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## The ideal-variety correspondence

I guess I didn’t plan this very well! Instead of completing one series I ended one and am right in the middle of another. Well, I’d really like to continue this series, but seeing as how finals are coming up I probably won’t be able to maintain the one-a-day pace. So I’ll just stop tagging MaBloWriMo.

Let’s summarize the story so far. $R$ is a commutative ring, and $X = \text{MaxSpec } R$ is the set of maximal ideals of $R$ endowed with the Zariski topology, where the sets $V(f) = \{ x \in X | f \in m_x \}$ are a basis for the closed sets. Sometimes we will refer to the closed sets as varieties, although this is mildly misleading. Here $x$ denotes an element of $X$, while $m_x$ denotes the corresponding ideal as a subset of $R$; the difference is more obvious when we’re working with polynomial rings, but it’s good to observe it in general.

We think of elements of $R$ as functions on $X$ as follows: the “value” of $f$ at $x$ is just the image of $f$ in the residue field $R/m_x$, and we say that $f$ vanishes at $x$ if this image is zero, i.e. if $f \in m_x$. (As we have seen, in nice cases the residue fields are all the same.)

For any subset $J \subseteq R$ the set $V(J) = \{ m | J \subseteq m \}$ is an intersection of closed sets and is therefore itself closed, and it is called the variety defined by $J$ (although note that we can suppose WLOG that $J$ is an ideal). In the other direction, for any subset $V \subseteq X$ the set $I(V) = \{ f | \forall x \in V, f \in m_x \}$ is the ideal of “functions vanishing on $V$” (again, note that we can suppose WLOG that $V$ is closed).

A natural question presents itself.

Question: What is $I(V(-))$? What is $V(I(-))$?

In other words, how close are $I, V$ to being inverses?

Basic properties

$I, V$ have the following elementary properties for any subsets $X_1, X_2$ of $X$ and subsets $J_1, J_2$ of $R$:

1. $V(J_1) \supseteq X_1 \Leftrightarrow J_1 \subseteq I(X_1)$
2. $V(I(X_1) \supseteq X_1$
3. $I(V(J_1) \supseteq J_1$
4. $V(J_1 \cup J_2) = V(J_1) \cap V(J_2)$
5. $I(X_1 \cup X_2) = I(X_1) \cap I(X_2)$
6. $V(J_1 \cap J_2) \supseteq V(J_1) \cup V(J_2)$
7. $I(X_1 \cap X_2) \supseteq I(X_1) \cup I(X_2)$.

Note that since the image of $I$ is always an ideal, it’s usually assumed that $J_1, J_2$ are ideals, in which case one should replace the union by the sum.

The first property, unpacked, says that every element of $J_1$ vanishes on $X_1$ if and only if every point in $X_1$ is sent to zero by $J_1$, and can be restated abstractly as saying that $I, V$ form an antitone Galois connection between the lattice of subsets of $R$ and the lattice of subsets of $X$. As we observed, this means equivalently that they are adjoint functors between the corresponding categories. $I, V$ restrict to a Galois connection between the smaller lattices of ideals and closed subsets, and even between the lattices of prime ideals and irreducible closed subsets.

The first property actually implies all the others; they’re true of any antitone Galois connection, and similar statements are true of any adjoint functors. The second and third properties can be strengthened into the statement that that $I(V(-)), V(I(-))$ are closure operators on their respective lattices.

The easy case

First we need to discuss a trivial lemma that I never explicitly stated. It should look very familiar from our discussion of locally compact Hausdorff and completely regular spaces.

Lemma: Let $x \in X$ be a point and let $Y \subset X$ be a closed set not containing $x$. Then there exists $f \in R$ such that $f \in I(Y)$ but $f \not \in m_x$, i.e. $R$ separates points and closed sets.

Proof. To say that $Y$ is closed is exactly say that $Y$ is the intersection of the zero sets of every element of $I(Y)$, so every $x \not \in Y$ is not contained in this intersection, which means there is some $f$ such that $f \not \in m_x$.

In other words, relative to $R$ the topology of $X$ behaves as if it were completely regular.

Corollary: For any $Y \subseteq X$, we have $V(I(Y)) = \overline{Y}$.

So the closure operator $V(I(-))$ is in fact the closure operator for the Zariski topology; in hindsight, this is true more or less by definition and can be taken to be the definition of the Zariski topology via the Kuratowski closure axioms.

The less easy case

Recall that if $g^n \in J$ for $g \in R$ and an ideal $J$, then $g^n$ lies in every maximal ideal containing $J$, hence so does $g$. In other words, $I(V(J))$, the intersection of all the maximal ideals containing $J$, contains the radical $\text{rad}(J)$ of $J$, which you’ll sometimes see written $\sqrt{J}$.

Proposition: The radical of an ideal is an ideal.

Proof. Closure under multiplication by $R$ is obvious, so it remains to check addition. Suppose $a^n \in J, b^m \in J$. By the binomial theorem, $(a + b)^{n+m-1}$ consists of only terms in which either $a^n$ appears or $b^m$ appears, so $(a + b)^{n+m-1} \in J$.

