Nullstellensatz equivalence question In an assignment question, I have to prove (among other things) that the following two are equivalent:


*

*Nullstellensatz

*If $F$ is an algebraically closed field, and $I$ is a proper ideal of $F[\vec x]$, then $I(V(I))$ is a radical ideal. 
I have the following "proof" of (1)$\implies$ (2). 
Let $S:=V(I)$, and $M:=I(S)$.  Since every ideal is a subset of its radical, proving (2) comes down to showing that $\sqrt M\subseteq M$. So let $f(\vec x)\in\sqrt M$, with $m>0$ s.t. $f^m(\vec x)\in M$. Then, since $f^m(\vec x)\in M=I(S)$, it follows that $\forall a \in S, f^m(a)=0$.  However, $F[\vec x]$ has no zero divisors, since neither did $F$. That implies then that $f^m(a)=0\implies f(a)=0$.  So $\forall a \in S, f(a)=0$, so $f(\vec x)\in I(S)=M$.  Hence, $\sqrt M\subseteq M$. 
However, this proof didn't use the Nullstellensatz at any point.  Nor the fact that $I$ is proper. 
Is there a problem with my "proof"? Or is (2) true regardless of (1)? How about the fact that $I$ had to be proper? Am I missing something in that regard?
 A: Both (1) and (2) can be shown to be true each in their own right without need for the other. 
Actually, they don't really imply each other. To show $(2) \implies (1)$, you essentially have to prove the Nullstellensatz (or some form of it). The only reason this statement is true is because each of the individual statements are true in their own right.
Perhaps this is the argument they were expecting for $(1) \implies (2)$:

Proof: Suppose the Nullstellensatz, and let $J$ be a subset of $k[x_1, \ldots, k_n]$, then this means that $I(V(J)) \subseteq \sqrt J$. (Note: this is the version of the Nullstellensatz you said you had available to you.)
That $\sqrt J \subseteq I(V(J))$ follows from: Let $f \in \sqrt J$, then $f^n \in J$ for some $n > 0$. But $V(J) = \{P \in \mathbb A^n : g(P) = 0$ for all $g \in J\}$ and $f^n \in J$, so $f^n(P) = 0$ for all $P \in V(J)$. But $0 = f^n(P) = \big(f(P)\big)^n$ and since $k$ is an integral domain, then $f(P) = 0$. Hence $f \in I(V(J))$.
Conclude that $I(V(J)) = \sqrt J$ and hence $I(V(\sqrt J))$ is radical.

Also, notice that we just proved a more general result than what was being asked: we showed this is true for all ideals $J$, not just those that are proper.
