# This ideal is prime. But is it also maximal?

Could someone please give me a hint how to prove (preferably directly, without finding clever homomorphisms) that $(X,Y)$ is a maximal ideal in $\mathbb{C}[X,Y]$ ?

This ideal is prime, since it contains all polynomials of all degree that don't have a constant term, so my guess was, it's maybe also maximal.

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Recall that an ideal is maximal if and only if quotienting by it gets you a field. –  Qiaochu Yuan Oct 23 '12 at 21:12

If $P\notin (X,Y)$, identifying $P$ with its polynomial function, $P(0,0)\neq 0$. So if $I$ is a ideal which contains strictly $(X,Y)$, it contains a polynomial $P=\sum_{0\leq k,j\leq n}a_{k,j}X^jY^k$. As $\sum_{0\leq k\leq j\leq n,(k,j)\neq (0,0)}a_{k,j}X^jY^k\in (X,Y)$, $a_{0,0}\in I$ and is different from $0$. Hence $1\in I$, and $I=\Bbb C[X,Y]$, proving the maximality of $(X,Y)$.
An ideal (of a commutative ring with identity) is maximal if and only if quotienting by it gives you a field. $\mathbb{C} [X,Y]/(X,Y) \cong \mathbb{C}$, where the isomorphism maps every polynomial to its constant term.