## Section9.2Field Automorphisms

Our goal now is to establish a link between group theory and field theory by examining automorphisms of fields.

Suppose $\alpha$ and $\beta$ are two roots of the same irreducible polynomial $p(x) \in F[x]\text{.}$ Since $F[x]/\langle p(x)\rangle$ is isomorphic to both $F(\alpha)$ and $F(\beta)\text{,}$ we get that there is an isomorphism from $F(\alpha)$ to $F(\beta)$ by following the isomorphism from $F(\alpha) \to F[x]/\langle p(x) \rangle$ and then from $F[x]/\langle p(x) \rangle \to F(\beta)\text{.}$ This isomorphism sends $\alpha$ to $\beta$ but sends every element of $F$ to itself.

Now we can continue to add roots of $p(x)$ until we arrive at the splitting field. If we extend the isomorphism we had above to the splitting field, then it will be an isomorphism from the splitting field to itself. An isomorphism $\varphi:E \to E$ is called an automorphism (a “self-isomorphism”). Since the domain and codomain are equal, we can easily compose automorphism. We can think of automorphism as special kinds of permutations of the set, which we already know together form a group.

###### Proof.

If $\sigma$ and $\tau$ are automorphisms of $F\text{,}$ then so are $\sigma \tau$ and $\sigma^{-1}\text{.}$ The identity is certainly an automorphism; hence, the set of all automorphisms of a field $F$ is indeed a group.

In the context of field extensions, we can consider only those automorphisms of a field that preserve of fix elements from the base field. That is, for every element $\alpha in F\text{,}$ we require that $\varphi(\alpha) = \alpha\text{.}$ Restricting to just these special automorphisms also gives us a group:

###### Proof.

We need only show that the set of automorphisms of $E$ that fix $F$ elementwise is a subgroup of the group of all automorphisms of $E\text{.}$ Let $\sigma$ and $\tau$ be two automorphisms of $E$ such that $\sigma( \alpha ) = \alpha$ and $\tau( \alpha ) = \alpha$ for all $\alpha \in F\text{.}$ Then $\sigma \tau( \alpha ) = \sigma( \alpha) = \alpha$ and $\sigma^{-1}( \alpha ) = \alpha\text{.}$ Since the identity fixes every element of $E\text{,}$ the set of automorphisms of $E$ that leave elements of $F$ fixed is a subgroup of the entire group of automorphisms of $E\text{.}$

These $F$-fixing automorphisms will be very useful, so we give the group of them a special name. Let $E$ be a field extension of $F\text{.}$ We will denote the full group of automorphisms of $E$ by $\aut(E)\text{.}$ We define the Galois group of $E$ over $F$ to be the group of automorphisms of $E$ that fix $F$ elementwise; that is,

\begin{equation*} G(E/F) = \{ \sigma \in \aut(E) : \sigma(\alpha) = \alpha \text{ for all } \alpha \in F \}\text{.} \end{equation*}

We will often want to consider the Galois group relative to a particular polynomial. If $f(x)$ is a polynomial in $F[x]$ and $E$ is the splitting field of $f(x)$ over $F\text{,}$ then we define the Galois group of $f(x)$ to be $G(E/F)\text{.}$

###### Example9.11.

Complex conjugation, defined by $\sigma : a + bi \mapsto a - bi\text{,}$ is an automorphism of the complex numbers. Since

\begin{equation*} \sigma(a) = \sigma(a + 0i) = a - 0i = a\text{,} \end{equation*}

the automorphism defined by complex conjugation must be in $G( {\mathbb C} / {\mathbb R} )\text{.}$

