# How to show $\mathbb{Q}[x, y]/(x^2 - 2, xy - 2) \cong \mathbb{Q}(\sqrt{2})$?

I am embarrassingly stuck on trying to show these two rings are isomorphic. This isn't a homework assignment, it's just something I saw mentioned with no proof in some notes I was reading, and i'm struggling to prove it.

Of course $\mathbb{Q}[x]/(x^2 - 2) \cong \mathbb{Q}(\sqrt{2})$, and since I already know the solution i'm guessing there is some argument in between to show that $\mathbb{Q}[x, y]/(x^2 - 2, xy - 2) \cong \mathbb{Q}[x]/(x^2 - 2)$. Thank you.

From $\dots / (x^2-2)$, you already have $x = \sqrt{2}$ (or $-\sqrt{2}$; pick one). Then, from $x y = 2$, you get $y = \sqrt{2}$ (or $-\sqrt{2}$, if you picked the other one), so, in fact $y=x$ and $\Bbb{Q}[x,y] = \Bbb{Q}[x]$.
It is, perhaps, better to say \begin{align} xy = 2 &\implies x^2 y = 2x \\ &\implies 2 y = 2x \\ &\implies y = x \text{,} \end{align} meticulously detailing the resolution involved. But, really, that borders on pedantry.
Argue you can replace $Y$ by $Y/2$, and hence you are taking the quotient by $(X^2-2,XY-1)$. The $XY-1$ has the effect of adding a multiplicative inverse to $X$. But in $\Bbb Q[X]/(X^2-2)$, $X$ already has a multiplicative inverse, namely $X/2$. The idea is that if $R$ is a commutative ring, then for $a\in R$, $R[X]/(aX-1)$ is the ring obtained by formally adjoining a multiplicative inverse to $a$. If $a$ is invertible, then $$R[X]/(aX-1)=R[X]/(X-a^{-1})$$ is canonically isomorphic to $R$ via the evaluation at $a^{-1}$. In your case, $\Bbb Q[\sqrt 2]$ is already a field, in particular $\sqrt 2$ is invertible and hence the above holds.