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Let $K[X_1,...,X_n]$ polynomial ring in $n$ variables and denote by $S_1, S_2,..., S_n$ the elementary symmetric polynomials:

$S_1= \sum X_i$

$ S_2 = \sum_{1 \le j < k \le n}X_j X_k$

...

$S_n= X_1 X_2 ... X_n$

The question is why is the polynomial ring $K[X_1,...,X_n]$ finite module over $K[S_1,...,S_n]$?

Guess, that clever induction on number of indeterminants does the job although I not see how I can perform the induction step. The start with $n=1$ is trivial. Assume the claim is true for $n-1$ and we want it for $n$. Any hint?

Beside nice answer using Vieta's formulas I would like to know if is also possible to go ahead inductively as follows:

it's obvious that $K[X_1,...,X_n]=K[X_2,..., X_n][S_1]$ holds. By IH $K[X_2,..., X_n]$ is finite over $K[S_2,..., S_n]$. The most important question is if tensoring preserves finiteness. If yes, then $K[X_2,..., X_n] \otimes K[S_1]=K[X_2,..., X_n][S_1]=K[X_1,X_2,..., X_n]$ is finite over $K[S_2,..., S_n]\otimes K[S_1]= K[S_1, S_2,..., S_n]$.

Is my "proof" correct?

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    $\begingroup$ Vieta's formulas: $P(X)=X^n-S_1X^{n-1}+S_2X^{n-2}+...+(-1)^nS_n$ has roots $X_1, X_2,...,X_n$. This means that $X_k^m$, for $m>n-1$, can be expressed as a $K[S_1,...,S_n]$-combination of $1,X_k,X_k^2,...,X_k^{n-1}$. So, you can generate all of $K[X_1,...,X_n]$ with $K[S_1,...,S_n]$-linear combinations of the finite set $\{X_k^r:\ k=1,2,...,n, r = 0,1,...,n-1\}$. $\endgroup$ – user752802 Mar 4 '20 at 21:27
  • $\begingroup$ @flan: Yes, this argument is smart. Although this definitely answers my question I'm still keen curious how one could go ahead if we assume that we not know the used Vieta's formulas and try to more conceptually using induction. Do you have an idea? $\endgroup$ – user7391733 Mar 4 '20 at 22:22
  • $\begingroup$ Tensoring preserves being finitely generated: if $a_i$ generate $A$ and $b_j$ generate $B$, then $a_i\otimes b_j$ generate $A\otimes B$. However, your induction step is not clear. $\endgroup$ – Berci Mar 5 '20 at 0:23
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    $\begingroup$ Vieta's formulas are proven by induction. An argument structured as a proof by induction is just going to obtain the $\forall n$ at the end instead of at the beginning. What is of substance is the existence of the polynomal relation $P(X_k)=0$ and how it is interpreted as a $K[S_1,...,S_n]$-linear relation. The rest is only presentation. $\endgroup$ – user752802 Mar 5 '20 at 1:47
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    $\begingroup$ More generally, if a finite group $G$ acts on a ring $R$, it is integral over $R^G$. Notice then that $K[S_1,\ldots,S_n]=K[X-1,\ldots, X_n]^G$ where $G$ is the symmetric group. $\endgroup$ – Mohan Mar 5 '20 at 2:48

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