\mathbb{E}=\mathbb{Q}\left(r_1\mathrm{,}{r}_2\mathrm{,}\cdots \mathrm{,}{r}_n\right)=\mathbb{Q}\left[r_1\mathrm{,}{r}_2\mathrm{,}\cdots ,r_n\right]$

We claim that $D=\mathbb{Z}\left[r_1\mathrm{,}{r}_2\mathrm{,}\cdots \mathrm{,}{r}_n\right]$ is a finitely generated (additive) Abelian group. Since each $r_i$ is a root of the monic polynomial $f\left(x\right)\in \mathbb{Z}\left[x\right]$ of degree n any positive power of $r_i,1\le i\le n$ can be expressed as an integral linear combination of $\mathrm{1,}{r}_i\mathrm{,}{r}_i^2\mathrm{,}\cdots r_i^{n-1}$ . It follows that

$D={\displaystyle \underset{0\le e_i\le n-1}{\sum }}\mathbb{Z}{r}_1^{e_1}{r}_2^{e_2}\cdots r_n^{e_n}.$

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 $\left\{u_1\mathrm{,}{u}_2\mathrm{,}\cdots \mathrm{,}{u}_N\right\}$ be a set consisting of an independent generating system of D. We have

$D=\mathbb{Z}{u}_1\oplus \mathbb{Z}{u}_2\oplus \cdots \oplus \mathbb{Z}{u}_N\mathrm{,}N\le n^n\mathrm{.}$

We claim that $\left\{u_1\mathrm{,}{u}_2\mathrm{,}\cdots \mathrm{,}{u}_N\right\}$ is a basis of $\mathbb{E}\mathrm{<mo>\; /\; </mo>}\mathbb{Q}$ . Obviously $\left\{u_1\mathrm{,}{u}_2\mathrm{,}\cdots \mathrm{,}{u}_N\right\}$ is linearly independent over $\mathbb{Q}$ . Let $\mathbb{Q}D={\displaystyle {\sum }_{1\le i\le N}}\mathbb{Q}{u}_i$ . Then $\mathbb{Q}D$ is a ring and $\mathbb{Q}\subset \mathbb{Q}D\subset \mathbb{E}$ therefore $\mathbb{Q}D$ is a field. Since $r_i\in D$ for $1\le i\le n$ , by (4) $\mathbb{Q}D=\mathbb{E}$ and $\left\{u_1\mathrm{,}{u}_2\mathrm{,}\cdots \mathrm{,}{u}_N\right\}$ is a basis of $E\mathrm{<mo>\; /\; </mo>}\mathbb{Q}$ . As $D=\mathbb{Z}{u}_1\oplus \mathbb{Z}{u}_2\oplus \cdots \oplus \mathbb{Z}{u}_N$ ,

$pD=\mathbb{Z}\left(p{u}_1\right)\oplus \mathbb{Z}\left(p{u}_2\right)\oplus \cdots \oplus \mathbb{Z}\left(p{u}_N\right)$

is an ideal of D and

$D/pD=\left\{\overline{a_1{u}_1+a_2{u}_2+\cdots +a_N{u}_N}:0\le a_i\le p-1\right\}.$

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 ${\mathbb{Z}}_p=\mathbb{Z}/p\mathbb{Z}$ which we will identify as ${\mathbb{Z}}_p$ 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

$\nu \mathrm{<mo>\; :\; </mo>}D\to D/M$

whose kernel is M and $p\mathbb{Z}\subset M$ . Therefore $\nu \left(\mathbb{Z}\right)={\mathbb{Z}}_p$ . We note that as $D=\mathbb{Z}\left[r_1\mathrm{,}{r}_2\mathrm{,}\cdots \mathrm{,}{r}_n\right]$ we have for $1\le i\le n$

$\nu \left(r_i\right)=r_i+M=\overline{r_i},\nu \left(D\right)={\mathbb{Z}}_p\left[\overline{r_1},\overline{r_2},\cdots ,\overline{r_n}\right]$

As $\nu$ is an epimorphism we have

$\nu \left(D\right)=D/M={\mathbb{Z}}_p\left[\overline{r_1},\overline{r_2},\cdots ,\overline{r_n}\right]$

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

$\varphi \mathrm{<mo>\; :\; </mo>}D/M\to {\mathbb{Z}}_p$

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

2) Let $\psi \mathrm{<mo>\; :\; </mo>}D\to {\mathbb{E}}_p$ be a homomorphism. So $\psi \left(1\right)=1$ . As $\mathbb{Z}\subset D$ , and ${\mathbb{E}}_p$ has characteristic p, $\psi \left(p\right)=0$ , so $\psi \left(\mathbb{Z}\right)={\mathbb{Z}}_p\subset {\mathbb{E}}_p$ . $\psi$ 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

