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

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.

• Is your question : "Is ... an already known field" ? Aug 12, 2016 at 8:38
• 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. Aug 12, 2016 at 10:09
• $\mathbb Q(x,y)=\mathbb Q(\frac xy,y)=\mathbb Q(\frac xy)(y)=\mathbb Q(i)(y)$ since $(\frac xy)^2+1=0$ Aug 13, 2016 at 8:54
• 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 ? Nov 25, 2016 at 20:56
• 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 ! Nov 25, 2016 at 21:54

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 $\mathbb Q(i)(y)$.