(Possibly) alternative statement of Hilbert's Nullstellensatz In my notes, there is a statement entitled "Nulstellensatz version 2":

If $k = \bar{k}$, and $\mathfrak{m} \subseteq k[x_1,\ldots,x_n]$ is a maximal ideal, then $k[x_1,\ldots,x_n]/\mathfrak{m} \cong k$.

I assume "version 2" implies it is equivalent to the usual version, which states:

If $k = \bar{k}$, $\mathfrak{a} \subseteq k[x_1,\ldots,x_n]$ is an ideal and $f \in k[x_1,\ldots,x_n]$ is a polynomial which vanishes on all points in $Z(\mathfrak{a})$, then $f^r \in \mathfrak{a}$ for some positive integer $r$. 

I can see that "version 2" follows from the "usual version" as follows: 
Usual version $\implies$ $I(Z(\mathfrak{a})) = \sqrt{\mathfrak{a}}$ for ideals $\mathfrak{a}$ of $k[x_1,\ldots,x_n]$. We also have the following easily verified facts:
i) $Y_1 \subseteq Y_2 \subseteq k^n \implies k[x_1,\ldots,x_n] \supseteq I(Y_1) \supseteq I(Y_2) $; 
ii)$T_1 \subseteq T_2 \subseteq k[x_1,\ldots,x_n] \implies k^n \supseteq Z(T_1) \supseteq Z(T_2)$ and
iii) $Y \subseteq k^n$ is irreducible if and only if $I(Y) \subseteq k[x_1\ldots x_n]$ is prime.
iv) prime ideals are radical
Combining this information gives that $Z$ and $I$ give an inclusion-reversing correspondence between (irreducible) affine subvarieties of $k^n$ and prime ideals in $k[x_1,\ldots,x_n]$. So if $\mathfrak{m}$ is a maximal ideal, then it corresponds to a minimal irreducible closed subset of $k^n$ (since $\mathfrak{m}$ is prime and maximal), which must be a point $\{p = (p_1,\ldots,p_n)\}$. So $\mathfrak{m} = I(p)$, which one can see is just $\langle (x_1-p_1)\cdots(x_n-p_n) \rangle$. So $k[x_1,\ldots,x_n]/\mathfrak{m} \cong k$ via the "evaluate at $p$" map.
My question is this: Does the "usual version" follow from "version 2" and, if so, how?
Thanks.
 A: Your statement is definitely one of the things which go under the label "Nullstellensatz".  As a student, I was taught that this was the weak Nullstellensatz and was counseled that if I were asked for a statement of the Nullstellensatz on an exam, this "weak" version would not be sufficient.
Then I grew up and, in particular, wrote some notes on commutative algebra.  The Nullstellensatz is treated in some detail in $\S 11$.  Hilbert's Nullstellensatz is Theorem 267, and you can see that part a) of that result is precisely the weak version you are asking about.  Part c) of this result is the statement that for any ideal $J$ of $\overline{k}[t_1,\ldots,t_n]$, $I(V(J)) = \operatorname{rad} J$.
Now this part is proven using part a)...and also a little preliminary ring-theoretic analysis, especially Rabinowitsch's Trick.   None of this other stuff is very hard or very lengthy, so I am tempted to say yes, the "weak" Nullstellensatz is more or less equivalent to Hilbert's Nullstellensatz, especially if you know what you're doing.  But don't write this on an exam -- you might lose points!
