# Understanding of connection between $\operatorname{Spec}k[x_1,\dots,x_n]$ and $k[x_1,\dots,x_n]$

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?

• 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. Mar 19, 2017 at 13:10

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 $I(Z(J))=\sqrt J$.
• 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? Mar 19, 2017 at 13:27
• 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. Mar 19, 2017 at 13:31
• @user419669 nope. For any ideal $I$ of a ring $A$, the intersection of all prime ideals containing $I$ is equal to $\sqrt I$. Mar 19, 2017 at 13:34