Prove that $\sqrt I\subset \sqrt J\iff Z(J)\subset Z(I)$

Question

Let $R=F[x_1,...,x_n]$ where $F$ is a field. I'm trying to prove that $$\sqrt I\subset \sqrt J\iff Z(J)\subset Z(I),$$ where $Z(I)=\{P\in F^n\mid \forall f\in I, f(P)=0 \}$.

For the implication, suppose $\sqrt I\subset \sqrt J$. Let $P\in Z(J)$. Then, for all $f(P)=0$ for all $f\in J$. Suppose there is a polynomial $g\in I$ s.t. $g(P)\neq 0$, in particular $g\in \sqrt I$ and thus $g\in \sqrt J$. Therefore $g^n(P)=0$ for a certain $n$ and thus $g(P)=0$ what is a contradiction.

1) Is it correct ?

For the converse, suppose $Z(J)\subset Z(I)$. Let $f\in I(Z(I))$, then $f(P)=0$ for all $P\in Z(I)$. Since $Z(I)\supset Z(J)$, then $f(P)=0$ for all $P\in Z(J)$, i.e. $f\in I(Z(J))$. Therefore $I(Z(I))\subset I(Z(J))$, and thus, by nullstellensatz theorem, $\sqrt I\subset \sqrt J$.

2) Is it correct ?

Answer 1

1)

Yes, this is correct! (You can do this without proceeding by contradiction, however).

2)

Provided that $F$ is algebraically closed, your proof is correct. Well done!

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In other words, we have

$$\sqrt I \subset \sqrt J \iff Z(\sqrt I) \supset Z(\sqrt J) \iff Z(I) \supset Z(J)$$ where the first "$\Longleftarrow$" is given by HNS.

Answer 2

This from the book: Algebraic geometry and arithemtic curves, p.28

Theorem: Let $A$ be a ring. Let $I,J$ be two ideals of $A$. The following properties are true:

$1)$ The radical $\sqrt I$ equals the intersection of the ideals $p\in V(I)$.

$2)$ We have $V(I)\subseteq V(J) $ if and only if $J\subseteq \sqrt I$.

proof $2)$ if $V(I)\subseteq V(J) \Rightarrow ZV(J)\subseteq ZV(I)$ then $J \subseteq ZV(J)\subseteq ZV(I) \subseteq \sqrt I$

if $J\subseteq \sqrt I$ then $V(\sqrt I)\subseteq V(J)\Rightarrow V(I)\subseteq V(J); V(\sqrt I)= V(I)$