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Section 7.3 Applications to Field Theory

Lagrange’s Theorem alone has striking consequences. Here we use it to prove a classical fact from number theory.

Proof.

By Example 7.1.2, \(F^{\ast} = F \setminus \{0\}\) is a group of order \(q - 1\) under multiplication. By Corollary 7.2.11, \(a^{q-1} = 1\) for every \(a \in F^{\ast}\text{.}\) Multiplying both sides by \(a\) gives \(a^q = a\text{,}\) which also holds trivially when \(a = 0\text{.}\)
Taking \(F = {\mathbb Z}_p\) for a prime \(p\) recovers the classical statement: if \(p\) does not divide \(a\text{,}\) then \(a^{p-1} \equiv 1 \pmod{p}\text{,}\) and \(b^p \equiv b \pmod p\) for every integer \(b\text{.}\)
We will next prove some result about abelian groups. That will have non-obvious consequences for fields.
The exponent, \(e(G)\) of a group is the smallest integer such that for all \(g\in G\) we have \(g^{e(G)}=e\text{.}\)

Proof.

First note that if a positive integer \(k\) divides \(|g|\) so that \(|g| = kr\) then \(|g^k|=r\text{.}\) Indeed, since \((g^k)^r=e\) we must have that the order of \(g^k\) divides \(r\text{.}\) Now if \(|g^k|=r'\lt r\) then \(g^{kr'}=e\text{,}\) which contradicts the definition of order since \(|g| = kr > kr'\text{.}\)
In particular this enables us to construct for \(g\) an element of any order dividing \(|g|\text{.}\)
Suppose now that \(|g|,|h|\) are relatively prime. Say their orders are \(n,m\) respectively. On the one hand \(|gh|=q\) must divide \(nm\text{.}\) Now note that
\begin{equation*} e = (gh)^q = g^qh^q \Rightarrow g^q = h^{-q}. \end{equation*}
which means that
\begin{equation*} g^{qm} = (g^q)^m=(h^{-q})^m=(h^m)^{-q} \end{equation*}
So \(n\text{,}\) the order of \(g\text{,}\) must divide \(qm\) and since \(\gcd(m,n)=1\) we must have that \(n\) divides \(q\text{.}\) A symmetric argument tells us that \(m\) also divides \(q\text{,}\) so altogether we must have \(q=nm\text{.}\)
We finally construct the element that realizes the order. Consider the set of all orders of elements of \(A\text{.}\) For every prime divisor \(p\) of some order let \(p^r\) be the maximal \(p\)-power that divides the order of some element. By taking powers of elements we can ensure that for every such maximal prime power there is an element of \(A\) that has that prime power order. Let \(g_1,\ldots,g_n\) be a complete list of such elements. Since these elements all have relatively prime orders we must have that for
\begin{equation*} g=g_1\cdots g_n \end{equation*}
that
\begin{equation*} |g| = |g_1|\cdots |g_n| = p_1^{r_1}\cdots p_n^{r_n} =N. \end{equation*}
By construction, the order of every element of \(A\) divides \(N\text{,}\) and since we have an explicit element \(g\) with order \(N\) we have that the exponent \(e(A)=N\) and the result follows.
This strange seeming result has the following immediate corollary.
We can use this to prove the following result
With each field \(F\) we have a multiplicative group of nonzero elements of \(F\) which we will denote by \(F^*\text{.}\) The multiplicative group of any finite field is cyclic. This result follows from the more general result that we will prove in the next theorem.

Proof.

Let \(G\) be a finite subgroup of \(F^\ast\) of order \(n\) and suppose towards a contradiction that \(G\) was not cyclic. By Corollary 7.3.3 and Proposition 7.3.2, the exponent \(e(G)=k\) must be strictly less than the order \(|G|=n\text{.}\)
Since \(G\) is a multiplicative subgroup of a field \(F\text{,}\) the exponent gives rise to the polynomial equation \(x^k-1\) and we have that every element of \(G\subset F\) is a root of this polynomial, i.e. this polynomial has at least \(n\) roots, but \(n>k\text{.}\) This leads to a contradiction since \(x^k-1\) can have at most \(k\) roots. It follows that \(G\) must be cyclic.

Proof.

Let \(\alpha\) be a generator for the cyclic group \(E^{\ast}\) of nonzero elements of \(E\text{.}\) Then \(E = F( \alpha )\text{.}\)

Example 7.3.7.

The finite field \(\gf(2^4)\) is isomorphic to the field \({\mathbb Z}_2[x]/ \langle 1 + x + x^4 \rangle\text{.}\) Therefore, the elements of \(\gf(2^4)\) can be taken to be
\begin{equation*} \{ a_0 + a_1 \alpha + a_2 \alpha^2 + a_3 \alpha^3 : a_i \in {\mathbb Z}_2 \text{ and } 1 + \alpha + \alpha^4 = 0 \}\text{.} \end{equation*}
Remembering that \(1 + \alpha +\alpha^4 = 0\text{,}\) we add and multiply elements of \(\gf(2^4)\) exactly as we add and multiply polynomials. The multiplicative group of \(\gf(2^4)\) is isomorphic to \({\mathbb Z}_{15}\) with generator \(\alpha\text{:}\)
\begin{align*} & \alpha^1 = \alpha & & \alpha^6 = \alpha^2 + \alpha^3 & & \alpha^{11} = \alpha + \alpha^2 + \alpha^3 &\\ & \alpha^2 = \alpha^2 & & \alpha^7 = 1 + \alpha + \alpha^3 & & \alpha^{12} = 1 + \alpha + \alpha^2 + \alpha^3 &\\ & \alpha^3 = \alpha^3 & & \alpha^8 = 1 + \alpha^2 & & \alpha^{13} = 1 + \alpha^2 + \alpha^3 &\\ & \alpha^4 = 1 + \alpha & & \alpha^9 = \alpha + \alpha^3 & & \alpha^{14} = 1 + \alpha^3 &\\ &\alpha^5 = \alpha + \alpha^2 & & \alpha^{10} = 1 + \alpha + \alpha^2 & & \alpha^{15} = 1. & \end{align*}
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