Exercise 2.2.20

Proof. Suppose that ${\sigma }_2=f(s_1,s_2,\cdots ,s_n)$, where $f$ is a polynomial with coefficients in ${𝔽}_2$. If we use the evaluation defined by $x_1=x_2=\cdots =x_n=0$, we obtain $0=f(0,\cdots ,0)$.

With the evaluation defined by $x_1=x_2=1$ and $x_i=0,i>2$, as ${\sigma }_2={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i<j}{x}_i{x}_j$, then ${\sigma }_2(1,1,0,\dots ,0)=1\times 1=1$ and $s_k(1,1,0,\dots ,0)=1^k+1^k=1+1=0$, so $1=f(0,\cdots ,0)$. As $1\ne 0$ in ${𝔽}_2$, this is a contradiction. So it is impossible to express ${\sigma }_2$ as a polynomial in $s_1,\dots ,s_n$ when $n\ge 2$. □