An ideal obtained in this way is called a radical ideal, and they are distinguished by the fact that $R/J$ is a reduced ring, i.e. has no nilpotents. Indeed, if we want to think of $R/J$ as the ring of functions on $V(J)$ like we did in the case $R = \mathbb{C}[x_1, ... x_n]$ earlier, then it shouldn’t have any nilpotents. Note that primes are radical.

We’ve already seen that $I(V(J))$ can be strictly larger than $\text{rad}(J)$, for example if $R$ is a local ring with non-maximal non-zero prime ideals. The problem with local rings geometrically is that a function can “vanish everywhere” (i.e. be contained in the maximal ideal) but not be nilpotent. The intersection of all the maximal ideals of a ring $R$ is called its Jacobson radical, which following the above one should think of as the set of “functions vanishing everywhere,” which for now are pathological and which we don’t care about.

Since any nilpotent element is contained in every maximal ideal, it follows that the Jacobson radical contains the nilradical $\text{rad}(0)$. So the problem with local rings algebraically is that the Jacobson radical can be larger than the nilradical. For Jacobson rings, it turns out that this is not true. More precisely, we have the following.

Proposition: a) The intersection of every prime ideal of a ring $R$ is the nilradical. In other words, the only elements of $R$ which “vanish on $\text{Spec } R$” are the nilpotent elements.

b) The intersection of every prime ideal containing an ideal $J$ is $\text{rad}(J)$.

c) $R$ is Jacobson if and only if the Jacobson radical of $R/J$ is equal to its nilradical for all $J$, if and only if $I(V(J)) = \text{rad}(J)$ for all $J$.

Proof. a) Any nilpotent element of $R$ will be in every prime ideal, so one inclusion is clear. The proof of the other inclusion is an example of a classic strategy in commutative algebra. Suppose $f \in R$ is not nilpotent, and consider the set $\mathcal{J}$ of ideals $J$ such that $f^n \not \in J$ for any positive integer $n$, which is not empty since it contains $(0)$. The union of any elements of a chain in $\mathcal{J}$ is again in $\mathcal{J}$, so Zorn’s lemma applies and there is a maximal such ideal $P$.

We claim that $P$ is prime. Indeed, suppose $x, y \not \in P$. The ideals $P + (x), P + (y)$ contain some power of $f$ by the assumption that $P$ was maximal, so $f^n \in P + (x), f^m \in P + (y)$, hence $f^{m+n} \in P + (xy)$, so $P + (xy)$ cannot lie in $\mathcal{J}$. It follows that $xy \not \in P$.

b) By the correspondence theorem, the ideals of $R/J$ are the ideals of $R$ containing $J$. If $P$ is a prime ideal containing $J$ then $R/P$ is an integral domain, hence so is the quotient of $R/J$ by the image of $P$. It follows that the prime ideals of $R/J$ are the prime ideals of $R$ containing $J$, and now we can apply a), noting that the nilradical of $R/J$ is precisely the image of $\text{rad}(J)$.

c) $R$ is Jacobson if and only if the intersection of the prime and maximal ideals containing a given ideal always coincide.

Interpretation

The relation $I(V(J)) = \text{rad}(J)$ is a little abstract as stated. Unpacked, it says the following: let $R$ be a Jacobson ring, let $f_1, f_2, ... f_r \in R$ generate an ideal $J$, and let $g \in R$. Then exactly one of the following is true:

• $g \in \text{rad}(J)$, i.e. there exist $h_1, ... h_r \in R$ and $n \in \mathbb{N}$ such that $\displaystyle g^n = \sum_{i=1}^{r} h_i f_i$.
• $g \not \in I(V(J))$, i.e. there exists a point $x \in V(J)$ at which $g$ doesn’t vanish.

If $R$ is also Noetherian then this is true of every ideal. This is very useful because if $r$ is large then $J$ gets more complicated but $V(J)$ gets simpler, and it is usually easier to check if $g$ vanishes on $V(J)$ than to check if $g^n$ is a linear combination of a large number of $f_i$.

The introduction of radical ideals allows us to take a first stab at factorization. Define a minimal prime over an ideal $J$ to be the preimage of a minimal prime of $R/J$, or equivalently a prime containing $J$ which is minimal with respect to this property. Then if $J$ is radical, the intersection of the prime ideals of $R/J$, hence the intersection of the minimal prime ideals of $R/J$, is zero. It follows that $J$ is the intersection of the minimal prime ideals over $J$, and this representation is necessarily unique because it involves the entire set of minimal prime ideals. If $R$ is Noetherian and Jacobson, there are only finitely many such primes, and they are precisely the irreducible components of $V(J)$! So at least for radical ideals there is a more-or-less clear geometric picture of what’s going on.

At this point we run into the issue that, in a setting where nonzero prime ideals are allowed to contain each other, prime factorization can only make sense if we restrict it to minimal primes, which have the unfortunate property that they depend on the ideal being considered. This is an issue of dimension, which we’ll discuss later.

Now, as nice as these results are, they don’t help us until we actually show that the rings we care about are Jacobson. The standard way of doing this from the weak Nullstellensatz makes the most sense in the context of localization, which we’ll discuss next.