Exercise 1.2.2 (Galois Group as a Subgroup)

Proof. Let $f$ be a polynomial with coefficients in $\mathbb{Q}$. Write ${\alpha }_1,\dots ,{\alpha }_k$ for its distinct roots in $ℂ$. By definition,

so $\mathrm{Gal}(f)\subset S_k$.

$\text{∙}$ $\mathrm{Id}\in \mathrm{Gal}(f)$, so $\mathrm{Gal}(f)\ne \varnothing$.

$\text{∙}$ Let $\sigma \in \mathrm{Gal}(f)$. For every polynomial $p\in \mathbb{Q}[t_1,\dots ,t_k]$,

Let $q\in \mathbb{Q}[t_1,\dots ,t_k]$ be any polynomial. Define $p(t_1,\dots ,t_k)=q(t_{{\sigma }^{-1}(1)},\dots ,t_{{\sigma }^{-1}(k)})$. Then $p({\alpha }_1,\dots ,{\alpha }_k)=q({\alpha }_{{\sigma }_{-1}(1)},\dots ,{\alpha }_{{\sigma }_{-1}(k)})$, and

(If we write ${\beta }_i={\alpha }_{{\sigma }^{\text{(}}i)}$ for $i\in \{1,\dots ,k\}$, then ${\beta }_{{\sigma }^{-1}(j)}={\alpha }_{{\sigma }^{\text{(}}{\sigma }^{-1}(j))}={\alpha }_j$ for $j\in \{1,\dots ,n\}$. Thus $p({\alpha }_{\sigma (1)},\dots ,{\alpha }_{\sigma (k)})=p({\beta }_1,\dots ,{\beta }_k)=q({\beta }_{{\sigma }^{-1}(1)},\dots ,{\beta }_{{\sigma }^{-1}(k)})=q({\alpha }_1,\dots ,{\alpha }_k)$)

Therefore the equivalence (1) applied to the polynomial $p(t_1,\dots ,t_k)=q(t_{{\sigma }^{-1}(1)},\dots ,t_{{\sigma }^{-1}(k)})$ gives

Since this is true for every $q\in \mathbb{Q}[t_1,\dots ,t_k]$, this proves that ${\sigma }^{-1}\in \mathrm{Gal}(f)$.

$\text{∙}$ Let $\tau ,\sigma \in \mathrm{Gal}(f)$. Let $p\in \mathbb{Q}[t_1,\dots ,t_k]$ such that $p({\alpha }_1,\dots ,{\alpha }_k)=0$. Then

For every polynomial $q\in \mathbb{Q}[t_1,\dots ,t_k]$,

Consider the polynomial $q(t_1,\dots ,t_k)=p(t_{\sigma (1)},\dots ,t_{\sigma (k)})$. Then $q({\alpha }_1,\dots ,{\alpha }_k)=0$, thus $q({\alpha }_{\tau (1)},\dots ,{\alpha }_{\tau (k)})=0$, that is

(Put ${\beta }_i={\alpha }_{\tau (i)},1\le i\le k$. Then ${\beta }_{\sigma (j)}={\alpha }_{\tau (\sigma (j))},1\le j\le k$, and $q({\alpha }_{\tau (1)},\dots ,{\alpha }_{\tau (k)})=q({\beta }_1,\dots ,{\beta }_k)=p({\beta }_{\sigma (1)},\dots ,{\beta }_{\sigma (k)})=p({\alpha }_{\tau (\sigma (1))},\dots ,{\alpha }_{\tau (\sigma (k))})=p({\alpha }_{(\tau \circ \sigma )(1))},\dots ,{\alpha }_{(\tau \circ \sigma )(k))})$. We have proved that, for every $p\in \mathbb{Q}[t_1,\dots ,t_k]$,

To prove the converse, note that from the first part, ${\sigma }^{-1},{\tau }^{-1}$ are elements of $\mathrm{Gal}(f)$, so that we have also, by the same proof given in the second part, for every polynomial $q$,

Apply this to $q(t_1,\dots ,t_k)=p(t_{(\tau \circ \sigma )(1)},\dots ,t_{(\tau \circ \sigma )(k)})$. Then $q({\alpha }_1,\dots ,{\alpha }_k)=p({\alpha }_{(\tau \circ \sigma )(1)},\dots ,{\alpha }_{(\tau \circ \sigma )(k)})$, and

(Put ${\beta }_i={\alpha }_{({\sigma }^{-1}\circ {\tau }^{-1})(i)})$, then ${\beta }_{(\tau \circ \sigma )(j)}={\alpha }_j$, and $q({\alpha }_{({\sigma }^{-1}\circ {\tau }^{-1})(1)},\dots ,{\alpha }_{({\sigma }^{-1}\circ {\tau }^{-1})(k)})=q({\beta }_1,\dots ,{\beta }_k)=p({\beta }_{(\tau \circ \sigma )(1)},\dots ,{\beta }_{(\tau \circ \sigma )(k)}=p({\alpha }_1,\dots ,{\alpha }_k)$.)

Then (2), applied to $q$, gives

so that

This proves that $\tau \circ \sigma \in \mathrm{Gal}(f)$.

So $\mathrm{Gal}(f)$ is a subgroup of $S_k$. □