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What is the field of fractions of $\mathbb{Q}[x,y]/(x^2+y^2)$?

Remarks:

(1) I think it is clear that $\mathbb{Q}[x,y]/(x^2+y^2)$ is an integral domain; indeed, $x^2+y^2 \in \mathbb{Q}[x,y]$ is irreducible (by considerations of degrees) hence prime.

(2) The field of fractions of $\mathbb{Q}[x,y]/(x^2+y^2-1)$ is isomorphic to $\mathbb{Q}(t)$, see this question and also this question.

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  • $\begingroup$ Is your question : "Is ... an already known field" ? $\endgroup$ – Jean Marie Aug 12 '16 at 8:38
  • $\begingroup$ Please, what is your definition of "an already known field"? I am guessing that the field of fractions is something like $R(t)$, where $R$ is an algebraic field extension of $\mathbb{Q}$; but I may be wrong. $\endgroup$ – user237522 Aug 12 '16 at 10:09
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    $\begingroup$ $\mathbb Q(x,y)=\mathbb Q(\frac xy,y)=\mathbb Q(\frac xy)(y)=\mathbb Q(i)(y)$ since $(\frac xy)^2+1=0$ $\endgroup$ – user26857 Aug 13 '16 at 8:54
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    $\begingroup$ Very nice comment, @user26857. The same idea proves that the normalization of $\mathbb{R}[x,y]/(x^2+y^2)$ is $\mathbb C|y]$. But this is very strange (to me): it says that the normalization of the real "conic" $\operatorname {Spec}\mathbb{R}[x,y]/(x^2+y^2) $ is the affine line $\mathbb A^1_\mathbb C=\operatorname {Spec}\mathbb{C}[y]$. Do I remember correctly that you already wrote something about this normalization ? $\endgroup$ – Georges Elencwajg Nov 25 '16 at 20:56
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    $\begingroup$ Dear @user26857, yes exactly: The normalization of $A=\mathbb R[X,Y]/(X^2+Y^2)$ is $\mathbb C[Y]$. I computed this long ago and I remember being quite amazed by the answer . Everything hangs together beautifully: the fraction field $\mathbb R(i)(Y)$, which you calculated, of $A$ is the same as that of its normalization $\overline A=\mathbb C[Y]$. By the way, I hope you will soon return to this site, so that we can again enjoy your beautifully crisp and elegant answers ! $\endgroup$ – Georges Elencwajg Nov 25 '16 at 21:54
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First do the substitution $x'=x/y$. Then the equation $x^2+y^2=0$ transforms to $x'^2+1=0$. Hence we are looking at the (isomorphic) field $\text{Frac}(\mathbb Q[x',y]/(x'^2+1))$.

This is just $\mathbb Q(i)(y)$.

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  • $\begingroup$ Very nice! Thank you very much user26857, Fredrik Meyer and arctic tern. $\endgroup$ – user237522 Aug 14 '16 at 19:44
  • $\begingroup$ Very nice: +1. ${}$ $\endgroup$ – Georges Elencwajg Nov 25 '16 at 21:00

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