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.

  • 2
    $\begingroup$ "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 $xyg(x,y)$. $\endgroup$ – Lord Shark the Unknown May 3 at 6:04
  • $\begingroup$ Oh ok that makes sense - but within that set of polynomials, does $xy = 0$? So for any $xyg(x,y)$ then $xy =0$? $\endgroup$ – Masha May 3 at 6:06
  • $\begingroup$ 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 $f(x,y)\in k[x,y]$. $\endgroup$ – Elliot G May 3 at 6:10
  • $\begingroup$ 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? $\endgroup$ – Dirk May 3 at 6:16
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    $\begingroup$ "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 $\bar 0$ $\endgroup$ – Arthur May 3 at 6:24

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.


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