Hilbert's Nullstellensatz implies $I(Z(I)) = \sqrt{I}$

Let $k$ be an algebraically closed field. Let $I$ be the map that takes algebraic sets in $k^n$ to the ideal generated by them: $I : \{$ algebraic sets $\} \to \{$ ideals of $k[x_1,\dots, x_n] \}$, $I(X) = \{ f \in k[x_1, \dots, x_n] : f(X) = \{0\}\}$. Similarly let $Z$ be the map that takes ideals in $k[x_1,\dots, x_n]$ to algebraic sets. Then $I(Z(I)) = \sqrt{I}$.

How does that follow from Hilbert's Nullstellensatz?

This version of Nullstellensatz: If $k$ is an algebraically closed field then, the maximal ideals of $k[x_1,\dots, x_n]$ are of the form $(x_1 - a_1, \dots, x_n - a_n)$ where $a = (a_1, \dots, a_n)$ is a point in $k^n$.

• This cannot be answered properly unless you give some formulation of the Nullstellensatz. For instance, one way to state it is to say precisely that $I(Z(I))= \sqrt{I}$ for any ideal $I$. Another way is for example that in a finitely generated $k$-algebra, every prime ideal is the intersection of all maximal ideals containing it. – Pavel Čoupek Dec 8 '13 at 20:03
• This is Hilbert's Nullstellensatz as I know it. – egreg Dec 8 '13 at 20:03
• Teh noes. I mean the version that says that over an algebraically closed field $k$, the maximal ideals in $k[x_1, \dots, x_n]$ are in bijective correspondence with points of $k^n$ and are given by the form $\hat{m} = (x_1 - a_1, \dots, x_n - a_n)$ for the corresponding point $a = (a_1, \dots, a_n) \in k^n$. – BananaCats Category Theory App Dec 8 '13 at 20:06
• You have to use Rabinowitsch trick, you can find a proof in almost any commutative algebra textbook, see for instance Zariski's book volume 1. – user40276 Dec 8 '13 at 20:41
• @EnjoysMath Sorry, but I'm not here for points, I'm here because I like math. – user40276 Dec 8 '13 at 20:45

Let $I$ be an ideal of $k[x_1, \dots, x_n]$. It is easy to see that $\sqrt{I}\subseteq IZ(I)$ ($f^l \in I$ implies $f^l(P)=0$ for every $P \in Z(I)$, which implies $f(P)=0$ for every $P \in Z(I)$ ).
The other inclusion is a bit more tricky. Assume that $f \in IZ(I)$ and take some generators $f_1, f_2, \dots, f_k$ of $I$. Then the set $Z(\{f_1, f_2, \dots, f_k,x_{n+1}f-1\})$ is empty, since $f(P)=0$ whenever $f_1(P)=f_2(P)=\dots =f_k(P)=0$ (note that here we operate "one dimension higher", i.e. in $k^{n+1}$; $x_{n+1}$ is a new indeterminate). Hence $$J:=(f_1, f_2, \dots, f_k,x_{n+1}f-1)=k[x_1,\dots , x_n,x_{n+1}]$$ (if $J$ were a proper ideal, it would be contained in some maximal ideal $m$, but $Z(m)$ is a point, hence nonempty, and $Z(m)\subseteq Z(J)$). Therefore, one can write $$1=\sum_{i=1}^{k}f_i(x_1, \dots x_{n})h_i(x_1, \dots x_n,x_{n+1})+(x_{n+1}f(x_1, \dots x_{n})-1)h(x_1, \dots x_n,x_{n+1})$$
by using the substitution $x_{n+1}=\frac{1}{f}$ (formally, using an evaluation homomorphism), one has (in the localisation $k[x_1, \dots x_n]_f$)
$$1=\sum_{i=1}^{k}f_i(x_1, \dots x_{n})h_i(x_1, \dots x_n,\frac{1}{f})+(\frac{1}{f}f(x_1, \dots x_{n})-1)h(x_1, \dots x_n,\frac{1}{f}),$$ but the last term clearly vanishes, so $$1=\sum_{i=1}^{k}f_i(x_1, \dots x_{n})h_i(x_1, \dots x_n,\frac{1}{f}).$$
One can then easily see that the terms $h_i(x_1, \dots x_n,\frac{1}{f})$ are of the form $\frac{g_i}{f^{m_i}}$, where $g_i$ are polynomials in indeterminates $x_1, \dots, x_n$. Take $M:=\max_i m_i$. Then $$f^M=\sum_{i=1}^{k}f_ig_if^{M-m_i},$$ which is an equality in $k[x_1, \dots x_n]$. Hence, $f \in \sqrt{(f_1, \dots, f_k)}=\sqrt{I}$.