I am following Michael Atiyah commutative algebra. In one of the proposition he proved $\operatorname{nil}(R) = \cap P$ where $P$ is prime ideal. One can easily see the inclusion $\operatorname{nil}(R) \subset P$. However, for the other inclusion we used Zorn lemma. I understand the proof, but I don't really get it conceptually speaking. Is there a geometrical way to think about it?
Nilradical geometrical way of thinking about it
4
$\begingroup$
abstract-algebra
algebraic-geometry
commutative-algebra
-
1Considerations of geometry aside, I found the proof of this theorem much easier to remember after I had learned the tools of localization. In particular, there is an elegant, short proof of this theorem that relies on the relationship between prime ideals of a ring $A$ and the prime ideals of a localization $S^{-1}A$. You might (as I did) find this proof more memorable/intuitive, so it may be worth coming back to this theorem after you have learned about localizations. – 2017-02-22
1 Answers
3
At least in the case when $R$ is a finitely generated algebra over an algebraically closed field, there is a nice way of seeing this geometrically. In this setting, the proposition says that an affine variety is the union of its irreducible subvarieties.
Consider an ideal $J$ in the polynomial ring $S=k[x_1,...,x_n]$. By looking in the quotient ring $S/J$, the statement becomes: $\sqrt{J}$ is the intersection of all primes containing $J$.
Now thinking about classical affine algebraic geometry, by Nullstellensatz we have $I(V(J))=\sqrt{J}$. Thus using the "geometric version" of the proposition, we have $$\sqrt{J}=I(V(J))=I(\bigcup_{V(\mathfrak{p})\subseteq V(J)} V(\mathfrak{p}))=\bigcap_{\mathfrak{p}\supseteq J} \mathfrak{p}$$
-
0$V(J)=V(\sqrt J)$ is not the Nullstellensatz, it ist just a trivial observation from the definition of $V(-)$. – 2017-02-22
-
0@MooS True. I suppose what I meant to imply by invoking Nullstellensatz was the somewhat intuitive idea that $V(\sqrt{J})$ is the "correct" form of $V(J)$, and hence the fact that a variety is the union of its irreducible subvarieties is equivalent to the fact that $\sqrt{J}$ is the intersection of the prime ideals containing it. – 2017-02-22
-
0Yes, one can deduce the statement (for the case of affine $k$-algebras with $k$ algebraically closed) from Nullstellensatz. But I am still confused from your answer, since you never used Nullstellensatz. Basically you should at some point say $$\sqrt{I}=I(V(I)) = I(\bigcup_{P \supset I} V(P)) \supset \bigcap_{P \supset I} P.$$ That would have nailed it. – 2017-02-22
-
0@MooS Good point. That probably would have been a better way of phrasing it. Thank you for the tip. – 2017-02-22
-
0Maybe you just add it. I will delete my answer. – 2017-02-22