prove that $(xy-2,x^2-2)$ is a maximal ideal of $\mathbb Q[x,y]$ I want to prove that $(xy-2,x^2-2)$ is a maximal ideal of $\mathbb Q[x,y]$.
To begin with, I constructed a map $\phi:\mathbb Q[x,y] \to \mathbb Q[\sqrt 2]$ by $f\mapsto f(\sqrt 2, \sqrt 2)$, hoping to show that its kernel is $(xy-2,x^2-2)$.
Clearly, $\phi$ kills $(xy-2,x^2-2)$. To show the other containment. We take $f$ vanishing at $(\sqrt 2, \sqrt 2)$. Here is where I get stuck. How to express $f$ as $f=u(xy-2)+v(x^2-2)$, where $u,v\in \mathbb Q[x,y]$?
Any other method will be appreciated if this one doesn't work. It would be great if there is a general method to deal with many other similar problems.
 A: The ideal $I$ contains $xy-2-(x^2-2)=x(y-x)$ and so also $x^2(y-x)$.
Then also it contains $x^2(y-x)-(x^2-2)(y-x)=2(y-x)$, and so $y-x$ too.
Therefore $f(x,y)\equiv f(x,x)\pmod{I}$ for any polynomial $f$ over $\Bbb Q$. If $f(\sqrt2,\sqrt2)=0$ then $g(x)=f(x,x)$ is divisible by $x^2-2$
so $g(x)\in I$. As $f(x,y)\equiv g(x)\pmod I$ then $f(x,y)\in I$ also.
A: We can find $\{ x-y,x^2-2 \}$ is a minimal Grobner basis for $I=(xy-2,x^2-2)$, so $$ I=(x-y,x^2-2) $$
then $ \mathbb Q[x,y]/(x-y,x^2-2) \cong \mathbb Q[x]/(x^2-2) \cong \mathbb Q(\sqrt2)  $
A: Lord Shark the Unknown's answer is very beautiful, but I also find it very clever. I thought I might post an answer which is decidedly less clever, but which uses very broadly applicable ideas. 
First, note that by the third isomorphism theorem for rings, we have $\mathbb{Q}[X, Y]/\langle XY-2, X^{2}-2 \rangle \cong (\mathbb{Q}[X, Y]/\langle XY-2)/\langle \overline{X^{2}-2} \rangle$, where $\overline{X^{2}-2}$ denotes the image of $X^{2}-2$ in $\mathbb{Q}[X, Y]/\langle XY-2\rangle$. I claim that $\mathbb{Q}[X, Y]/\langle XY-2\rangle \cong \mathbb{Q}[Z, Z^{-1}] \cong S^{-1}\mathbb{Q}[Z]$, where $S = \{Z^{n} \mid n \in \mathbb{N}\} \subset \mathbb{Q}[Z]$. 
Indeed, consider the morphism of $\mathbb{Q}$-algebras $\varphi \colon \mathbb{Q}[X, Y] \to \mathbb{Q}[Z, Z^{-1}]$ which is defined on generators by $X \mapsto Z, Y \mapsto 2Z^{-1}$. Then $XY-2 \in \ker(\varphi)$, so by the universal property of the quotient, $\varphi$ descends to a well-defined morphism of $\mathbb{Q}$-algebras $\tilde{\varphi} \colon \mathbb{Q}[X, Y]/\langle XY - 2 \rangle \to \mathbb{Q}[Z, Z^{-1}]$ sending $\overline{X}$ to $Z$ and $\overline{Y}$ to $2Z^{-1}$. 
On the other hand, we can define a morphism of $\mathbb{Q}$-algebras $\rho \colon \mathbb{Q}[Z] \to \mathbb{Q}[X, Y]/\langle XY - 2 \rangle$ by $Z \mapsto \overline{X}$. Since $\overline{X} \cdot \overline{\frac{1}{2}Y} = \overline{1}$, $\overline{X} \in (\mathbb{Q}[X, Y]/\langle XY - 2 \rangle)^{\times}$. In particular, by the universal property of localization, $\rho$ extends to a morphism $\tilde{\rho} \colon \mathbb{Q}[Z, Z^{-1}] \to \mathbb{Q}[X, Y]/\langle XY - 2 \rangle$ which sends $Z$ to $\overline{X}$ and $Z^{-1}$ to $\overline{\frac{1}{2}Y}$. It is straightforward to check that $\tilde{\rho} \circ \tilde{\varphi}$ and $\tilde{\varphi} \circ \tilde{\rho}$ are the identities on their respective domains; it is sufficient to check this on generators, e.g. that $(\tilde{\rho} \circ \tilde{\varphi})(\overline{X}) = \overline{X}$. Thus, via the isomorphism $\tilde{\varphi}$, we have
$$\mathbb{Q}[X, Y]/\langle XY-2, X^{2}-2 \rangle \cong (\mathbb{Q}[X, Y]/\langle XY-2)/\langle \overline{X^{2}-2} \rangle \cong \mathbb{Q}[Z, Z^{-1}]/\langle Z^{2}-2\rangle$$
It remains to show that $\mathbb{Q}[Z, Z^{-1}]/\langle Z^{2}-2\rangle \cong \mathbb{Q}[\sqrt{2}]$. There are many ways to do this, but one is the following: since taking quotients "commutes" with localization, we have $\mathbb{Q}[Z, Z^{-1}]/\langle Z^{2}-2\rangle \cong (\mathbb{Q}[Z]/\langle Z^{2}-2\rangle)_{\overline{Z}} \cong (\mathbb{Q}[\sqrt{2}])_{\sqrt{2}} \cong \mathbb{Q}[\sqrt{2}]$, where the last isomorphism follows since $\sqrt{2} \in (\mathbb{Q}[\sqrt{2}])^{\times}$.
