Exercise 13.4.2

Proof.

Let $f(x)=x^n-c_1{x}^{n-1}+...+{(-1)}^n{c}_n\in K[x]$. The universal version of $s_u(y)$ is given by :

$$S(y)=\underset{\sigma \in S_n}{\prod }(y-(u_1{x}_{\sigma (1)}+...+u_n{x}_{\sigma (n)}))\in K[x_1,...,x_n,u_1,...,u_n,y]$$ For any $\tau \in S_n$, $$\begin{array}{llll}\hfill \tau \cdot S(y)& =\underset{\sigma \in S_n}{\prod }(y-(u_1{x}_{(\mathit{\tau \sigma })(1)}+...+u_n{x}_{(\mathit{\tau \sigma })(n)}))& \hfill & \\ \hfill & =\underset{{\sigma }^{\prime }\in S_n}{\prod }(y-(u_1{x}_{{\sigma }^{\prime }(1)}+...+u_n{x}_{{\sigma }^{\prime }(n)}))=S(y).& \hfill & \end{array}$$ Thus the application of $\tau$ has merely permuted the roots of $S(y)$ leaving the coefficients fixed. It means that the coefficients of $S(y)$ are symmetric and are polynomials in ${\sigma }_1,...,{\sigma }_n$ (cf. Ex.9.1.6), i.e., $S(y)\in K[{\sigma }_1,...,{\sigma }_n,u_1,...,u_n,y]$. The application of evaluation map ${\sigma }_i\mapsto c_i$ to $S(y)$ gives $s_u(y)\in K[c_1,...,c_n,u_1,...,u_n,y]=K[u_1,...,u_n,y]$. □