The ideal $(X,Y)$ is prime in $\mathbb{C}[X,Y]$. But is it also maximal?

Question

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.

Answer 1

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 $(X,Y)$.

Answer 2

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.

Answer 3

Here is my answer which is in some sense same as the answer of Davide Giraudo.

We will use the direct meaning of maximality (instead of using quotient ring). The following is obvious:

FACT 1: In this ring if an ideal contains a constant (polynomial) then the ideal is the whole ring.

Consider a polynomial $f$, which is NOT in your ideal $I$. We have to show that any ideal $J$ that contains $I$ and $f$ is the whole ring. We will show that such an ideal contains a constant (polynomial) which will prove maximality of $I$ as an ideal.

As $f$ is not in $I$, it has a nonzero constant term, call it $a_0$.

Now define $f_0=f-a_0$. This has zero constant term, and hence $f_0\in I$.

So the ideal containing $f$ and $I$ contains $f-f_0=a_0$, a constant and hence by FACT 1 we are done.