$f\left(x\right)=\left(x-r_1\right)\left(x-r_2\right)\cdots \left(x-r_n\right)$

$\psi \left(f\left(x\right)\right)=f_p\left(x\right)=\left(x-\psi \left(r_1\right)\right)\left(x-\psi \left(r_2\right)\right)\cdots \left(x-\psi \left(r_n\right)\right)\mathrm{,}$

$\psi \left(r_i\right),1\le i\le n$ are the roots of the $f_p\left(x\right)$ in ${\mathbb{E}}_p$ and therefore the restriction of $\psi$ to R

${\psi }_{\mathrm{<mo>\; |\; </mo>}R}\mathrm{<mo>\; :\; </mo>}\left\{r_1\mathrm{,}{r}_2\mathrm{,}\cdots \mathrm{,}{r}_n\right\}\to \left\{\overline{r_1}\mathrm{,}\overline{r_2}\mathrm{,}\cdots \mathrm{,}\overline{r_n}\right\}$

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 $\psi \mathrm{<mo>\; :\; </mo>}D\to {\mathbb{E}}_p$ , and $\sigma \in Gal\left(f\right)=Aut\left(\mathbb{E}\mathrm{<mo>\; /\; </mo>}\mathbb{Q}\right)$ , ${\psi }^{\prime }=\psi \cdot \sigma$ is also a homomorphism from D to ${\mathbb{E}}_p$ . We note that the restriction of $\sigma \in Aut\left(\mathbb{E}\mathrm{<mo>\; /\; </mo>}\mathbb{Q}\right)$ to $D=\mathbb{Z}\left[r_1,r_2,\cdots ,r_n\right]$ is also an automorphism of the ring D. Since $\left[\mathbb{E}:\mathbb{Q}\right]=N$ , the group $Aut\left(\mathbb{E}\mathrm{<mo>\; /\; </mo>}\mathbb{Q}\right)$ has order N. Let

$Aut\left(\mathbb{E}/\mathbb{Q}\right)=\left\{{\sigma }_1,{\sigma }_2,\cdots ,{\sigma }_N\right\}$

So given a non-trivial homomorphism $\psi \mathrm{<mo>\; :\; </mo>}D\to {\mathbb{E}}_p$ , 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 $\left\{{\psi }_1\mathrm{,}{\psi }_2\mathrm{,}\cdots \mathrm{,}{\psi }_N\mathrm{,}{\psi }_{N+1}\right\}$ 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 $\left\{x_1\mathrm{,}{x}_2\mathrm{,}\cdots \mathrm{,}{x}_N\mathrm{,}{x}_{N+1}\right\}$ ,

${\displaystyle \underset{i=1}{\overset{N+1}{\sum }}}{x}_i{\psi }_i\left(u_j\right)=0,1\le j\le N.$

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

${\displaystyle \underset{i=1}{\overset{i=N+1}{\sum }}}{a}_i{\psi }_i\left(u_j\right)=0,1\le j\le N.$

Let $y\in D=\mathbb{Z}{u}_1\oplus \mathbb{Z}{u}_2\oplus \cdots \oplus \mathbb{Z}{u}_N$ . So $y=n_1{u}_1+n_2{u}_2+\cdots +n_N{u}_N$ , $n_k\in \mathbb{Z}$ , $1\le k\le N$ . Then for $1\le i\le N+1$ we have

${\psi }_i\left(y\right)=\overline{n_1}{\psi }_i\left(u_1\right)+\overline{n_2}{\psi }_i\left(u_2\right)+\cdots +\overline{n_N}{\psi }_i\left(u_N\right)={\displaystyle \underset{j=1}{\overset{j=N}{\sum }}}\overline{n_j}{\psi }_i(uj)$

where $\overline{n_j}=n_j+\left(p\right)$ . We shall show that

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

which will contradict the linear independence of $\left\{{\psi }_1\mathrm{,}{\psi }_2\mathrm{,}\cdots \mathrm{,}{\psi }_N\mathrm{,}{\psi }_{N+1}\right\}$ over ${\mathbb{E}}_p$ .

