Showing that $(x+1)$ is prime but not maximal in $\mathbb{C} [x,y]$

Question

I'm trying to show that $(x+1)$ is prime but not maximal in $\mathbb{C} [x,y]$.

Is is not maximal because $(x+1)\subset (x)$. But why is it prime? I know that it is enough to show: x + 1 is prime, as element in $\mathbb{C} [x,y]$.

Thanks

Answer 1

The ideal $(x+1)$ is indeed not maximal, but it is false that $(x+1)\subset(x)$. A maximal ideal containing it is, for instance, $(x+1,y)$.

A way to show that $(x+1)$ is prime is to observe that it is the kernel of the surjective homomorphism $$ f:\Bbb C[x,y]\longrightarrow\Bbb C[y],\qquad f(P(x,y))=P(-1,y). $$ Thus $(x+1)$ is prime because $\Bbb C[y]$ is a domain.

Answer 2

Here's a hint extracted from one of my old sci.math posts:

Even easier: the ideals $\rm\ (x\!+\!1)\ <\ (x\!+\!1,y)\ < 1\ $ are distinct primes$\ \ $ (or $1$)
since their residue rings $\rm\ \ C[y]\ \ >\ \ \ \ \ \ \Bbb C\ \ \ \ < \ \ \ 0\ $ are distinct domains (or $0$)