Saturday, February 20, 2016

proof verification - Brauer characters: independence over overlinemathbbQ implies independence over mathbbC




The proof of theorem in title has already been sketched in a question on MathStack (here); I tried to write the detailed proof, and I want to check it.



Let ϕ1,,ϕn be all the Brauer characters of a finite group G (so they have domain Gp={gG:po(g)}, and values in OC, the ring of algebraic integers; p is the characteristic of field F over which the inequivalent irreducible F-representations of G are defined with above Brauer characters.



Claim: Suppose ϕ1,,ϕn are independent over ¯Q; then they are independent over C.



Proof: (1) First if |Gp|=m, then m should be n, otherwise ϕi's can-not be independent.



(2) Let Gp={g1,g2,,gm}. For each gi we associate a row-vector
[ϕ1(gi),ϕ2(gi),,ϕn(gi)]¯Qn.

We get m vectors {vg1,vg2,,vgm} in ¯Qn, and mn.



(3) The ¯Q-independence of ϕi's is equivalent to say that these m rows in ¯Qn are independent over ¯Q. Extend this set to a basis of ¯Qn:
{vg1,vg2,,vgm,wm+1,,wn}.
(4) The matrix P formed by these n vectors as rows of a matrix Mn(¯Q) will be invertible.



(5) Hence the matrix P as an element of Mn(C) will be invertible.



(6) Hence ϕ1,,ϕn as functions into C are C-independent.




Is this proof correct?


Answer



Your proof looks fine to me. One thing I'll point out is that this really has nothing to do with the fact that you are dealing with Brauer characters.



You have some vectors v1,v2,...,vn in a finite dimensional vector space V (here class functions on Gp) over one field k (in this case ˉQ) that are linear independent and you want to make sure they remain linearly independent when you extend scalars to a field extension K (C in this case). More precisely you want to know that v11,v21,... are linearly independent inside VK. Your proof works fine in this level of generality.



I'll note that in fact V need not be finite dimensional for this result to hold, your proof would need to be modified a bit for the infinite dimensional case (as you complete to a basis and show a certain matrix is invertible) but it's not that hard to fix.


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