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?