6
$\begingroup$

In Wikipedia it is claimed that "A ring is called a Jacobson ring if the nilradical coincides with the Jacobson radical." Here the word "ring" means a commutative ring.

However, I remember that Jacobson ring is defined to be a ring with the property "every prime ideal is equal to the intersection of some maximal ideals".

Are these two definitions really equivalent?

$\endgroup$
  • 1
    $\begingroup$ I went ahead and fixed the wiki article, btw. It turns out that it was a relatively recent contribution. $\endgroup$ – rschwieb Apr 3 '14 at 15:34
12
$\begingroup$

No. What abx said is exactly correct.

Consider the ring of polynomials in countably many variables over $\mathbb Q$. Consider a surjective homomorphism to the localization of $\mathbb Q[x]$ at $x=0$. The kernel is prime, but is not an intersection of maximal ideals, so the ring is not Jacobson.

However, the Jacobson radical is the zero ideal, which is the nilradical. Indeed, any nonzero element of this ring is a polynomial in finitely many variables. Take a $\mathbb Q$-rational point in affine space that is not a solution to that polynomial equation, and assign those variables to the coordinates of that point. Then assign the other variables to $0$. This gives a homomorphism to $\mathbb Q$ where the polynomial is nonzero, hence a maximal ideal that does not contain that element.

$\endgroup$
6
$\begingroup$

Say a Jacobson ring is one in which each prime ideal is an intersection of maximal ideals. And also let's call $R$ "rad-nil" if $J(R)=nil(R)$. (This is something I picked up in Lam's First course in noncommutative rings, and I don't know if it's used elsewhere, really.)

"Jacobson" indeed implies that rad-nil, since it implies $J(R)\subseteq nil(R)$.

However, the converse isn't true. To show this, just pick a countable local domain which isn't a field (say, $S=\Bbb Q[x]_{(x)}$) and map onto it from $R=\Bbb Q[x_1,x_2,\ldots]$, the polynomial ring in countably many variables. For this homomorphism $f$, we have $R/P\cong S$ where $P=\ker f$ is a prime ideal.

Now $R$ certainly has the property $J(R)=nil(R)$ since $J(R)=\{0\}$.

On the other hand, since $R/P$ is local but not a field, there is only one maximal ideal in $R/P$ containing $P$, so $P$ is not an intersection of maximal ideals of $R$.

So despite $R$ being rad-nil, it is not Jacobson.


The problem is fixed if we say "$R$ is Jacobson iff $R/P$ is rad-nil for every prime ideal $P$ of $R$." This is more or less obvious since "$R/P$ rad-nil" says exactly that $P/P$, the nilradical, is the intersection of $M/P$ where $M$ ranges over all maximal ideals of $R$ lying above $P$, and so $P$ is an intersection of maximal ideals.

Here's the statement that appears in Bourbaki's Algebra Chs 1-7 book in section 3.4 on Jacobson rings visible in googlebooks

Notice they say in Proposition 3 "every ideal":

enter image description here

This is also true since each prime lying over $I$ can be replaced with an intersection of maximal ideals of $R$ over $I$.

So, we have the following nice equivalences:

  1. $R$ is Jacobson

  2. $R/P$ is Jacobson-semisimple (=semiprimitive) for all prime ideals $P\lhd R$

  3. $R$ is "totally rad-nil" (=all quotients are rad-nil)

$\endgroup$
6
$\begingroup$

$\newcommand{\mm}{\mathfrak{m}}$The counterexamples given so far seem perhaps more elaborate than necessary. Here's another:

Let $A$ be a domain which is not a field and which has finitely many maximal ideals $\mm_1,\ldots,\mm_n$. (So e.g. take $\mathbb{Z}_p$ or $\mathbb{C}[[x_1,\ldots,x_n]]$ or $\mathbb{F}_{17}[[t^2,t^3]]$ or...you have some leeway here.) Let $R = A[t]$. Then:

$\bullet$ Like every domain, $R$ has zero nilradical.
$\bullet$ Like every polynomial ring over a domain, $R$ has zero Jacobson radical: if $0 \neq x \in R$, then $1+tx \notin A^{\times} = R^{\times}$.
$\bullet$: The ideal $\mathfrak{p} = \langle t \rangle$ of $R$ is prime. The maximal ideals of $R$ containing $\mathfrak{p}$ correspond to the maximal ideals $\mm_1,\ldots,\mm_n$ of $R/\mathfrak{p} = A$, so $\bigcap_{i=1}^n \mm_i \supset \mm_1 \cdots \mm_n \supsetneq (0)$. Thus $\mathfrak{p}$ is not the intersection of the maximal ideals containing it.

It may be helpful to mention the (easy) fact that a ring $R$ is Jacobson iff for all ideals $I$ of $R$, the nilradical of $R/I$ equals the Jacobson radical of $R/I$. This simplifies the last bullet point and is really the point of the construction. In fact for any ring $A$, the Jacobson radical of $A[t]$ is equal to the nilradical of $A[t]$ (Atiyah-Macdonald, Exercise 1.4), so taking any $A$ with nilradical smaller than its Jacobson radical works to give a counterexample: e.g. a domain which not a field and which has only finitely many maximal ideals.

$\endgroup$
  • $\begingroup$ It seems to me that, what you said in the last paragraph contradicts what's said in the wiki page about Amitsur's theorem. $\endgroup$ – Yai0Phah Apr 22 '16 at 22:06
-1
$\begingroup$

Use the fact that nilradical is intersection of prime ideals

$\endgroup$
  • 3
    $\begingroup$ So what? I believe the Wikipedia article just forgot for every quotient ring the nilradical coincides with the Jacobson radical. $\endgroup$ – abx Dec 6 '13 at 13:14

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.