galois

[logistic_guy]

\(\displaystyle \bold{(a)}\) Show that if the field \(\displaystyle K\) is generated over \(\displaystyle F\) by the elements \(\displaystyle \alpha_1,\cdots,\alpha_n\) then an automorphism \(\displaystyle \sigma\) of \(\displaystyle K\) fixing \(\displaystyle F\) is uniquely determined by \(\displaystyle \sigma(\alpha_1),\cdots,\sigma(\alpha_n)\). In particular show that an automorphism fixes \(\displaystyle K\) if and only if it fixes a set of generators for \(\displaystyle K\).

\(\displaystyle \bold{(b)}\) Let \(\displaystyle G \leq \text{Gal}(K/F)\) be a subgroup of the Galois group of the extension \(\displaystyle K/F\) and suppose \(\displaystyle \sigma_1,\cdots,\sigma_n\) are generators for \(\displaystyle G\). Show that the subfield \(\displaystyle E/F\) is fixed by \(\displaystyle G\) if and only if it is fixed by the generators \(\displaystyle \sigma_1,\cdots,\sigma_n\).

 

[logistic_guy]

\(\displaystyle \bold{(a)}\)

Suppose that we have only two elements, \(\displaystyle \alpha_1 \ \text{and} \ \alpha_2\) such that:

\(\displaystyle \alpha_1 = \sqrt{2}\)
\(\displaystyle \alpha_2 = \sqrt{3}\)
\(\displaystyle F = \mathbb{Q}\)
\(\displaystyle K = \mathbb{Q}(\sqrt{2},\sqrt{3})\)

Then, any element \(\displaystyle x \in K\) can be written in the form:

\(\displaystyle x = a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}\)

 

[logistic_guy]

\(\displaystyle x = a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}\)

where \(\displaystyle a,b,c,d \in \mathbb{Q}\)

 

[logistic_guy]

Let \(\displaystyle \sigma \in \text{Aut}(K/F)\).

Since \(\displaystyle \sigma\) is an automorphism fixing \(\displaystyle F\)

Then,

\(\displaystyle \sigma\left(\sqrt{2}\right) \in \left\{\sqrt{2}, -\sqrt{2}\right\}\)

And

\(\displaystyle \sigma\left(\sqrt{3}\right) \in \left\{\sqrt{3}, -\sqrt{3}\right\}\)

 

[logistic_guy]

Let us calculate \(\displaystyle \sigma(x)\).

\(\displaystyle \sigma(x) = \sigma\left(a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}\right)\)


\(\displaystyle = a + b\sigma\left(\sqrt{2}\right) + c\sigma\left(\sqrt{3}\right) + d\sigma\left(\sqrt{6}\right)\)

Or

\(\displaystyle \sigma(x) = a + b\sigma\left(\sqrt{2}\right) + c\sigma\left(\sqrt{3}\right) + d\sigma\left(\sqrt{2}\right)\sigma\left(\sqrt{3}\right)\)


Since we already know \(\displaystyle \sigma\left(\sqrt{2}\right)\) And \(\displaystyle \sigma\left(\sqrt{3}\right)\)

Then,

\(\displaystyle \sigma(x)\) is entirely determined.

Therefore, once we specify the rule of \(\displaystyle \sigma\left(\sqrt{2}\right)\) And \(\displaystyle \sigma\left(\sqrt{3}\right)\), we have completely determined the action of \(\displaystyle \sigma\) on all the elements of \(\displaystyle K\).

 

[logistic_guy]

\(\displaystyle \bold{(a)}\)

Now I have to prove it in general.

Let \(\displaystyle K = F(\alpha_1,\cdots,\alpha_n)\)
Let \(\displaystyle \sigma \in \text{Aut}(K/F)\).
Let \(\displaystyle x \in K\)

Then, any element \(\displaystyle x \in K\) can be written in the form:

\(\displaystyle x = f(\alpha_1,\cdots,\alpha_n)\) where \(\displaystyle f \in F\)

Since \(\displaystyle \sigma\) is an automorphism fixing \(\displaystyle F\) and we know how it acts on \(\displaystyle \alpha_n\), then we can calculate \(\displaystyle \sigma(x)\). That is:

\(\displaystyle \sigma(x) = \sigma\bigg(f(\alpha_1,\cdots,\alpha_n)\bigg) = f(\sigma(\alpha_1),\cdots,\sigma(\alpha_n))\)

Then, \(\displaystyle \sigma\) is uniquely determined by its values on \(\displaystyle \alpha_1,\cdots,\alpha_n\).

If \(\displaystyle \sigma \in \text{Aut}(K/F)\) fixes the generators \(\displaystyle \alpha_1,\cdots, \alpha_n\)

then, for all \(\displaystyle x \in K\), we have \(\displaystyle \sigma(x) = x\).

Therefore, \(\displaystyle \sigma\) fixes all of \(\displaystyle K\).

This means that if \(\displaystyle \sigma\) is the identity, then \(\displaystyle \sigma(\alpha_n) = \alpha_n\) for all \(\displaystyle n\).

 

[logistic_guy]

\(\displaystyle \bold{(b)}\)

Suppose \(\displaystyle E\) is fixed by every \(\displaystyle \sigma \in G\).

Then in particular, each generator \(\displaystyle \sigma_i \in G\) fixes every \(\displaystyle x \in E\), so

\(\displaystyle \sigma_i(x) = x, \quad \forall x \in E, \quad \forall i=1, \ldots, n.\)

Suppose each generator \(\displaystyle \sigma_i\) fixes \(\displaystyle E\), i.e.,

\(\displaystyle \sigma_i(x) = x, \quad \forall x \in E, \quad \forall i=1, \ldots, n.\)

Since \(\displaystyle G = \langle \sigma_1, \ldots, \sigma_n \rangle\), any element \(\displaystyle \tau \in G\) can be expressed as a finite product

\(\displaystyle \tau = \tau_1 \tau_2 \cdots \tau_k,\)

where each \(\displaystyle \tau_j\) equals some generator \(\displaystyle \sigma_i\).

For any \(\displaystyle x \in E\), we have

\(\displaystyle \tau(x) = (\tau_1 \tau_2 \cdots \tau_k)(x) = (\tau_1 \tau_2 \cdots \tau_{k-1})(\tau_k(x)).\)

Since \(\displaystyle \tau_k\) fixes \(\displaystyle x\), \(\displaystyle \tau_k(x) = x\). Applying this repeatedly yields

\(\displaystyle \tau(x) = x.\)

Thus, every element of \(\displaystyle G\) fixes every element of \(\displaystyle E\).

The subfield \(\displaystyle E\) is fixed by all elements of \(\displaystyle G\) if and only if it is fixed by the generators of \(\displaystyle G\).