Let $R$ be a commutative ring. Let's say that the Jacobson radical $J(R)$ of $R$ is uninteresting if

  1. $J(R)$ coincides with the nilradical, or

  2. $J(R)$ is the intersection of a finite number of maximal ideals.

It seems as if most rings used in algebraic geometry have uninteresting Jacbson radical:

  • Every finitely-generated commutative algebra over a field or over a Dedekind domain is Jacobson, so its Jacobson radical coincides with its nilradical, and so is uninteresting by (1).

  • Every local or semilocal commutative ring has finitely many maximal ideals, and so has uninteresting Jacobson radical by (2).

For a nonuninteresting example, take the localization $R = \mathbb Z[x]_S$ where $S = \{f(x) \in \mathbb Z[x] \mid f(0) = 1\}$. The maximal ideals of $R$ are of the form $(p,x)$ where $p \in \operatorname{Spec} \mathbb Z$. So the Jacobson radical is $(x)$, which is not uninteresting.

But this example seems rather artificial to me; for example I don't know anywhere a ring like this would show up in algebraic geometry.

  • 8
    $\begingroup$ Sure, tons of these arise when completing along a non-maximal non-nilpotent ideal. Let $O$ be a complete local noetherian ring with non-nilpotent maximal ideal $m$ and $m$-adically complete $O[x_1,\dots,x_n]$ with $n>0$. We get the ring $O\{x_1,\dots,x_n\}$ of formal power series $\sum a_I x^I$ whose coefficients $a_I$ tend to 0 in $O$ as $|\!|I|\!| \to \infty$. The Jacobson radical is $m\{x_1,\dots,x_n\}$, which is not the nilradical (since $m$ isn't nilpotent), whereas finite intersections of maximal ideals correspond to finite sets of closed points in $\mathbf{A}^n_{O/m}$ (with $n>0$). $\endgroup$
    – nfdc23
    Mar 7 '18 at 6:30
  • 1
    $\begingroup$ Oh wow -- I had discounted such examples out of the misconception that the completion of a ring at a (prime) ideal always factors through the localization at that ideal (and hence is local, with uninteresting Jacobson radical), which I now see is very much false when the ideal is not maximal. Rings like this are primarily important in deformation theory, right? $\endgroup$
    – Tim Campion
    Mar 7 '18 at 6:59
  • 2
    $\begingroup$ @nfdc23: Why not make that comment an answer? $\endgroup$ Sep 13 '19 at 8:13

Rings similar to the ring $R$ as you consider above, "show up" in real algebraic geometry. If $k:=\mathbb{R}, A:=k[x]$ is the ring of polynomials with real coefficients, and $S\subseteq A$ is the set of polynomials with no real roots ($f(x):=x^2+1$ is such a polynomial), then $B:=S^{-1}A$ may be seen as the global sections of the "structure sheaf" of $X(k):=\mathbb{A}^1_k(k):=Spec(A)(k)$ - the "affine real line" in the sense of real algebraic geometry (this is "vague"). You may construct a locally ringed space $(X(k), \mathcal{O}_{X(k)})$, where $X(k)$ is the $k$-rational points of $X$ and where $\mathcal{O}:=\mathcal{O}_{X(k)}$ i a sheaf on $X(k)$ with global sections equal to $B$.

Example 1. The rational function $f(x):=\frac{a(x)}{x^2+1}$ is a regular function on $X(k)$ and gives rise to a global section of $\mathcal{O}$ for any polynomial $a(x) \in A$. If $\mathfrak{m}$ is a maximal ideal in the localized ring $B$, it follows the residue field $\kappa(\mathfrak{m})\cong k$ is the field of real numbers. Hence when you localize at $S$ you have removed all maximal ideals in $A$ with residue field the complex numbers.

Example 2. The global sections $\Gamma(X, \mathcal{O}_X)=A$ is the polynomial ring, hence the real algebraic variety $X(k)$ has "more global sections" than the affine scheme $X$. Again this is "vague" since $X(k)$ is a subspace of $X$.

You may similarly construct the real projective line $\mathbb{P}(1):=\mathbb{P}^1_k(k)$, its structure sheaf $\mathcal{O}_{\mathbb{P}(1)}$ and the real affine space $\mathbb{A}(n)$ and you may embed $\mathbb{P}(1)$ as a "closed subvariety" (in the sense of real algebraic geometry)

$$ \phi: \mathbb{P}(1) \rightarrow \mathbb{A}(3)$$

of affine $3$-space.

The above approach to "real algebraic geometry" uses the classical language of "schemes" and introduce a real algebraic variety using the $k$-rational points of the scheme $\mathbb{A}^1_k$. There are other approaches and you find references at the zentralblatt site under "real algebraic geometry". The apporoach indicated above gives for any scheme $X$ of finite type over $k$, a "real algebraic variety" $(X(k), \mathcal{O}_{X(k)})$, which is a locally ringed space where the structure sheaf $\mathcal{O}_{X(k)}$ has "more" local sections. In the case of projective space $\mathbb{P}^n_k(k)$ you get a space with properties similar to the underlying smooth manifold. You may (similar to the situation with differentiable manifolds and the Whitney embedding theorem) embed $\mathbb{P}^n_k(k)$ into "real affine space" $\mathbb{A}^d_k(k)$. In fact I believe there is an endofunctor $F$ of the category of schemes over $k$ with the property that for any projective scheme $X \subseteq \mathbb{P}^n_k$ it follows the scheme $F(X):=X(k)$ is affine: There is a closed embedding $F(X) \subseteq \mathbb{A}^d_k(k)$. The scheme $F(X)$ is no longer of finite type over $k$.

Example 2. A motivation for doing this is the study of the Weil restriction $W(X)$ of a complex projective manifold $X \subseteq \mathbb{P}^n$. In the case when the Weil restriction $W(X) \subseteq \mathbb{P}^m_k$ is projective over $k$ you get in a functorial way an "affine real algebraic variety" $F(W(X))=Spec(B)$, and any complex holomorphic vector bundle $E$ on $X$ gives a finite rank projective $B$-module $E(k)$.

You may have heard of the "Jouanolou-Thomason trick", which to any quasi projective variety $X$ of finite type over a field $k$ associates an affine variety $Y$ of finite type over $k$ with $K_i(Y)=K_i(X)$. With the above construction you get a similar construction for complex projective varieties which is functorial.

  • $\begingroup$ Thanks! One of my favorite things about MO is that quite apart from getting answers to the questions I think I have, it gives me a chance to air out the assumptions going into why I have those questions, when folks do exactly this and correct my assumptions! You emphasize that these considerations in real algebraic geometry are "vague", but the way you put it, it seems totally precise to me, at least in these examples. Is there anywhere this perspective is developed some more? $\endgroup$
    – Tim Campion
    Feb 10 at 14:20
  • $\begingroup$ I was getting this question confused with another question of mine -- I now see that in addition to the preconception-busting, this directly answers the question! $\endgroup$
    – Tim Campion
    Feb 10 at 14:23

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.