# Basis for $\Bbb Z[x_1,\cdots,x_n]$ over $\Bbb Z[e_1,\cdots,e_n]$

I'm reading the introductory bits in Procesi's Lie Groups, and on p. 22 we have (paraphrasing)

Theorem 2. $\mathcal{B}=\{x_1^{\large h_1}\cdots x_n^{\large h_n}: 0\le h_k\le n-k\}$ is a basis for the ring $\Bbb Z[x_1,\cdots,x_n]$ considered over $\Bbb Z[e_1,\cdots,e_n]$, where $e_i$ are the elementary symmetric polynomials in the $x_i$.

I haven't been able to see why this is though. The previous theorem was the fundamental theorem of symmetric polynomials, which was proven inductively with a recursive algorithm:

If $x_n|f$ then $x_1\cdots x_n|f$, and dividing out we are left with a symmetric polynomial of smaller degree than before. Otherwise, write $f(x_1,\cdots,x_{n-1},0)$ as a polynomial $p$ in the elementary symmetric polynomials $\hat{e}_i$ of the first $n-1$ variables, $p(\hat{e}_1,\cdots,\hat{e}_{n-1})$. Now the polynomial $$f(x_1,\cdots,x_n)-p(e_1,\cdots,e_{n-1})$$ is symmetric in all of $x_1,\cdots,x_n$ and evaluates to $0$ at $x_n=0$ ie is divisible by $x_n$. Induct.

Is there a straightforward adaptation of this with which we can argue for theorem 2? Or is there perhaps another way to see that it must be true? I feel I am missing something simple here.

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Dear anon: You may look at Part 2 of Section G of Galois Theory: Lectures Delivered at the University of Notre Dame, by Emil Artin, freely and legally available here. – Pierre-Yves Gaillard Apr 16 '12 at 9:27
@Pierre-Yves Cool, thanks. – anon Apr 16 '12 at 9:41
Dear anon: In fact Artin proves the statement over $\mathbb Q$. Some extra work is needed to prove it over $\mathbb Z$. It is proved in Bourbaki (Alg. IV.6.5, Prop. 5). I wrote a short text about this. – Pierre-Yves Gaillard Apr 16 '12 at 12:04
The above link will expire soon. Here is the new link. – Pierre-Yves Gaillard Sep 8 '14 at 17:23

I suppose I should actually come back to answer this. Main idea: for $1\le d<n$ we have
$$\Bbb Z[x_1,\cdots,x_n]^{S_d}=\bigoplus_{j=0}^d \Bbb Z[x_1,\cdots,x_n]^{S_{d+1}}x_{d+1}^j. \tag{\circ}$$
They're both equal to $\Bbb Z[x_1,\cdots,x_n]^{S_{d+1}}[x_{d+1}]$. Using $(\circ)$ we can begin unpeeling $\Bbb Z[x_1,\cdots,x_n]$ by setting $d=1,2,\cdots,n-1$. What I found interesting about this argument is that the outer layers of the onion are the smaller indices rather than the larger - initially I believed it'd be the reverse.