Exercise 12.2.3

Proof. We know from Section 12.2.B the definition of the Galois resolvent

Let $j\in \text{[[}1,n\text{]]}$ be a fixed integer. We show that ${\alpha }_j\in F(V)$, where $V=V_e=t_1{\alpha }_1+\cdots +t_n{\alpha }_n$.

Let

If $p_{\sigma }={\prod <!--\; FUNCTION\; APPLICATION\; -->}_{\tau \ne \sigma }(y-V_{\tau })$, then for $\phi \in S_n$, $p_{\sigma }(V_{\phi })=0$ if $\phi \ne \sigma$, and ${\prod <!--\; FUNCTION\; APPLICATION\; -->}_{\tau \ne \sigma }(V_{\sigma }-V_{\tau })$ if $\phi =\sigma$. Therefore, for $\phi =e$,

Moreover $s^{\prime }(y)={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{\sigma \in S_n}{\prod }_{\tau \ne \sigma }(y-V_{\tau })$, so

Since the Galois resolvent $s(y)$ is separable, $s^{\prime }(V)={\prod <!--\; FUNCTION\; APPLICATION\; -->}_{\tau \ne e}(V-V_{\tau })\ne 0$, so

We know that $s(y),s^{\prime }(y)$ are in $F[y]$. It remains to prove that ${\psi }_j(y)\in F[y]$. We use $V(x_1,\dots ,x_n)=t_1{x}_1+\cdots +t_n{x}_n\in F[x_1,\dots ,x_n]$, so that

and

so that ${\psi }_j(y)={\mathrm{\Psi }}_j(y,{\alpha }_1,\dots ,{\alpha }_n)$. Then, for all $\phi \in S_n$,

Therefore the coefficients of ${\mathrm{\Psi }}_j$ lie in the field $F({\sigma }_1,\dots ,{\sigma }_n)$, and the evaluation $x_i\mapsto {\alpha }_i$ gives

Therefore ${\alpha }_j=\frac{{\psi }_j(V)}{s^{\prime }(V)}\in F(V),j=1,\dots ,n$, and $V\in F({\alpha }_1,\dots ,{\alpha }_n)$, so

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