Show that $(x,y)$ is prime and maximal ideal in $\mathbb Q[x,y]$.

Question

My attempt: Define a mapping $\phi:\mathbb Q[x,y] \rightarrow\mathbb Q$ by $\phi(f(x,y))= f(0,0)$. Homomorphism and onto is easy to show. Claim: $\ker(\phi)=I$, where $I=(x,y)$. clearly, $I \subset \ker(\phi)$. Now I want to show that $\ker(\phi) \subset I$.

Let $f \in \ker(\phi)$, $f(x,y)=\sum_{i=0}^{i=n} a_{i}(x) y^{i}$.

$\phi(f(x,y)=a_{0}$. Subtract the above two equations,

$$f(x,y)=\sum_{i=0}^{i=n}a_{i}(x)y^{i} - \phi(f(x,y)=\sum_{i=0}^{i=n}a_{i}(x)y^{i} - a_{0}$$ This implies $f(x,y) \in (x,y)$. This shows that $\ker(\phi) \subset (x,y)=I.$

Is this attempt is right?

Answer 1

You proved that $Q[x,y]/I$ is isomorphic to $Q$, and for commutative rings, an ideal is maximal if and only if the the quotient is a field. Note also that maximal implies prime. So you are correct.

Answer 2

Let $f\in \mathbb{Q}[x, y]$. Then $f$ can be written in the form $f(x,y)=q+xg(x)+\sum_{i=0}^{i=n} a_{i}(x) y^{i}$, where $q\in \mathbb{Q}$ and $g(x), a_{i}(x)\in \mathbb{Q}[x]$. Now $f\not\in (x, y)$ if and only if $q\not=0$ if and only if $(f)+(x, y)=\mathbb{Q}[x, y]$ if and only if $(x, y)$ is a maximal ideal of $\mathbb{Q}[x, y]$.