###### Example9.12.

Consider the fields ${\mathbb Q} \subset {\mathbb Q}(\sqrt{5}\, ) \subset {\mathbb Q}( \sqrt{3}, \sqrt{5}\, )\text{.}$ Then for $a, b \in {\mathbb Q}( \sqrt{5}\, )\text{,}$

\begin{equation*} \sigma( a + b \sqrt{3}\, ) = a - b \sqrt{3} \end{equation*}

is an automorphism of ${\mathbb Q}(\sqrt{3}, \sqrt{5}\, )$ leaving ${\mathbb Q}( \sqrt{5}\, )$ fixed. Similarly,

\begin{equation*} \tau( a + b \sqrt{5}\, ) = a - b \sqrt{5} \end{equation*}

is an automorphism of ${\mathbb Q}(\sqrt{3}, \sqrt{5}\, )$ leaving ${\mathbb Q}( \sqrt{3}\, )$ fixed. The automorphism $\mu = \sigma \tau$ moves both $\sqrt{3}$ and $\sqrt{5}\text{.}$ It will soon be clear that $\{ \identity, \sigma, \tau, \mu \}$ is the Galois group of ${\mathbb Q}(\sqrt{3}, \sqrt{5}\, )$ over ${\mathbb Q}\text{.}$ The following table shows that this group is isomorphic to ${\mathbb Z}_2 \times {\mathbb Z}_2\text{.}$

\begin{equation*} \begin{array}{c|cccc} & \identity & \sigma & \tau & \mu \\ \hline \identity & \identity & \sigma & \tau & \mu \\ \sigma & \sigma & \identity & \mu & \tau \\ \tau & \tau & \mu & \identity & \sigma \\ \mu & \mu & \tau & \sigma & \identity \end{array} \end{equation*}

We may also regard the field ${\mathbb Q}( \sqrt{3}, \sqrt{5}\, )$ as a vector space over ${\mathbb Q}$ that has basis $\{ 1, \sqrt{3}, \sqrt{5}, \sqrt{15}\, \}\text{.}$ It is no coincidence that $|G( {\mathbb Q}( \sqrt{3}, \sqrt{5}\, ) /{\mathbb Q})| = [{\mathbb Q}(\sqrt{3}, \sqrt{5}\, ):{\mathbb Q})] = 4\text{.}$

As illustrated in the example above, automorphisms in the Galois group send roots of polynomials somewhere. But what sorts of elements can be the image of such roots? The next proposition proves that roots can only be sent to other roots of the same polynomial!

###### Proof.

Let

\begin{equation*} f(x) = a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n \end{equation*}

and suppose that $\alpha \in E$ is a zero of $f(x)\text{.}$ Then for $\sigma \in G(E/F)\text{,}$

\begin{align*} 0 & = \sigma( 0 )\\ & = \sigma( f( \alpha ))\\ & = \sigma(a_0 + a_1\alpha + a_2 \alpha^2 + \cdots + a_n \alpha^n)\\ & = a_0 + a_1 \sigma(\alpha) + a_2 [\sigma(\alpha)]^2 + \cdots + a_n [\sigma(\alpha)]^n; \end{align*}

therefore, $\sigma( \alpha )$ is also a zero of $f(x)\text{.}$

Let $E$ be an algebraic extension of a field $F\text{.}$ We say that two elements $\alpha, \beta \in E$ are conjugate over $F$ if they have the same minimal polynomial. For example, in the field ${\mathbb Q}( \sqrt{2}\, )$ the elements $\sqrt{2}$ and $-\sqrt{2}$ are conjugate over ${\mathbb Q}$ since they are both roots of the irreducible polynomial $x^2 - 2\text{.}$

The previous proposition says that automorphisms in $G(E/F)$ must send elements to their conjugates. (But note that an element is always conjugate with itself, and can be sent to itself by an automorphism; after all, the identity automorphism is an element of $G(E/F)\text{.}$) A sort of converse of this would say that if two elements of conjugate, then there exists an automorphism that sends one to the other. This is true, and the proof follows directly from Lemma 9.4.

In Example 9.12, we noticed that the size of the Galois group was 4, and this was also the degree of the field extension $E$ over $F\text{.}$ We can now prove that this is no coincidence.

