Editing Invariant theory (section)

==Hilbert's theorems==

{{harvtxt|Hilbert|1890}}  proved that if ''V'' is a finite-dimensional representation of the complex algebraic group ''G'' = SL<sub>''n''</sub>(''C'') then the [[ring of invariants]] of ''G'' acting on the ring of polynomials ''R'' = ''S''(''V'') is finitely generated. His proof used the [[Reynolds operator]] ρ from ''R'' to ''R''<sup>''G''</sup> with the properties
*''ρ''(1) = 1
*''ρ''(''a'' + ''b'') = ''ρ''(''a'') + ''ρ''(''b'')
*''ρ''(''ab'') = ''a'' ''ρ''(''b'')  whenever ''a'' is an invariant.
Hilbert constructed the Reynolds operator explicitly using [[Cayley's omega process]] Ω, though now it is more common to construct ρ indirectly as follows: for compact groups ''G'', the Reynolds operator is given by taking the average over ''G'', and non-compact reductive groups can be reduced to the case of compact groups using Weyl's [[unitarian trick]].

Given the Reynolds operator, Hilbert's theorem is proved as follows. The ring ''R'' is a polynomial ring so is graded by degrees, and the ideal ''I'' is defined to be the ideal generated by the homogeneous invariants of positive degrees. By [[Hilbert's basis theorem]] the ideal ''I'' is finitely generated (as an ideal). Hence, ''I'' is finitely generated ''by finitely many invariants of G'' (because if we are given any – possibly infinite – subset ''S'' that generates a finitely generated ideal ''I'', then ''I'' is already generated by some finite subset of ''S''). Let ''i''<sub>1</sub>,...,''i''<sub>''n''</sub> be a finite set of invariants of ''G'' generating ''I'' (as an ideal). The key idea is to show that these generate the ring ''R''<sup>''G''</sup> of invariants. Suppose that ''x'' is some homogeneous invariant of degree ''d''&nbsp;&gt;&nbsp;0. Then
:''x'' = ''a''<sub>1</sub>''i''<sub>1</sub> + ... + ''a''<sub>n</sub>''i''<sub>n</sub>
for some ''a''<sub>''j''</sub> in the ring ''R'' because ''x'' is in the ideal ''I''. We can assume that ''a''<sub>''j''</sub> is homogeneous of degree ''d'' &minus; deg ''i''<sub>''j''</sub> for every ''j'' (otherwise, we replace ''a''<sub>''j''</sub> by its homogeneous component of degree ''d'' &minus; deg ''i''<sub>''j''</sub>; if we do this for every ''j'', the equation ''x'' = ''a''<sub>1</sub>''i''<sub>1</sub> + ... + ''a''<sub>''n''</sub>''i''<sub>n</sub> will remain valid). Now, applying the Reynolds operator to ''x'' = ''a''<sub>1</sub>''i''<sub>1</sub> + ... + ''a''<sub>''n''</sub>''i''<sub>n</sub> gives
:''x'' = &rho;(''a''<sub>1</sub>)''i''<sub>1</sub> + ... + ''&rho;''(''a''<sub>''n''</sub>)''i''<sub>''n''</sub>

We are now going to show that ''x'' lies in the ''R''-algebra generated by ''i''<sub>1</sub>,...,''i''<sub>''n''</sub>.

First, let us do this in the case when the elements ρ(''a''<sub>''k''</sub>) all have degree less than ''d''. In this case, they are all in the ''R''-algebra generated by ''i''<sub>1</sub>,...,''i''<sub>''n''</sub> (by our induction assumption). Therefore, ''x'' is also in this ''R''-algebra (since ''x'' = ''ρ''(''a''<sub>1</sub>)''i''<sub>1</sub> + ... + ρ(''a''<sub>n</sub>)''i''<sub>n</sub>).

In the general case, we cannot be sure that the elements ρ(''a''<sub>''k''</sub>) all have degree less than ''d''. But we can replace each ρ(''a''<sub>''k''</sub>) by its homogeneous component of degree ''d'' &minus; deg ''i''<sub>''j''</sub>. As a result, these modified ρ(''a''<sub>''k''</sub>) are still ''G''-invariants (because every homogeneous component of a ''G''-invariant is a ''G''-invariant) and have degree less than ''d'' (since deg ''i''<sub>''k''</sub> &gt; 0). The equation ''x'' = ρ(''a''<sub>1</sub>)''i''<sub>1</sub> + ... + ρ(''a''<sub>n</sub>)''i''<sub>n</sub> still holds for our modified ρ(''a''<sub>''k''</sub>), so we can again conclude that ''x'' lies in the ''R''-algebra generated by ''i''<sub>1</sub>,...,''i''<sub>''n''</sub>.

Hence, by induction on the degree, all elements of ''R''<sup>''G''</sup> are in the ''R''-algebra generated by ''i''<sub>1</sub>,...,''i''<sub>''n''</sub>.