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One of the equivalent forms in definition of a Noetherian Jacobson ring $R$ is that $R$ has no prime ideals $P$ such that $R/P$ is a 1-dimensional semi-local ring.

When $R/P$ has dimension 1, it means that between $R$ and $P$ there is at most one prime ideal, and as $R/P$ is semi-local, there are only finitely many maximal ideals of $R$ containing $P$. Do not these two imply that $P$ lies only in one maximal ideal of $R$?

Incidentally, what is a proof for equivalence of the above-mentioned statement with $R$ being Jacobson? Thanks for any cooperation.

karparvar
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There are Noetherian semi-local domains of dimension $1$ that are not local, e.g. $S^{-1}\mathbb{Z}$, where $S = \mathbb{Z} \setminus ((2) \cup (3))$. Probably the line of reasoning you meant is this: a Jacobson domain of dimension $1$ cannot be semilocal. If it were, then $0$ would be an intersection of maximal ideals, and if the intersection is finite, then $0$ would equal one of the maximal ideals, contradicting dimension $1$.

Since quotients of Jacobson rings are Jacobson, this shows that $R$ Jacobson $\implies R$ has no primes $P$ with $R/P$ semi-local of dimension $1$. For the converse, assuming $R$ is Noetherian, one can use the equivalence (Lemma 4.20 in Eisenbud, called the Rabinowitch trick)

i) $R$ is Jacobson
ii) If $P$ is a prime in $R$ with $(R/P)[f^{-1}]$ a field for some $f \in R \setminus p$, then $R/P$ is a field

along with the following proposition (see this question):

Proposition: Let $R$ be a Noetherian domain, not a field. Then $R$ is semi-local of $\dim 1$ iff $R_f$ is a field for some $f \in R$.

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zcn
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Let $R$ be a noetherian ring. Then $R$ is Jacobson if and only if for every prime ideal $P$ of $R$ such that $\dim R/P=1$ the ring $R/P$ has infinitely many maximal ideals.

"$\Rightarrow$" Actually any homomorphic image of dimension $\ge 1$ of a Jacobson ring has infinitely many maximal ideals. Since homomorphic images of Jacobson rings are also Jacobson, we can start with a Jacobson ring $R$ of dimension $\ge1$, and prove that $R$ has infinitely many maximal ideals. Suppose the contrary, and let $P$ be a prime ideal of $R$ which is not maximal. Then $P$ is a finite intersection of maximal ideals, and therefore $P$ itself is maximal, a contradiction.

"$\Leftarrow$" If $R$ isn't Jacobson, then there is a prime ideal which is not an intersection of maximal ideals. Since $R$ is noetherian we can choose a prime $P$ maximal with this property. In particular, $P$ isn't maximal, so $\dim R/P\ge 1$.
If $\dim R/P=1$, then, by hypothesis, the intersection of its maximal ideals is $(0)$. (Otherwise, taking a non-zero element $x$ in the intersection, and using the primary decomposition of $(x)$ we find that $x$ is contained in a finite number of maximal ideals, contradiction.) Thus $P$ is an intersection of maximal ideals, and this contradicts the choice of $P$.
If $\dim R/P>1$, then there are infinitely many height one prime ideals in $R/P$ and (by a similar argument to the one in the previous paragraph) their intersection is $(0)$, that is, $P$ is an intersection of larger primes. From the choice of $P$ we deduce that $P$ is an intersection of maximal ideals, contradiction.

user26857
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  • An alternative approach for $\Leftarrow$: In a domain $R$, if $x \ne 0$ is contained in every height $1$ prime, then ${\text{ht 1 primes}} \subseteq \operatorname{Min}(x)$. If $R$ is Noetherian of dim $> 1$, then ${\text{ht 1 primes}}$ is infinite while $\operatorname{Min}(x)$ is finite, and if $\dim R = 1$, then ${\text{ht 1 primes}} = \operatorname{mSpec}(R)$. – zcn Jun 20 '14 at 18:49
  • @zcn Why $Min(x)$ is finte? And what is the contradiction in the case $dim R=1$? – karparvar Jun 02 '15 at 08:26
  • @karparvar In a noetherian ring the set of minimal primes over an ideal is always finite (they are among the associated primes). (Unfortunately user zcn is no longer active on M.SE.) – user26857 Jun 02 '15 at 13:52