I am trying to solve that, given the ring of polynomials $\mathbb{C}[x,y]$, $\langle x\rangle \cap \langle y\rangle \space = \space \langle xy \rangle $. I have no problem showing that $\langle xy\rangle \space \subseteq \space \langle x\rangle \cap \langle y \rangle $
My doubts arises with the other part, proving that : $\langle x\rangle \cap \langle y\rangle \space \subseteq \space \langle xy \rangle $
My idea goes like this, I want you to comment if my resolution is right : I start with a polynomial $p (x,y)$ that is in the intersection between $\langle x \rangle$ and $\langle y\rangle$. Therefore I can say that $p(x,y) = x q(x,y)$. Since $p(x,y)$ is also in $\langle y \rangle$ by hypothesis, and $\langle y \rangle $ is a prime ideal of $\mathbb{C}[x,y]$ (This is my crucial doubt and the reason for the title : Is this last affirmation correct ?) , I deduce by definition of prime ideal that $q(x,y$) is in $\langle y\rangle$ (because $x$ is not). Then , if $q(x,y) = yh(x,y)$ I can say that $p(x,y) = xyh(x,y)$ and this is in $\langle xy \rangle$.
I would appreciate some comments (and corrections if its necessary) of my resolution.
Thanks you.
EDIT : A comment has proposed that since the homomorphism $F : \mathbb{C}[x,y] \to \mathbb{C}[x]$ has a kernel $\langle y \rangle$ therefore $\langle y \rangle$ is a prime ideal of $\mathbb{C}[x,y]$
So basically I define the homomorphism $F(p(x,y)) = p(x,0)$ and then the kernel of F is $p(x,y) \in \langle y \rangle$ because otherwise $p(x,0) =/= 0$ . How can I conclude that kerF = $\langle y \rangle$ is a prime ideal of $C[x,y]$ ?
Best Answer
As far as I can tell, your argument is correct. As Arturo has mentioned in the comments, a useful characterization of an ideal being prime is the following:
Assume $A$ is a commutative unital ring. An ideal $\mathfrak{p}\subseteq A$ is prime if and only $A/\mathfrak{p}$ is an integral domain.
It is pretty easy to see that $\Bbb{C}[X]$ is an integral domain (indeed, the product of nonzero polynomials is nonzero).