$k[x_1,.\dots,x_n]$ is a free module over the ring of symmetric polyomials

Question

I know that this question has been already discussed (e.g here: shorturl.at/coFQX) but I dont understand the proofs given there, and I found a different one.

I want to show that $R[x_1,\dots,x_n]$ is free over $R[\sigma_1,\dots,\sigma_n]$ (with $\sigma_i$ is the elementary symmetric polynommial). I found the following proof:

We do induction over $n$. For $n=1$ there is nothing to show. Suppose this is true for $n-1$ variables. By induction hypothesis we know that $R[x_1,\dots,x_n]$ is free over $R[\sigma_1,\dots,\sigma_n,x_n]$. We just have to show that $R[\sigma_1,\dots,\sigma_n,x_n]$ is free over $R[\sigma_1,\dots,\sigma_n]$. The claim is now that $(x_n^0,\dots,x_n^{n-1})$ is a basis for this. I understand that this generates $x_n$ (hence $R[\sigma_1,\dots,\sigma_n,x_n]$).

Im Struggeling trying to show that this are linearly independent. So let $\lambda_i \in R[\sigma_1,\dots,\sigma_n]$ and: $$ \lambda_0 +\lambda_1 z^1+ \dots + \lambda_{n-1}z^{n-1}$$ be a polynomial with $x_n$ as root. The claim which I think is right (because it seems similiar to other similar contexts) would be that then also $x_1,\dots,x_{n-1}$ have to be roots and so this polynomial can't have degree $n-1$. But Im not able to show this. Is this the right approach and how can I show this?

Answer 1

For the linear independence point: this is supposed to be well-known if $R$ is a field so I'll assume that $R$ is an arbitrary ring.

Letting $\sigma_{n,i}$ be the $X^i$ coefficient of $\prod_{j=1}^n (X-x_j)$.

If $\sum_{l=0}^{n-1} c_lx_n^l=0$ with $c_l\in R[\sigma_{n,0},\dots,\sigma_{n,n-1}]$ not all zero,

the $c_l$ are also $n$-variables polynomials $\in R[x_1,\ldots,x_n]$, let $A\subset R$ be the finite set made from all their coefficients,

$A$ generates a finitely generated $\Bbb{Z}$-module $M\subset (R,+)$, there exists some prime number $p$ such that $M/pM\ne 0$,

so there exists a non-zero $\Bbb{Z}$-module homomorphism $\phi:M\to M/pM\to \Bbb{F}_p$.

Let $c_l^{\phi}$ mean applying $\phi$ to the coefficients of the polynomials $c_l\in M[x_1,\ldots,x_n] $,

so that $c_l^\phi \in \Bbb{F}_p[x_1,\ldots,x_n]$.

You'll get that the $c_l^\phi$ belong to $\Bbb{F}_p[\sigma_{n,0},\dots,\sigma_{n,n-1}]$ and that they are not all zero,

so that $\sum_{l=0}^{n-1} c_l^\phi x_n^l=0$ will contradict that the $\Bbb{F}_p(\sigma_{n,0},\dots,\sigma_{n,n-1})$-minimal polynomial of $x_n$ is $\prod_{j=1}^n (X-x_j)=X^n+\sum_{i=0}^{n-1} \sigma_{n,i} X^i$ which has degree $n$.