I'm trying to express the ideal $I = \langle xy\rangle \in k[x,y]$ as an intersection of prime ideals. I understand that the radical of an ideal is the intersection of prime ideals, but we haven't actually covered that in class yet so I don't want to use that in my explanation.
I understand that $\langle xy\rangle \rangle $ means that $xy=0$. This means that either $x = 0$ or $y=0$. Geometrically, I can see that this is just the set of the x and y axis. So my idea was that the intersection would be $\langle x\rangle \cap \langle y\rangle = \langle xy\rangle $. But 1) I'm not really sure how to prove that $\langle x\rangle $ and $\langle y\rangle $ are prime ideals and 2) then how to prove that their intersection is $\langle xy\rangle $.
For 1) I know that a prime ideal is a proper ideal, $p$ of a ring such that $\forall f,g \in p$ then either $f \in p$ or $g \in p$. But I'm a little confused on what this means still and how it relates here.
Any hints or suggestions would be appreciated, thank you.
Indeed within $k[x,y]$ ($k$ a field), $$\left<xy\right>=\left<x\right>\cap\left<y\right>$$ and $\left<x\right>$ and $\left<y\right>$ are prime ideals. To see this, observe that $\left<x\right>$ consists of the polynomials $\sum_{i,j}a_{i,j}x^iy^j$ with all $a_{0,j}=0$. Similarly with $\left<y\right>$. The intersection consists of all $\sum_{i,j\ge 1}a_{i,j}x^iy^j=xy\sum_{i,j\ge 1}a_{i,j}x^{i-1}y^{j-1}$.
To see $\left<x\right>$ is a prime ideal, observe that it's the kernel of the ring homomorphism $\phi: k[x,y]\to k[y]$ taking $f(x,y)$ to $f(0,y)$, and the kernel of a homomorphism to an integral domain is prime.