Exercise 6.4.3

Proof.

(a)

Let $\tau \in S_n$. As $f\mapsto \tau .f$ is the evaluation map that evaluates $f(x_1,\dots ,x_n)$ at $(x_{\tau (1)},\dots ,x_{\tau (n)})$, Theorem 2.1.2 shows that this application is a ring homomorphism, thus $$\begin{array}{llll}\hfill \tau \cdot (f+g)& =\tau \cdot f+\tau \cdot g& \hfill & \\ \hfill \tau \cdot (\mathit{fg})& =(\tau \cdot f)(\tau \cdot g)& \hfill & \end{array}$$

(b)

Let $f=f(x_1,\cdots ,x_n)\in F(x_1,\cdots ,x_n)$, and $\tau ,\gamma \in S_n$. Define $g$ by

$$g(x_1,\cdots ,x_n)=\gamma \cdot f=f(x_{\gamma (1)},\cdots ,x_{\gamma (n)}).$$

Then $\tau \cdot (\gamma \cdot f)=\tau \cdot g=g(x_{\tau (1)},\dots ,x_{\tau (n)})$ is obtained by substituting each $x_i$ in the expression of $g$ by $x_{\tau (i)}$, thus $x_{\gamma (j)}$ becomes $x_{\tau (\gamma (j))}=x_{(\mathit{\tau \gamma })(j)}$:

$$\tau \cdot (\gamma \cdot f)=f(x_{(\mathit{\tau \gamma })(1)},\dots ,x_{(\mathit{\tau \gamma })(n)})=(\mathit{\tau \gamma })\cdot f.$$

Conclusion:

$$\tau \cdot (\gamma \cdot f)=(\mathit{\tau \gamma })\cdot f.$$