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I need to find the radicals of the following ideals:

i) $\mathfrak{a} = (xy^3, x(x-y))$

ii) $\mathfrak{b} = (xy^3, x^2(y-3))$

iii) $\mathfrak{c} = (x^2(y-z), xy(y-z), xz(y-z)^2)$

Can I just use the Nullstellensatz? My working below seems a bit too easy, which makes me think I'm doing something horrendously awful.

Let $k$ be an algebraically closed field.

i) It's pretty obvious that $Z(\mathfrak{a}) = \{ (0,t) \ | \ t \in k \} = Z(x) $. So by the Nullstellensatz, $\sqrt{\mathfrak{a}} = I(Z(\mathfrak{a})) = I(Z(x)) = (x) $.

ii) Isn't this the same as above?

iii) $Z(\mathfrak{c}) = \{ (0,s,t) \ | \ s,t \in k \}\cup \{(s,t,t) \ | \ s,t \in k\} = Z(x) \cup Z(y-z)$. So $I(Z(\mathfrak{c})) = I(Z(x)) \cap I(Z(y-z)) = (x) \cap (y-z) = (x(y-z))$

Am I breaking any laws?

Thanks!

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@AlexBecker No, it isn't. But $x^4$ is... –  Jonathan Apr 27 '12 at 22:58
    
Your computations make sense to me. –  user29743 Apr 27 '12 at 23:07
3  
Everything you write is correct, Jonathan, and it is an ingenious way to solve the problem: congratulations! –  Georges Elencwajg Apr 27 '12 at 23:24
    
Thanks for checking! –  Jonathan Apr 27 '12 at 23:36
    
Hmn, Hilbert's Nullstellensatz is only true if your underlying field is algebraically closed. –  user1412 Jan 6 at 18:59
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1 Answer

up vote -1 down vote accepted

You are right!

However, you can employ the symbol $V$. Then $$\begin{eqnarray}V(\mathfrak{a})&=&V(xy^3,x(x-y))=V(x,y^3)\cap V(x(x-y))\\ &=&(V(x)\cup V(y))\cap (V(x)\cup V(x-y))\\ &=&V(x)\cup V(x,y)=V(x)\end{eqnarray}$$

So, $\sqrt{\mathfrak{a}}=\sqrt{(x)}$ in any commutative ring. If our ring is the polynomial ring $k[x,y]$ over a field $k$, then $(x)$ is a prime ideal and $\sqrt{(x)}=(x)$.

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