I want to find all prime and maximal ideals of $\mathbb C[x,y]$ that contain $I=\langle x^2 + 1, y + 3\rangle$.

My approach is that I know that if $f(x)$ is irreducible then $ < f(x) > $ is a prime ideal. So, if I factor $x^2+1$ into $(x+i)(x-i)$ then those are irreducible factors in $C[x,y]$. And $y+3$ is also irreducible over $C[x,y]$. So then the prime ideals that contain $I$ should be: $< x+i, y+3 >$ $< x-i, y+3 >$

Is this true? Can I say that $< x-i, y+3 >$ is prime because $f(x,y) = (x-i) + (y+3)$ is irreducible over $C[x,y]$ or how do I motivate that? I'm kind of unfamiliar with working with ideals in two variabels, and ideals generated by two elements, so I'm not quite sure how I can use the theorems that I know from one-variable.

I also don't know how I can show that these ideals are also maximal. Since $C[x,y]$ is not a PID then I can't draw the conclusion that just because the ideals are prime that they are also maximal.

Also, just for verification that I understood it right. The elements of the ideal $I=< x^2 + 1, y + 3 >$ are all elements of the form $f(x,y) \times (x^2+1) + g(x,y)\times(y+3)$. So therefor it is true that the ideals that I suggested both contain $I$?

Thankful for any help!

  • $\begingroup$ Your last question is those are the elements of your ideal. One way to show that ideals are prime is by looking at the residue field and showing it is an integral domain. The ideals you suggested indeed contain $I$. $\endgroup$ – LASV Dec 4 '13 at 21:32

Let $P$ be a prime ideal of $\mathbb C[x,y]$ containing $I=\langle x^2 + 1, y + 3\rangle$. Then $x^2+1\in P$ and $y+3\in P$. Since $x^2+1=(x+i)(x-i)$ we have $x+i\in P$ or $x-i\in P$. In conclusion, $\langle x + i, y + 3\rangle\subseteq P$ or $\langle x - i, y + 3\rangle\subseteq P$. Since $\langle x \pm i, y + 3\rangle$ are maximal ideals (why?) then $P=\langle x \pm i, y + 3\rangle$.

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