A nice generating set for the symmetric power of an algebra

I'm looking for a reference for the following fact.

Suppose $A$ is a finitely generated associative commutative unital algebra over an algebraically closed field of characteristic zero. Let $S^n(A)$ be the $n$-th symmetric power of $A$. That is, $S^n(A)$ is the subalgebra of $A^{\otimes n}$ fixed by the natural action of $S_n$ on this tensor product by permuting the factors. I would like to know a reference for the fact that $S^n(A)$ is generated by elements of the form

$(a \otimes 1 \otimes \cdots \otimes 1) + (1 \otimes a \otimes 1 \otimes \cdots \otimes 1) + \cdots + (1 \otimes \cdots \otimes 1 \otimes a)$

for $a$ in $A$.

Note that if $A = \mathbb{C}[t]$, then $S^n(A)$ is the ring of symmetric polynomials in $n$ variables and then the result follows from the fact that the power sum polynomials generate this ring.

I call the above a "fact" since I've seen it used in papers (without proof -- and I've contacted the authors and they don't know of a reference) and a colleague of mine says he could write out a proof. However, I can't help but think that there must be a reference for this in some text. But I'm unable to find one...

Alternatively, if someone knows of a very short proof, that would be the next best thing to a reference. I want to use this fact in a paper I'm writing and I'd like to avoid having to digress from the main thread of the exposition by including a long proof of this fact.

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Pedantic remark: Your definition of $S^n(A)$ does not coincide with the usual definition of the symmetric algebra when $k$ has positive characteristic. But this definition is obviously useful here because it inherits $S^n(A)$ with an algebra structure. – Martin Brandenburg Feb 18 '13 at 21:10
see my comment to MO 122287. – Abdelmalek Abdesselam Feb 19 '13 at 14:15
@Martin: Thanks. I meant to say that the field has characteristic zero. I've updated the question to include this assumption. – Alistair Savage Feb 19 '13 at 14:51
It's Lemma 4.56 (ii) in Etingof's "Introduction to representation theory" ( www-math.mit.edu/~etingof/replect.pdf ), though the proof of part (i) given there is not the easiest and relies on representation theory (unsurprisingly for a representation theory text). I believe this is one of those facts better proven than cited away for the sake of reader... – darij grinberg Feb 19 '13 at 16:22
@darij: Thanks. Since the paper I'm writing is a representation theory paper, this is quite a good reference for me. – Alistair Savage Feb 20 '13 at 14:02

1 Answer

Since your algebra is finitely generated, you only really need this result in the case when $A=\mathbb{C}[t_1,\ldots,t_n]$.

It seems that this result first appeared in

F. Junker, Über symmetrische Funktionen von mehreren Reihen von Veränderlichen, Math. Ann. 43 (1893), 225-270

(see here for a scanned version of it).

A much more modern reference (from which I know about Junker's paper) is

J. Dalbec, Multisymmetric functions, Beiträge zur Algebra und Geometrie, Vol. 40 (1999), No. 1, pp. 27-51.

(see here for an electronic version).

Edit: as Abdelmalek Abdesselam suggests, this result goes back to

L. Schläfli, Über die Resultante eines Systemes mehrerer algebraischer Gleichungen Ein Beitrag zur Theorie der Elimination. (A. d. IV Bd. der Denkschriften der math.-naturw. Classe der kais. Acad. d. Wiss. bes. abgedruckt), Wien 1852

(see here for a scanned version).

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it's even older than that, it is in the 1852 paper by Schlafli on resultants. – Abdelmalek Abdesselam Feb 19 '13 at 19:34
Thanks! My command of German is not enough to conclude that Schlafli's paper contains a proof. I shall add a reference in the body of the answer. – Vladimir Dotsenko Feb 19 '13 at 23:02
Thanks, Vladmir and Abdelmalek. Of course, I think one would need to include the proof that if the result is true for A then it is true for any homomorphic image of A, in order to deduce the result for finitely generated algebras. Luckily that isn't so hard (although not immediately obvious, I think). – Alistair Savage Feb 20 '13 at 14:09