I have a question about ideals generated by two elements. I've searched MathStackexchange and found some related posts, but I haven't been able to understand how it all works.

The question is in general:

  1. How can I show that an ideal generated by two prime elements, is indeed prime in a UFD.

  2. More specific I want to show that in $\Bbb C[x,y]$ the ideals generated by $(x-i, y+3)$ and $(x+i,y+3)$ are in fact prime ideals.

My initial approach was: $\Bbb C[x,y]$ is a UFD. In a UFD every irreducible element is prime. In a UFD every prime element generates a prime ideal. But I'm not sure I can use that argument when my ideal is generated by two prime elements.

I know that I can check that $I$ is a prime ideal by checking that $R/I$ is an integral domain, but I'm not sure how to do that in my example. Can anyone show me?

I would be really grateful for any help!


Consider the map $\mathbb{C}[x,u] \to \mathbb{C}$ given by $f(x,y) \mapsto f(i, -3)$, i.e. evaluating the polynomial at $x=i, y=-3$.

Show this is a surjective homomorphism. Show the kernel is exactly the ideal $I=(x-i, y+3)$. Then you can use the first isomorphism theorem to show that $\mathbb{C}[x,y]/I$ is a field, and thus $I$ is a maximal (and therefore prime) ideal.

In general, though, we cannot assume that starting with two prime ideals will give us a new prime ideal. For example, the ideals $(3)$ and $(5)$ are both prime in the integers, but their product is not.

  • $\begingroup$ @user: Thank you for letting me know! $\endgroup$ – davidlowryduda Dec 7 '13 at 21:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.