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!
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...