1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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 $(1)$.

$\endgroup$
3
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Nocturne Oct 30 '14 at 13:13
  • 2
    $\begingroup$ @Nocturne Form $g$ from the terms with nonzero power of $x$ and form $h$ from the other terms, which necessarily have nonzero powers of $y$ $\endgroup$ – user2345215 Oct 30 '14 at 13:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.