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==Dickson invariant==
When ''G'' is the finite general linear group GL<sub>''n''</sub>('''F'''<sub>''q''</sub>) over the finite field '''F'''<sub>''q''</sub> of order a prime power ''q'' acting on the ring '''F'''<sub>''q''</sub>[''X''<sub>1</sub>, ... ,''X''<sub>''n''</sub>] in the natural way, {{harvtxt|Dickson|1911}} found a complete set of invariants as follows. Write [''e''<sub>1</sub>, ... ,''e''<sub>''n''</sub>] for the determinant of the matrix whose entries are ''X''{{su|b=''i''|p=''q''<sup>''e''<sub>''j''</sub></sup>}}, where ''e''<sub>1</sub>, ... ,''e''<sub>''n''</sub> are non-negative integers. For example, [0,1,2] is
:<math>\begin{vmatrix} x_1 & x_2 & x_3\\x_1^q & x_2^q & x_3^q\\x_1^{q^2} & x_2^{q^2} & x_3^{q^2} \end{vmatrix}</math>
Then under the action of an element ''g'' of GL<sub>''n''</sub>('''F'''<sub>''q''</sub>) these determinants are all multiplied by det(''g''), so they are all invariants of SL<sub>''n''</sub>('''F'''<sub>''p''</sub>) and the ratio [''e''<sub>1</sub>, ... ,''e''<sub>''n''</sub>]/[0,1,...,''n''−1] are invariants of GL<sub>''n''</sub>('''F'''<sub>''q''</sub>), called '''Dickson invariants'''. Dickson proved that the full ring of invariants '''F'''<sub>''q''</sub>[''X''<sub>1</sub>, ... ,''X''<sub>''n''</sub>]<sup>GL<sub>''n''</sub>('''F'''<sub>''q''</sub>)</sup> is a polynomial algebra over the ''n'' Dickson invariants [0,1,...,''i''−1,''i''+1,...,''n'']/[0,1,...,''n''−1] for ''i''=0, 1, ..., ''n''−1.
{{harvtxt|Steinberg|1987}} gave a shorter proof of Dickson's theorem.
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