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?

  • 1
    $\begingroup$ 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

1 Answer 1


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 $I(Z(J))=\sqrt J$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.