###### Proof.

We will use mathematical induction on the degree of $f(x)\text{.}$ If the degree of $f(x)$ is $0$ or $1\text{,}$ then $E = F$ and there is nothing to show. Assume that the result holds for all polynomials of degree $k$ with $0 \leq k \lt n\text{.}$ Suppose that the degree of $f(x)$ is $n\text{.}$ Let $p(x)$ be an irreducible factor of $f(x)$ of degree $r\text{.}$ Since all of the roots of $p(x)$ are in $E\text{,}$ we can choose one of these roots, say $\alpha\text{,}$ so that $F \subset F( \alpha ) \subset E\text{.}$ Then

\begin{equation*} [E: F(\alpha)] = n/r \quad \text{and} \quad [F(\alpha): F] = r\text{.} \end{equation*}

If $\beta$ is any other root of $p(x)\text{,}$ then $F \subset F( \beta ) \subset E\text{.}$ By Lemma 9.4, there exists a unique isomorphism $\sigma: F( \alpha ) \rightarrow F( \beta )$ for each such $\beta$ that fixes $F$ elementwise. Since $E$ is a splitting field of $p(x)\text{,}$ there are exactly $r$ such isomorphisms. For each of these automorphisms, we can use our induction hypothesis on $[E: F(\alpha)] = n/r \lt n$ to conclude that

\begin{equation*} |G(E/F(\alpha))| = [E:F(\alpha)]\text{.} \end{equation*}

Consequently, there are

\begin{equation*} [E:F] = [E:F(\alpha)] [F( \alpha):F] = n \end{equation*}

possible automorphisms of $E$ that fix $F\text{,}$ or $|G(E/F)| = [E:F]\text{.}$

###### Proof.

Let $p$ be the characteristic of $E$ and $F$ and assume that the orders of $E$ and $F$ are $p^m$ and $p^n\text{,}$ respectively. Then $nk = m\text{.}$ We can also assume that $E$ is the splitting field of $x^{p^m} - x$ over a subfield of order $p\text{.}$ Therefore, $E$ must also be the splitting field of $x^{p^m} - x$ over $F\text{.}$ Applying Theorem 9.15, we find that $|G(E/F)| = k\text{.}$

To prove that $G(E/F)$ is cyclic, we must find a generator for $G(E/F)\text{.}$ Let $\sigma : E \rightarrow E$ be defined by $\sigma(\alpha) = \alpha^{p^n}\text{.}$ We claim that $\sigma$ is the element in $G(E/F)$ that we are seeking. We first need to show that $\sigma$ is in $\aut(E)\text{.}$ If $\alpha$ and $\beta$ are in $E\text{,}$

\begin{equation*} \sigma(\alpha + \beta) = (\alpha + \beta)^{p^n} = \alpha^{p^n} + \beta^{p^n} = \sigma(\alpha) + \sigma(\beta) \end{equation*}

by Lemma 17.3. Also, it is easy to show that $\sigma(\alpha \beta) = \sigma( \alpha ) \sigma( \beta )\text{.}$ Since $\sigma$ is a nonzero homomorphism of fields, it must be injective. It must also be onto, since $E$ is a finite field. We know that $\sigma$ must be in $G(E/F)\text{,}$ since $F$ is the splitting field of $x^{p^n} - x$ over the base field of order $p\text{.}$ This means that $\sigma$ leaves every element in $F$ fixed. Finally, we must show that the order of $\sigma$ is $k\text{.}$ By Theorem 9.15, we know that

\begin{equation*} \sigma^k( \alpha ) = \alpha^{p^{nk}} = \alpha^{p^m} = \alpha \end{equation*}

is the identity of $G( E/F)\text{.}$ However, $\sigma^r$ cannot be the identity for $1 \leq r \lt k\text{;}$ otherwise, $x^{p^{nr}} - x$ would have $p^m$ roots, which is impossible.

