The relation is on . It is not on the algebraic closure of , because as you noticed it would make the problem trivial.
Recall that a relation on a set is a subset of . An equivalence relation is a relation such that
(1)
(2)
(3)
Hi:
Let K be a field and, for a, b in K \ {0}, write a ~ b if ab is a sum of two squares in K. The author writes: why is ~ an equivalence relation?
One proof for transitivity is
ab = x^2 + y^2
bc = z^2 + w^2
acb^2 = (xz)^2 + (xw)^2 + (yz)^2 + (yw)^2 =
= [(xz + yw)^2 + (xw - yz)^2].
However, I see that for any a, b in K \ {0}, ab= ab + 0, where 0 is a square and ab is the square of an element in some extension of E of K and ab, 0 are in K. So a ~ b for every a, b in K \ {0}. That such an E exists amounts to saying that the polynomial x^2 - ab belonging to K[x] has a root in some extension of K, which is certainly true. What's wrong with this line of reasoning, if wrong at all?
Any hint will be welcome. Regards.
Then I'm a null entity at mathematics. Don't think I have comfortably recourse to MHF to solve the problem. I've tried hard at understanding the statement of the problem but it makes no sense to me. "ab is a sum of two squares in K": what does this mean? For me it is: "there exist x, y such that ab = x^2 + y^2, with x^2, y^2 in K". I understand the last statement doesn't make much sense, because x and y must exist somewhere. It is because of that I am forced to assumed an extension where to place x and y. So the last statement is left now in this form:
"there exist x,y in some extension of K such that ab = x^2 + y^2, with x^2 ,y^2 in K". But this makes the problem trivial. So, I do not see a way out.
Grateful for your reply.