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Let $\mathbb{A}^n_k$ be defined as the set $\{(a_1,\dots,a_n):a_i\in k\}$. There are standard maps $Z$ and $I$. The former assigns to an ideal its zero set and the latter assigns to a subset of $\mathbb{A}^n_k$ the ideal of the subset, i.e., the set of all polynomials which vanish at all points of that subset. If $k$ is algebraically closed, then the Nullstellensatz says that $I(Z(J))=\sqrt{J}$.

What if we replace $\mathbb{A}^n_k$ with $\operatorname{Spec}k[x_1,\dots,x_n]$? Then the map $Z$ can be replaced with another map $Z$ which assigns to an ideal $I$ the set of all prime ideals in $k[x_1,\dots,x_n]$ containing $I$, $Z(I)$. (And just like the sets $Z(I)$ in the previous paragraph are the closed sets for the Zariski topology on $\mathbb{A}^n$, the sets $Z(I)$ that I have just described, are the closed sets for the Zariski topology on $\operatorname{Spec}k[x_1,\dots,x_n]$.) But what about the map $I$ from the previous paragraph? Does it have an analogue in this case? That is, does there exist a map $I: \operatorname{Spec}k[x_1,\dots,x_n]\rightarrow k[x_1,\dots,x_n]$? How it is defined if so? Further, is it true that these new $I$ and $Z$ also satisfy $I(Z(J))=\sqrt{J}$? If so, does it follow from the analogous fact above?

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    $\begingroup$ The "maps" that you have defined do not have the correct domains/images. For example, the second map that you define $I:\mathbb{A}_k^n\rightarrow k[x_1,\cdots,x_n]$ is not a map because the result is not an element of $k[x_1,\cdots,x_n]$, but a subset. In fact, you have a map, right there, to Spec. $\endgroup$ Mar 19, 2017 at 13:10

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Yes, for $X\subset\operatorname{Spec}k[x_1,\dots,x_n]$, you take

$$I(X)=\bigcap_{\mathfrak p\in X}\mathfrak p.$$

In particular, when $X=Z(J)$, we see $I(Z(J))$ is the intersection of all prime ideals containing $J$, and if you know some commutative algebra then it's just about trivial that $I(Z(J))=\sqrt J$.

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  • $\begingroup$ Does one need the hypothesis that $k$ is algebraically closed to conclude that the last equality in your answer holds? I believe not because one only uses the definition of $I$ you gave and the corresponding theorem about radicals of ideals. And so this fact is not usually referred to as the Nullstellensatz, is it? $\endgroup$
    – user557
    Mar 19, 2017 at 13:27
  • $\begingroup$ No. $\sqrt{I} \subset \cap p$ since all primes are radical. The converse is shown by the prime avoidance lemma via contraposition. Suppose that $f \notin \sqrt{I}$, then there exists some prime ideal containing $I$, but nof $f$. Proposition 2.2 in this link will do the trick: jmilne.org/math/xnotes/CA200.pdf. One benefit of nullstellensatz is it tells us that these abstract definitions make sense and are correct in the case of our geometric intuition for $k$-algebras. $\endgroup$ Mar 19, 2017 at 13:31
  • $\begingroup$ @user419669 nope. For any ideal $I$ of a ring $A$, the intersection of all prime ideals containing $I$ is equal to $\sqrt I$. $\endgroup$ Mar 19, 2017 at 13:34

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