###### Example9.17.

We can now confirm that the Galois group of ${\mathbb Q}( \sqrt{3}, \sqrt{5}\, )$ over ${\mathbb Q}$ in Example 9.12 is indeed isomorphic to ${\mathbb Z}_2 \times {\mathbb Z}_2\text{.}$ Certainly the group $H = \{ \identity, \sigma, \tau, \mu \}$ is a subgroup of $G({\mathbb Q}( \sqrt{3}, \sqrt{5}\, )/{\mathbb Q})\text{;}$ however, $H$ must be all of $G({\mathbb Q}( \sqrt{3}, \sqrt{5}\, )/{\mathbb Q})\text{,}$ since

\begin{equation*} |H| = [{\mathbb Q}( \sqrt{3}, \sqrt{5}\, ):{\mathbb Q}] = |G({\mathbb Q}( \sqrt{3}, \sqrt{5}\, )/{\mathbb Q})| = 4\text{.} \end{equation*}
###### Example9.18.

Let us compute the Galois group of

\begin{equation*} f(x) = x^4 + x^3 + x^2 + x + 1 \end{equation*}

over ${\mathbb Q}\text{.}$ First let's check that $f(x)$ is irreducible over $\Q\text{.}$ To do this, consider a new polynomial $g(x) = f(x+1)$ (essentially doing a change of variables). We have

\begin{equation*} g(x) = x^4 + 5x^3 + 10x^2 + 10x + 5. \end{equation*}

This is irreducible by Eisenstein's criterion. But if $f(x)$ factored, then so would $g(x)\text{,}$ so we see that $f(x)$ is irreducible.

Furthermore, since $(x -1)f(x) = x^5 - 1\text{,}$ we can use DeMoivre's Theorem to determine that the roots of $f(x)$ are $\omega^i\text{,}$ where $i = 1, \ldots, 4$ and

\begin{equation*} \omega = \cos(2 \pi / 5 ) + i \sin(2 \pi / 5 )\text{.} \end{equation*}

Hence, the splitting field of $f(x)$ must be ${\mathbb Q}(\omega)\text{.}$ We can define automorphisms $\sigma_i$ of ${\mathbb Q}(\omega )$ by $\sigma_i( \omega ) = \omega^i$ for $i = 1, \ldots, 4\text{.}$ It is easy to check that these are indeed distinct automorphisms in $G( {\mathbb Q}( \omega) / {\mathbb Q} )\text{.}$ Since

\begin{equation*} [{\mathbb Q}( \omega) : {\mathbb Q}] = | G( {\mathbb Q}( \omega) / {\mathbb Q})| = 4\text{,} \end{equation*}

the $\sigma_i$'s must be all of $G( {\mathbb Q}( \omega) / {\mathbb Q} )\text{.}$ Therefore, $G({\mathbb Q}( \omega) / {\mathbb Q})\cong {\mathbb Z}_4$ since $\omega$ is a generator for the Galois group.

### ExercisesCollected Homework

###### 1.

Consider the polynomial $a(x) = x^4 - 10x^2 + 21$ in $\Q[x]\text{.}$

###### (a)

Find the splitting field $E$ over $\Q\text{.}$ Draw a tower diagram including all intermediate fields and their degrees.

Hint

Start by factoring $a(x)$/

###### (b)

Let $\Q(\alpha) \ne \Q(\beta)$ be different intermediate fields between $\Q$ and $E\text{.}$ Explain why there is NOT an isomorphism from $\Q(\alpha)$ to $\Q(\beta)$ that sends $\alpha$ to $\beta\text{.}$ Then explain why there is no isomorphism between these fields at all.

###### (c)

Describe a non-trivial automorphism of $E\text{.}$ Explain how you know your example works.

###### (d)

Describe the Galois group $G(E/\Q)\text{.}$ Be sure to explicitly say what each element of the group is, as well as say what “standard” group it is isomorphic to.

###### (e)

Bonus! The polynomial $x^4 - 62x^2 + 625$ has $\sqrt{3}+2\sqrt{7}$ as a root. Use your results from the previous parts to find the three other roots of this polynomial.