Finding $f= g(s_1,s_2,s_3)$ if $char(K) = p$

Question

Let $f$ be a symmetric polynomial in $K[x_1,x_2,x_3]$, where $K$ is a field and $f = x_1^5+x_2^5+x_3^5$.

I need to find g in $K[x_1,x_2,x_3]$ such that $f= g(s_1,s_2,s_3) $, where $ s_1,s_2,s_3$ are the elementary symmetric polynomials.

My attempt : I used Newton’s identities and obtained :

$g = s_1^5+5s_1^2s_3-5s_1^3s_2+5s_1s_2^2-5s_2s_3$.

I checked and this is really equal to $f$. But my doubt is that since $K$ is a field, we have $char(K)=0$ or $ char (K) = p$ where $p$ is a prime number. If $char(K) = p$ what changes in $g$ ? I think that I have to write $ 5$ $mod(p)$ on the coefficients but I am not sure that this is the only thing ...

Answer 1

In general, if you have a polynomial identity with integer coefficients, it is also an identity in any ring $R$ via the unique map $\phi_R : \mathbf{Z} \to R$. This map extends uniquely to $\Phi_R^{(n)} : \mathbf{Z}[x_1,\dots,x_n] \to R[x_1,\dots,x_n]$ via extension of scalars: $R[x_1,\dots,x_n] \cong R \otimes_{\mathbf Z} \mathbf{Z}[x_1,\dots,x_n]$.

Now if I have a polynomial identity like $$f = s_1^5+5s_1^2s_3-5s_1^3s_2+5s_1s_2^2-5s_2s_3,$$ then it must also be true that $$\Phi_R^{(3)}(g) = \Phi_R^{(3)}(s_1^5+5s_1^2s_3-5s_1^3s_2+5s_1s_2^2-5s_2s_3)$$ and this gives you the same identity in the ring $R$. This is a general fact:

A polynomial identity which is true with integer coefficients, is true in every ring.

For example, if $R = \mathbf{Z}/5\mathbf{Z}$, then we have

$$ f = s_1^5,$$

which says

$$ x_1^5 + x_2^5 + x_3^5 = (x_1 + x_2 + x_3)^5$$

(and I of course mean the image of all these polynomials in $\mathbf{Z}/5\mathbf{Z}$).