# Expressing an ideal as intersection of prime ideals

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.

• "I understand that $\langle xy\rangle$ means that $xy=0$." No. $\langle xy\rangle$ is not a proposition, it is the set of polynomials of the form $xyg(x,y)$. – Lord Shark the Unknown May 3 at 6:04
• Oh ok that makes sense - but within that set of polynomials, does $xy = 0$? So for any $xyg(x,y)$ then $xy =0$? – Masha May 3 at 6:06
• You might want to revisit what the definition of an ideal is. The ideal generated by $xy$ in $k[x,y]$ consists of all polynomials of the form $xy f(x,y)$, where $f(x,y)\in k[x,y]$. – Elliot G May 3 at 6:10
• Also, there is a nice theorem stating that an ideal is prime if and only if the factor ring doesn't contain zero divisors. Can you find the factor ring for your two ideals? – Dirk May 3 at 6:16
• "I understand that $\langle xy\rangle$ means that $xy=0$." What you are probably thinking of is that in the quotient ring $k[x,y]/I$, we do have $xy=0$. But that is not a statement about the values of $x$ and $y$. That's not what $x$ and $y$ are; they are not receptacles for ("unknown") elements of $k$. Instead, they are placeholders for coefficients, and that's that. What "$xy=0$" means is that if you take the element $\bar x\in k[x,y]/I$ (some times called $[x]$, but often enough just simplified to $x$), and you multiply that with $\bar y$, you end up with $\bar 0$ – Arthur May 3 at 6:24

Indeed within $$k[x,y]$$ ($$k$$ a field), $$\left=\left\cap\left$$ and $$\left$$ and $$\left$$ are prime ideals. To see this, observe that $$\left$$ consists of the polynomials $$\sum_{i,j}a_{i,j}x^iy^j$$ with all $$a_{0,j}=0$$. Similarly with $$\left$$. 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 $$\left$$ 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.