# $K[[x]]$ is not a Jacobson ring

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?

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