Calculate the radical of ideals

Question

Let $k$ be an algebraically closed field and consider $A=k[x,y,z]$. I am supposed to calculate $\text{rad}(x,y)= \{ f \in k[x,y,z] : f^n \in (x,y)$ $\text{for some n} \}$, $\text{rad}(x,z)$ and $\text{rad}(y,z,x^2)$. I feel I have done something wrong, I want to check if my arguments are ok.

For $\text{rad}(x,y)$, suppose that $f \notin (x,y)$ then I can write $f=g_1(x,y)+ g_2(z)$, that is, its not only a function of $x$ and $y$ but also contains a constant term or a term which is a function of $z$, however then we will never have $f^n \in (x,y)$ because $f^n$ always contains $g_{2}^n(z)$. So that $\text{rad}(x,y) = (x,y)$

With the same argument I have calculated $\text{rad}(x,z)=(x,z)$ and $\text{rad}(z,y,x^2)=(x,y,z)$. Is this correct?

A good thing to know is that prime ideals are their own radical, which often helps with checking in these sort of problems. — Tobias Kildetoft, Jan 08 '15 at 13:14
@TobiasKildetoft I see, and $(x,y)$ and $(x,z)$ are prime ideals? — , Jan 08 '15 at 13:19
@TobiasKildetoft, thanks. Was my calculation for $(z,y,x^2)$ wrong? — , Jan 08 '15 at 13:20
Related to http://math.stackexchange.com/questions/1095586/find-shortest-primary-decomposition. — user26857, Jan 08 '15 at 22:09

score 1 · Answer 1 · answered Jan 08 '15 at 13:18

1

Hint: (1). A prime ideal is always equal to it's own radical.

(2). For $I = (y, z, x^2),$ you can consider the ring $A/I.$ Then the nilradical of this ring will be the radical of $I$ in $A.$ Now $A/I \cong k[x]/x^2.$ What is the nilradical in this ring?

answered Jan 08 '15 at 13:18

Krish

7,262

Alternatively for the second one, he has already found a prime ideal contained in the radical. – Tobias Kildetoft Jan 08 '15 at 13:19
@TobiasKildetoft: yes!! OP already found $\text{rad}(y, z, x^2).$ I just gave an alternate way to this. – Krish Jan 08 '15 at 13:21

Calculate the radical of ideals

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