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?

  • $\begingroup$ Would be clearer with $\sigma_{i,n}$ the $i$-th coefficient of $\prod_{j=1}^n (X-x_j)$. Why is $R[x_1,\dots,x_n]$ free over $R[\sigma_1,\dots,\sigma_n,x_n]$? $\endgroup$
    – reuns
    Feb 17 at 16:07
  • $\begingroup$ ok I will write $\sigma_i = \sigma_{n,i}$ and $\sigma_i' = \sigma_{n-1,i}$. Than by induction Hypothesis we know that $R[x_1,\dots,x_{n-1}]$ is free over $R[\sigma_1',\dots,\sigma_{n-1}']$. From this we get directly that $R[x_1,\dots,x_{n-1},x_n]$ is free over $R[\sigma_1',\dots,\sigma_{n-1}',x_n]$. But with the identity $\sigma_i = x_n \sigma_{i-1}' + \sigma_i'$ we get that $R[\sigma_1',\dots,\sigma_{n-1}',x_n] = R[\sigma_1,\dots,\sigma_n,x_n]$ (just by checking that the generators are inside). $\endgroup$ Feb 17 at 17:12

1 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 $n$.

  • $\begingroup$ Im sorry but I dont understand the proof. First of all should $c_l$ not be in $R[\sigma_{n,0},\dots, \sigma_{n,n}]$? Why are you just going to the $n-1$th symmetric polynomial? When Why does $p$ exist such that $M/pM \not=0$ and how is $\phi$ working, and why are the images still polynomials in the elementary symmetric polynomilals and why does they preserve the identity at the end? $\endgroup$ Feb 18 at 10:07
  • $\begingroup$ I meant $\sigma_{n,i}$ the $i$-th coefficient of $\prod_{j=1}^n (X-x_j)$. Such a $p$ exists by the structure theorem for finitely generated $\Bbb{Z}$-modules. $\phi$ is any non-zero homomorphism $M/pM\to \Bbb{F}_p$ so it will be non-zero for some of the coefficients of the polynomials. It preserves the identity because it is a $\Bbb{Z}$-module homomorphism: $(r \sigma_{n,i})^\phi = \phi(r) \sigma_{n,i}$ @user1072285 $\endgroup$
    – reuns
    Feb 18 at 11:19
  • $\begingroup$ Ah now I understand a bit more the idea. I still not understand why $c_l^{\phi} \in F_p[\sigma_{n,0},\dots,\sigma_{n,n-1}].$ $\endgroup$ Feb 18 at 12:24
  • $\begingroup$ But thinking about your proof (which if I understand it right just tries to shift the problem into a field problem) I had the following Idea. Could we shorten your argument on the following way: if we have such a linearcombination like above $$\lambda_0 +\lambda_1 z^1+ \dots + \lambda_{n-1}z^{n-1}=0$$ than this is especcially a linear combination inside $Quot(R)(\sigma_1,\dots,\sigma_n)$. But like at the end of your prove the minimal polynomial of $x_n$ above this has degree $n$ so this is a contradiction. Does this work if at least R is a integral domain. Thank you for your help! $\endgroup$ Feb 18 at 12:27
  • $\begingroup$ and sorry one last thing about your proof, why does a morphism $M/pM \to F_P$ exist which is not trivial? $\endgroup$ Feb 18 at 12:57

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