Q:

The algebraic set of ideal

So this is a homework question just trying to see if my thinking is correct as I'm new to algebraic geometry.
"Let $$J=(x,xy^2)\subset \mathbb{C}\left[x,y\right]$$
1."Determine $$Z(J)\subset \mathbb{C}^2$$"
so this is what I've got, if $x^2=0$ then $x=0$ and $y$ can be anything thus $Z(J)=\{(0,z)\subset \mathbb{C}^2: z\in \mathbb{C}\}$
Is this correct?
2."Determine $I(Z(J))\subset \mathbb{C}[x,y]$"
Well if answer to 1 is correct then polynomials must vanish at $(0,z)\quad \forall\ z \in \mathbb{C}$ thus $I(Z(J))=(x,0)=(x)$
is this right ?

"Check if $I(Z(J))=\sqrt{J}$ "

well no because $xy^2\in\sqrt{J}$ but $xy^2\notin I(Z(J))$
Is my answers correct?I'm new to this so not sure if my reasoning is correct. If anyone could help me understand any failings in my logic or reasoning I'd greatly appreciate it!

A:

You're right that $xy^2$ is in $J$, and thus in $\sqrt{J}$, but it is also in $I(Z(J))$. Indeed $(x)$ contains all elements of the form $r x$ for $r \in k[x, y]$, so take $r = y^2$. Also, $\sqrt{J} = J = (x)$ since $(x)$ is a prime ideal ($k[x, y]/(x) \cong k[y]$ is a domain), and so it is radical, so $I(Z(J)) = \sqrt{J}$.
This problem is a bit tricky, actually, since $J$ itself is also already equal to $(x)$. Since you're new to the subject, here are perhaps some sanity checks you could have done to see that what you got is impossible. First of all, note that $I(Z(J))$ is always equal to $\sqrt{J}$, so you should be a bit suspicious if you get that they are not equal!  Even if you didn't know this, one always has $J \subset I(Z(J))$, since the right hand side are "polynomials which vanish at the points of $Z(J)$", and certainly elements of $J$ vanish at $Z(J)$.

P.S. Are you sure the ideal isn't $J = (x, y^2)$? This seems to be a much better exercise...

