• Oct 8th 2011, 04:41 PM
slevvio
Hello all,

I was wondering if anyone could help me answer the question: find ideals $\displaystyle I,J$ of $\displaystyle k[x_1,\ldots,x_n]$ such that $\displaystyle \sqrt{I+J} \not= \sqrt{I} + \sqrt{J}$, 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 $\displaystyle \mathbb{C}[x]$ but I can't really work anything out.

Any help would be appreciated, thank you
• Oct 9th 2011, 11:42 AM
NonCommAlg
Quote:

Originally Posted by slevvio
Hello all,

I was wondering if anyone could help me answer the question: find ideals $\displaystyle I,J$ of $\displaystyle k[x_1,\ldots,x_n]$ such that $\displaystyle \sqrt{I+J} \not= \sqrt{I} + \sqrt{J}$, 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 $\displaystyle \mathbb{C}[x]$ but I can't really work anything out.

Any help would be appreciated, thank you

in $\displaystyle \mathbb{C}[x,y]$ choose $\displaystyle I = (x)$ and $\displaystyle J=(x^2+y^2)$. then $\displaystyle \sqrt{I}=I$ because $\displaystyle I$ is prime and $\displaystyle \sqrt{J}=J$ because $\displaystyle J$ is the intersection of two primes. so $\displaystyle \sqrt{I}+\sqrt{J}=I+J=(x,y^2)$ but $\displaystyle \sqrt{I+J}=\sqrt{(x,y^2)}=(x,y)$.
• Oct 9th 2011, 02:47 PM
slevvio
Thank you,

could you give me a bit more information about why $\displaystyle \langle x^2 + y^2 \rangle$ is the intersection of 2 prime ideals?
• Oct 9th 2011, 03:04 PM
Deveno
isn't $\displaystyle \langle x^2 + y^2\rangle = \langle x + iy\rangle \cap \langle x - iy\rangle$ ?
• Oct 9th 2011, 03:06 PM
slevvio
I'm not sure. I'm very bad with ideals :(
• Oct 9th 2011, 03:25 PM
slevvio
$\displaystyle x + iy$ is irreducible in $\displaystyle \mathbb{C}[x,y]$ 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 $\displaystyle \varphi: \mathbb{C}[x,y] \to \mathbb{C}[y]$ by $\displaystyle \varphi(f(x,y))=f(-iy,y)$ and then see that $\displaystyle \ker \varphi = (x+iy)$.
also, to see why $\displaystyle (x^2+y^2)=(x+iy) \cap (x-iy)$, first we note that since $\displaystyle x^2+y^2=(x+iy)(x-iy)$, it is clear that $\displaystyle (x^2+y^2) \subseteq (x+iy) \cap (x-iy)$. for the converse, let $\displaystyle z =(x+i)u = (x-iy)v$ for some $\displaystyle u,v \in \mathbb{C}[x,y]$. then $\displaystyle x+iy$ divides $\displaystyle (x-iy)v$ and thus $\displaystyle x+iy$ divides $\displaystyle v$ because $\displaystyle \mathbb{C}[x,y]$ is a UFD and both $\displaystyle x+iy$ and $\displaystyle x-iy$ are irreducible. so $\displaystyle v=(x+iv)w$ for some $\displaystyle w \in \mathbb{C}[x,y]$ and hence $\displaystyle z = (x^2+y^2)w \in (x^2+y^2)$.