# Maximal ideal in $\mathbb{Q}[x,y]$

I am trying to prove that $(x,y)$ is a maximal ideal of $\mathbb{Q}[x,y]$. Since an ideal $I \subseteq R$ is maximal if and only if $R/I$ is a field, it suffices to prove that $\mathbb{Q}[x,y]/(x,y)$ is a field.
Let $\phi : \mathbb{Q}[x,y] \rightarrow \mathbb{Q}$ be the evaluation homomorphism that sends $p(x,y)$ to $p(0,0)$. $\phi$ is clearly surjective and if I prove that the kernel of $\phi$ is the ideal $(x,y)$, the result will follow by the First Isomorphism Theorem.

$(x,y) \subseteq Ker(\phi)$ is obvious, but to prove the reverse inclusion, my idea is to take a polynomial $p(x,y) \in Ker(\phi)$ and to perform Euclidean division of $p(x,y)$ with $y$ and $x$ in $(\mathbb{Q}[x])[y]$ and $(\mathbb{Q}[y])[x]$ respectively. The problem is that neither of them is a polynomial ring over field. There is a theorem saying that if R is commutative and if $R[x]$ is a PID, then $R$ is a field. Since neither $\mathbb{Q}[x]$, nor $\mathbb{Q}[y]$ are fields, it follows that neither $(\mathbb{Q}[x])[y]$, nor $(\mathbb{Q}[y])[x]$ are PIDs, hence are not Euclidean domains.

Obviously, Euclidean division does not solve the problem. If we consider $p(x,y)$ as an element in $(\mathbb{Q}[x])[y]$ and if we divide by $y$, then we get a quotient $q(x,y)$ and a remainder which is $0$ or of degree $0$ over $(\mathbb{Q}[x])[y]$, hence a polynomial $r(x)$ in $x$ only.

Then, $p(x,y) = yq(x,y) + r(x)$ and the hypothesis $p(0,0)=0$ does not guarantee that $r(x)=0$. So, we cannot conclude that $y$ divides $p(x,y)$.

Is there a way to remedy this proof, or is there another approach to this problem?

## 2 Answers

Take an element $p \notin (x, y)$. You can always represent it this way $p = xf + yg+ h$ where $f, g\in\mathbb{Q}[x,y], h \in \mathbb{Q}$. Then it follows that $h \neq 0$ since $p \notin (x, y)$. Then $h$, which is invertible is in $(x, y, p)$, which means that $(x, y, p) = (1)$. Then $(x, y)$ is maximal because the only ideal which strictly contains it is $(1)$.

$\varphi(p)=0\iff p(0,0)=0\iff a_{0,0}=0$, where $a_{0,0}$ is the contant coefficient of $p$. So both inclusions should be easy.

• Thanks! If the constant coefficient of $p(x,y)$ is $0$, I can "see" that $p(x,y)$ can be written as a linear combination $xg(x,y) + yh(x,y)$, but how does one prove this formally? – Nocturne Oct 30 '14 at 13:13
• @Nocturne Form $g$ from the terms with nonzero power of $x$ and form $h$ from the other terms, which necessarily have nonzero powers of $y$ – user2345215 Oct 30 '14 at 13:22