7
$\begingroup$

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!

  • 1
    Hmn, Hilbert's Nullstellensatz is only true if your underlying field is algebraically closed.2014-01-06

1 Answers 1

1

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 $\sqrt{(x)}=(x)$.