What is the maximal ideal of $\mathbb{Q}[x,y]$ corresponding to $(\sqrt{2},\sqrt{2})$? I was reading Ravi Vakil's notes on Algebraic Geometry and I encountered
this exercise:
Describe the maximal ideal of $\mathbb{Q}[x,y]$ corresponding to $(\sqrt{2},\sqrt{2})$
and $(-\sqrt{2},-\sqrt{2})$.
Describe the maximal ideal of $\mathbb{Q}[x,y]$ corresponding to $(\sqrt{2},-\sqrt{2})$
and $(-\sqrt{2},\sqrt{2})$.
Could someone give me some assistance with this question?
Thanks!
 A: I think the two ideals in question are $(x-y, x^2-2)$ and $(x+y, x^2-2)$.
They are both maximal ideals since the quotient is $\mathbb{Q}(\sqrt{2})$ , a field, and they are satisfied by the points.
A: The question is not very meaningful as stated: maximal ideals in $\mathbb Q[X,Y]$ are special sets of polynomials with coefficients in $\mathbb Q$ and do not "correspond" to  pairs of irrational numbers.
However here is a precise result in the requested vein:
Consider the inclusion of rings $R=\mathbb Q [X,Y]  \subset A=\mathbb Q(\sqrt 2)[X,Y]$
and the maximal ideals ${\frak m}=\langle X-\sqrt 2, Y-\sqrt 2\rangle,  {\frak m'}=\langle X+\sqrt 2, Y+\sqrt 2\rangle \subset A$.
They both have intersection with $R$: $${\frak m}\cap R={\frak m'}\cap R=\langle X^2-2,X-Y\rangle=: M \subset R$$  We then say that ${\frak m}, {\frak m'}\subset A$ lie over $M$ and the precise result is that ${\frak m}$, ${\frak m'}$ are the only maximal ideals in $A$ lying over $M$.
Similarly ${\frak n}=\langle X-\sqrt 2, Y+\sqrt 2\rangle \subset A$ and ${\frak n'}=\langle X+\sqrt 2, Y-\sqrt 2\rangle \subset A$ are the only maximal ideals in $A$ lying over the maximal ideal $N=\langle X^2-2, X+Y\rangle \subset R$.
Remark
The geometric picture is of course more satisfying: we have a canonical scheme morphism $$\pi:\operatorname {Spec}A=\mathbb A^2_{\mathbb Q(\sqrt 2)}\to \operatorname {Spec}R=\mathbb A^2_\mathbb Q$$ and we have studied two of its fibres:$$\pi^{-1}(M)=\{{\frak m}, {\frak m'}\},\pi^{-1}(N)=\{\frak n, \frak n'\}$$
