Radical Ideals

**slevvio** · October 8th 2011, 04:41 PM

Hello all,

I was wondering if anyone could help me answer the question: find ideals $I,J$ of $k[x_1,\ldots,x_n]$ such that $\sqrt{I+J} \not= \sqrt{I} + \sqrt{J}$ , where k is an algebraically closed field (i.e. find a counterexample) Can anyone give me a hint about how to go about this? I tried messing around with $\mathbb{C}[x]$ but I can't really work anything out.

Any help would be appreciated, thank you

**NonCommAlg** · October 9th 2011, 11:42 AM

Originally Posted by slevvio

Hello all,

I was wondering if anyone could help me answer the question: find ideals $I,J$ of $k[x_1,\ldots,x_n]$ such that $\sqrt{I+J} \not= \sqrt{I} + \sqrt{J}$ , where k is an algebraically closed field (i.e. find a counterexample) Can anyone give me a hint about how to go about this? I tried messing around with $\mathbb{C}[x]$ but I can't really work anything out.

Any help would be appreciated, thank you

in $\mathbb{C}[x,y]$ choose $I = (x)$ and $J=(x^2+y^2)$ . then $\sqrt{I}=I$ because $I$ is prime and $\sqrt{J}=J$ because $J$ is the intersection of two primes. so $\sqrt{I}+\sqrt{J}=I+J=(x,y^2)$ but $\sqrt{I+J}=\sqrt{(x,y^2)}=(x,y)$ .

**slevvio** · October 9th 2011, 02:47 PM

Thank you,

could you give me a bit more information about why $\langle x^2 + y^2 \rangle$ is the intersection of 2 prime ideals?

**Deveno** · October 9th 2011, 03:04 PM

isn't $\langle x^2 + y^2\rangle = \langle x + iy\rangle \cap \langle x - iy\rangle$ ?

**slevvio** · October 9th 2011, 03:06 PM

I'm not sure. I'm very bad with ideals

**slevvio** · October 9th 2011, 03:25 PM

why is <x +iy> prime? I tried to show the quotient of C[x,y] by it is an intergral domain but to no avail

**NonCommAlg** · October 9th 2011, 03:43 PM

$x + iy$ is irreducible in $\mathbb{C}[x,y]$ and we know that in a UFD, irreducible elements are prime and the ideal generated by a prime element is a prime ideal. another way is to define a homomorphism $\varphi: \mathbb{C}[x,y] \to \mathbb{C}[y]$ by $\varphi(f(x,y))=f(-iy,y)$ and then see that $\ker \varphi = (x+iy)$ .

**slevvio** · October 9th 2011, 03:45 PM

excellent thank you

**NonCommAlg** · October 9th 2011, 03:56 PM

also, to see why $(x^2+y^2)=(x+iy) \cap (x-iy)$ , first we note that since $x^2+y^2=(x+iy)(x-iy)$ , it is clear that $(x^2+y^2) \subseteq (x+iy) \cap (x-iy)$ . for the converse, let $z =(x+i)u = (x-iy)v$ for some $u,v \in \mathbb{C}[x,y]$ . then $x+iy$ divides $(x-iy)v$ and thus $x+iy$ divides $v$ because $\mathbb{C}[x,y]$ is a UFD and both $x+iy$ and $x-iy$ are irreducible. so $v=(x+iv)w$ for some $w \in \mathbb{C}[x,y]$ and hence $z = (x^2+y^2)w \in (x^2+y^2)$ .

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