$\begin{array}{l}{\displaystyle \underset{i=1}{\overset{i=N+1}{\sum }}}{a}_i{\psi }_i\left(y\right)\\ ={\displaystyle \underset{i=1}{\overset{i=N+1}{\sum }}}{a}_i{\displaystyle \underset{j=1}{\overset{j=N}{\sum }}}\overline{n_j}{\psi }_i\left(u_j\right)\\ ={\displaystyle \underset{i=1}{\overset{i=N+1}{\sum }}}{a}_i\left(\overline{n_1}{\psi }_i\left(u_1\right)+\overline{n_2}{\psi }_i\left(u_2\right)+\cdots +\overline{n_N}{\psi }_i\left(u_N\right)\right)\\ =\overline{n_1}{\displaystyle \underset{i=1}{\overset{i=N+1}{\sum }}}{a}_i{\psi }_i\left(u_1\right)+\overline{n_2}{\displaystyle \underset{i=1}{\overset{i=N+1}{\sum }}}{a}_i{\psi }_i\left(u_2\right)+\cdots +\overline{n_N}{\displaystyle \underset{i=1}{\overset{i=N+1}{\sum }}}{a}_i{\psi }_i\left(u_N\right)\\ =0.\end{array}$

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 $\pi \mathrm{<mo>\; :\; </mo>}{\mathbb{E}}_p\to {\mathbb{E}}_p$ 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 $\psi \mathrm{<mo>\; :\; </mo>}D\to {\mathbb{E}}_p$ 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 $\sigma \in Aut\left(\mathbb{E}\mathrm{<mo>\; /\; </mo>}\mathbb{Q}\right)$ such that $\pi \cdot \psi =\psi \cdot \sigma$ or ${\psi }^{-1}\cdot \pi \cdot \psi =\sigma$ . This proves that the action on $\sigma$ on $\left\{r_1\mathrm{,}{r}_2\mathrm{,}\cdots \mathrm{,}{r}_n\right\}$ is similar to the action of $\pi$ on $\left\{\overline{r_1}\mathrm{,}\overline{r_2}\mathrm{,}\cdots \mathrm{,}\overline{r_n}\right\}$ . Note: In the following diagram the mapping

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

is the restriction of $\sigma \in Aut\left(\mathbb{E}\mathrm{<mo>\; /\; </mo>}\mathbb{Q}\right)$ to D and we are only concerned with the effect of the mappings $\sigma$ , $\psi$ and $\pi$ on $\left\{r_1\mathrm{,}{r}_2\mathrm{,}\cdots \mathrm{,}{r}_n\right\}$ and $\left\{\overline{r_1}\mathrm{,}\overline{r_2}\mathrm{,}\cdots \mathrm{,}\overline{r_n}\right\}$ . Clearly

$\begin{array}{l}\left\{r_1\mathrm{,}{r}_2\mathrm{,}\cdots \mathrm{,}{r}_n\right\}\stackrel{\sigma }{\to }\left\{r_1\mathrm{,}{r}_2\mathrm{,}\cdots \mathrm{,}{r}_n\right\}\\ \left\{\overline{r_1}\mathrm{,}\overline{r_2}\mathrm{,}\cdots \mathrm{,}\overline{r_n}\right\}\stackrel{\pi }{\to }\left\{\overline{r_1}\mathrm{,}\overline{r_2}\mathrm{,}\cdots \mathrm{,}\overline{r_n}\right\}\\ \left\{r_1\mathrm{,}{r}_2\mathrm{,}\cdots \mathrm{,}{r}_n\right\}\stackrel{\psi }{\to }\left\{\overline{r_1}\mathrm{,}\overline{r_2}\mathrm{,}\cdots \mathrm{,}\overline{r_n}\right\}\end{array}$

As ${\psi }^{-1}\cdot \pi \cdot \psi =\sigma$ and $\psi \cdot \sigma \cdot {\psi }^{-1}=\pi$ the effect of $\sigma$ on $\left\{r_1\mathrm{,}{r}_2\mathrm{,}\cdots \mathrm{,}{r}_n\right\}$ is similar to the effect of $\pi$ on $\left\{\overline{r_1}\mathrm{,}\overline{r_2}\mathrm{,}\cdots \mathrm{,}\overline{r_n}\right\}$ . This is further illustrated by the following:

$\sigma \left(r_i\right)=r_j\Rightarrow \pi \left(\overline{r_i}\right)=\overline{r_j}$

$\overline{r_i}\stackrel{{\psi }^{-1}}{\to }{r}_i\stackrel{\sigma }{\to }{r}_j\stackrel{\psi }{\to }\overline{r_j}$

$\pi \left(\overline{r_i}\right)=\overline{r_j}\Rightarrow \sigma \left(r_i\right)=r_j$

$r_i\stackrel{\psi }{\to }\overline{r_i}\stackrel{\pi }{\to }\overline{r_j}\stackrel{{\psi }^{-1}}{\to }{r}_j$

Conflicts of Interest

The authors declare no conflicts of interest.

References

[1]	[1] Jacobson, N. (2014) Basic Algebra. 2nd Edition, Dover Publications, Inc., Mine-ola, New York.