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

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?

• 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 $R[\sigma_1,\dots,\sigma_n,x_n]$? Feb 17 at 16:07
• 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). Feb 17 at 17:12

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 $$n$$.

• 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? Feb 18 at 10:07
• 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 Feb 18 at 11:19
• Ah now I understand a bit more the idea. I still not understand why $c_l^{\phi} \in F_p[\sigma_{n,0},\dots,\sigma_{n,n-1}].$ Feb 18 at 12:24
• 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! Feb 18 at 12:27
• and sorry one last thing about your proof, why does a morphism $M/pM \to F_P$ exist which is not trivial? Feb 18 at 12:57