Recall that a ring is called Jacobson if the radical of an ideal in the intersection of the maximal ideals that contains it (this is always true with prime ideals).

$K[[x]]$ is not Jacobson.

I know that this ring is local with $(x)$ as its only maximal ideal, so I should find a prime ideal not containing $(x)$, any hint or suggestion?

  • 3
    $\begingroup$ $K[[x]]$ is an integral domain, so... $\endgroup$ – Qiaochu Yuan Jan 9 '13 at 23:52
  • $\begingroup$ Oh my bad, its quite trivial $\endgroup$ – user56741 Jan 9 '13 at 23:58
  • $\begingroup$ Still thank you so much $\endgroup$ – user56741 Jan 9 '13 at 23:59

To resolve this question: $\{0\}$ is what you're looking for.

Actually it's the only other prime ideal in the entire ring since the nontrivial ideals all look like $(x^n)$.

Thinking about this a little, you can see that a local ring is Jacobson if and only if its primes are maximal. In particular, local rings with Krull dimension of $1$ or more can't be Jacobson.

You can also generalize this away from this local principal ideal domain example to Dedekind domains. Since nonzero primes are maximal in Dedekind domains, the only question that remains to be answered is "is the zero ideal an intersection of maximal ideals or not?"

If a Dedekind domain has finitely many maximal ideals, the intersection of all of these ideals is nonzero, and so no combination of ideals can intersect to zero. In this case, it is not Jacobson.

On the other hand, suppose we have a Dedekind domain with infinitely many prime ideals. If the intersection of all primes is nonzero, then it has a unique factorization in terms of finitely many powers of prime ideals. But since "divides mean contains," and all prime ideals contain (divide) this ideal, all prime ideals would have to appear in its factorization (but there are too many.) Thus the intersection of all prime ideals would have had to have been zero in the first place, and the ring is Jacobson.


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