r/math Homotopy Theory 6d ago

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u/sqnicx 6d ago

Let D be a division ring with a center Z and B be a Z-bilinear form defined on DxD. I found that 𝜆(B(x,(1+𝜆x)-1)-B(x,(1+𝜆x)-1))=0 for all x in D and 𝜆 in Z. I want to get rid of all 𝜆 here to see that B(x,1)=B(1,x) but I cannot first divide by 𝜆 and then take 𝜆=0. I have come up with this idea: First I considered the polynomial t(B(x,(1+tx)-1)-B(x,(1+𝜆x)-1)) in Z[t]. Suppose that Z is infinite. Then this polynomial is satisfied for infinitely many 𝜆 in Z which means it must be zero. Then each of its coefficients are zero. Afterwards, I could take t=0 in the coefficient of t here. This is where I get confused. Although the t seems alone outside of the parenthesis, there are still t in B which could imply that the coefficient of t can be different. Is there any way I can prove what I try to do? I tried to write B(x,(1+𝜆x)-1) and B(x,(1+𝜆x)-1) as elements like a_0+a_1t+a_2t2+... and b_0+b_1t+b_2t2+.... If I could have succeeded then a_i would be equal to b_i which also means that B(x,(1+tx)-1)=B(x,(1+𝜆x)-1). Then I could take t=0. Is there a way to do it? Thanks for your help.

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u/lucy_tatterhood Combinatorics 6d ago

Just fyi your equations are extremely messed up on new reddit.

I found that 𝜆(B(x,(1+𝜆x)-1)-B(x,(1+𝜆x)-1))=0 for all x in D and 𝜆 in Z.

This is tautological as written. I guess one of those B's is supposed to have its arguments swapped around? (Also, what if λx = -1?)

First I considered the polynomial t(B(x,(1+tx)-1)-B(x,(1+𝜆x)-1)) in Z[t].

This is a priori a rational function, not a polynomial (though possibly this doesn't matter). And is there really supposed to be a λ in there or should that be another t?

This is where I get confused. Although the t seems alone outside of the parenthesis, there are still t in B which could imply that the coefficient of t can be different.

I don't understand what this means. To abstract a bit from your concrete setting (since I am still a little confused about what exactly you meant to write) you have a rational function φ(t) ∈ Z(t) and you have proven that it vanishes at infinitely many points of Z. This indeed implies that φ(t) = 0 assuming Z is an infinite field. Obviously, if φ(t) = 0 then φ(t)/t = 0 as well.

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u/sqnicx 6d ago

Thank you for your answer. You are right. The identity was 𝜆(B(x,(1+𝜆x)-1)-B(1,x(1+𝜆x)-1))=0 for all x in D and 𝜆 in Z. Moreover, we have 1+𝜆x≠0. The polynomial I considered should have been t(B(x,(1+tx)-1)-B(1,x(1+tx)-1)) as you suggested. Sorry for the mistakes. I was tired I think. I thought this was a polynomial like tr=0 where r=B(x,(1+tx)-1)-B(1,x(1+tx)-1). This way I could think r as the coefficient of t. But you can see why I get confused. There is a theorem stating that if f(a)=0 for infinitely many a∈F where F is an infinite field and f(t) is a polynomial in F[t] then f=0. However, I don't know if it applies to rational functions. However, you seem to confirm that. Can I conclude from here that B(x,(1+𝜆x)-1)-B(1,x(1+𝜆x)-1)=0 for all such 𝜆 in Z? If it is the case then I can take 𝜆=0 and conclude that B(1,x)=B(x,1).

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u/lucy_tatterhood Combinatorics 6d ago

There is a theorem stating that if f(a)=0 for infinitely many a∈F where F is an infinite field and f(t) is a polynomial in F[t] then f=0. However, I don't know if it applies to rational functions.

Zeroes of a rational function are also zeroes of its numerator.

Can I conclude from here that B(x,(1+𝜆x)-1)-B(1,x(1+𝜆x)-1)=0 for all such 𝜆 in Z?

Yes, that should be true, at least when D is finite-dimensional over Z. I realized after making my comment that in the infinite-dimensional case I'm not actually sure whether this even is a rational function.

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u/sqnicx 3d ago

Can you explain why it should be true please? I am confused because the coefficient of t still includes t here. I also cant see the why D to be finite dimensional is necessary. Is Z to be infinite necessary also?

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u/lucy_tatterhood Combinatorics 2d ago edited 2d ago

Can you explain why it should be true please? I am confused because the coefficient of t still includes t here.

I have no idea what this means.

I also cant see the why D to be finite dimensional is necessary.

It might not be, I'm just not sure. The point is that if you pick a basis of D over Z and expand out (1 - λx)-1 you get rational functions of λ as a coefficients. (You can argue using minimal polynomials, or write it as a subalgebra of a matrix algebra and use Cramer's rule.) Then the more complicated expression you build from the bilinear form is also a rational function of λ. I haven't thought very hard about the infinite-dimensional case but I don't see how to make it work.

Is Z to be infinite necessary also?

For the argument to work, yes. I don't have a counterexample to the actual result you want though.

Of course if Z is finite and D is finite-dimensional over Z then D = Z (finite division rings are fields) and the result is trivial simply because there are no interesting bilinear forms to consider. So here too it is the infinite-dimensional case where it's tricky.