[FOM] Hilbert's proof
colin.mclarty at case.edu
Tue Mar 9 12:56:39 EST 2010
> From: <addamo at wp.pl>
> To: "FOM" <fom at cs.nyu.edu>
> Date: Fri, 5 Mar 2010 00:38:31 +0100
> Subject: [FOM] hilbert's proof
> What was an essence of the "totally new method" by which Hilbert could
> prove his the existence of a finite basis for the invariant forms?
> with regards, adam
The proof was non-constructive, but that is not much of a description
of its method. And a lot of proofs in those days were
Actually, one difficulty at the time was that Hilbert's proof as
published in the Mathematische Annalen was wrong and the editors added
a footnote correcting it. But it did work with that correction.
The method is hard to isolate because Hilbert takes no care to show
exactly how general his result is. That is, he takes little care to
say exactly what his assumptions are. The "method" was never made
clear until Emmy Noether gave it a more advanced form 30 some years
later: any finitely generated algebra over a Noetherian ring is
I think it is fair to focus on two points apparent to readers when it
1) The proof was a pure finiteness result. It showed certain sets
are finite without showing anything more specific about them. This
is bread and butter today in many fields but was not easy to grasp at
2) The key step was to forget virtually everything about the problem.
Virtually the whole proof was contained in the first lemma which
apparently has nothing to do with algebras of invariants (of a given
form), and nothing to do with "forms" at all. From a modern point of
view we say it indeed has nothing to do with those but is pure
commutative algebra. Hilbert at the time merely neglected all the
particulars without saying how or why he was doing this.
Hilbert just points out a few banalities about addition and
multiplication, and this turns out to mean there is a finjte set B of
invariants (for a given form) such that any invariant can be written
as a sum of multiples for these with polynomial coefficients (today we
say B is a module base for all the invariants). And then a quick
averaging argument shows every invariant can be written as a
polynomial in the invariants in B (with coefficients in the field
which was never specified) (today we say B is an algebra base).
It was a stunning result and really not easy to grasp at the time --
largely *because* it was so incredibly simple.
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