Exercise 9.1.6

Proof.

(a)

The algorithm in the proof of Theorem 2.2.2 consists to replace the symmetric polynomial $f$, here with coefficients in $\mathbb{Z}$, by $f_1=f-\mathit{cg},f_2=f-\mathit{cg}-c_1{g}_1,\cdots$, until we obtain 0. The coefficient $c$ is the leading coefficient of $f$, so it is an integer, and $g={\sigma }_1^{a_1-a_2}\cdots {\sigma }_{n-1}^{a_{n-1}-a_n}{\sigma }_n^{a_n}\in \mathbb{Z}[{\sigma }_1,\dots ,{\sigma }_n]$, so $f_1\in \mathbb{Z}[x_1,\dots ,x_n]$. The same reasoning applied to $f_1$ and to the following terms shows that $c_i\in \mathbb{Z}$ for all $i$. Therefore $$f=\mathit{cg}+c_1{g}_1+\cdots +c_{m-1}{g}_{m-1}\in \mathbb{Z}[{\sigma }_1,\cdots ,{\sigma }_n].$$

In particular, the symmetric polynomial ${\sigma }_i(x_1^p,\cdots ,x_r^p)-{\sigma }_i{(x_1,\cdots ,x_r)}^p$ is a polynomial in ${\sigma }_1,\dots ,{\sigma }_r$ with integer coefficients:

$${\sigma }_i(x_1^p,\cdots ,x_r^p)-{\sigma }_i{(x_1,\cdots ,x_r)}^p=S({\sigma }_1,\cdots ,{\sigma }_r)\in \mathbb{Z}[{\sigma }_1,\cdots ,{\sigma }_r].$$

(b)

Let $h\in {𝔽}_p[x_1,\dots ,x_n]$. Write

$$h=\underset{(i_1,\dots ,i_n)\in A}{\sum }{a}_{i_1,\dots ,i_n}{x}_1^{i_1}\cdots x_n^{i_n},$$

where $A\subset {\mathbb{N}}^n$ is finite, and the coefficients $a_{i_1,\dots ,i_n}\in {𝔽}_p$. As the characteristic of the field ${𝔽}_p(x_1,\cdots ,x_r)$ is $p$, using the Little Fermat’s Theorem: $a^p=a$ for all $a\in {𝔽}_p$,

$$\begin{array}{llll}\hfill f{(x_1,\dots ,x_n)}^p& ={\left(\underset{(i_1,\dots ,i_n)\in A}{\sum }{a}_{i_1,\dots ,i_n}{x}_1^{i_1}\cdots x_n^{i_n}\right)}^p& \hfill & \\ \hfill & =\underset{(i_1,\dots ,i_n)\in A}{\sum }{a}_{i_1,\dots ,i_n}^p{x}_1^{p{i}_1}\cdots x_n^{p{i}_n}& \hfill & \\ \hfill & =\underset{(i_1,\dots ,i_n)\in A}{\sum }{a}_{i_1,\dots ,i_n}{x}_1^{p{i}_1}\cdots x_n^{p{i}_n}& \hfill & \\ \hfill & =f(x_1^p,\dots ,x_n^p)& \hfill & \end{array}$$

In particular, write ${\bar{\sigma }}_i$ the projection of ${\sigma }_i$ in ${𝔽}_p[x_1,\cdots ,x_r]$, and $\bar{S}$ the projection of $S$. As the characteristic of the field ${𝔽}_p(x_1,\cdots ,x_r)$ is $p$,

$$\begin{array}{llll}\hfill {\bar{\sigma }}_i{(x_1,\cdots ,x_r)}^p& ={\left(\underset{1\le j_1<j_2<\cdots <j_i\le r}{\sum }{x}_{j_1}\cdots x_{j_i}\right)}^p& \hfill & \\ \hfill & =\underset{1\le j_1<j_2<\cdots <j_i\le r}{\sum }{x}_{j_1}^p\cdots x_{j_i}^p& \hfill & \\ \hfill & ={\bar{\sigma }}_i(x_1^p,\cdots ,x_r^p)& \hfill & \end{array}$$ Hence $\bar{S}({\bar{\sigma }}_1,\dots ,{\bar{\sigma }}_r)={\bar{\sigma }}_i(x_1^p,\cdots ,x_r^p)-{\bar{\sigma }}_i^p=0$. Since ${\bar{\sigma }}_1,\dots ,{\bar{\sigma }}_r$ are algebraically independant over ${𝔽}_p$ (see Ex. 2.2.5), $\bar{S}=0$, so $S\equiv 0(\mathit{mod}p)$. Therefore $p$ divides all the coefficients of $S$.