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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 ?

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    Hilbert's Nullstellensatz (only) holds over **algebraically closed** fields. – 2017-01-29
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    It's not specified in my exercise, but they probably forgot to precise it. So, I think $F$ is algebraically closed (otherwise, this result may be not correct). – 2017-01-29
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    (The result is wrong otherwise, indeed, take $I=(x^2+1),J=(x^2+2)$ in $\Bbb R[x]$). – 2017-01-29

2 Answers 2

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1)

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

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

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.

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    Notice that [strong Hilbert's Nullstellensatz over $k \implies k$ is algebraically closed]. Indeed, let $f \in k[x]$ be irreducible, so that $(f)$ is a prime of $k[x]$. Then $I(V( (f) )) = \sqrt{(f)}=(f)$, which implies that $V( (f) ) \neq \varnothing$ (otherwise $I(V((f))) = I(\varnothing) = k[x]$). – 2017-01-29
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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 $V(\sqrt I)\subseteq V(J)\Rightarrow V(I)\subseteq V(J); V(\sqrt I)= V(I)$