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.