3
$\begingroup$

I'm trying to prove that a local ring $(R,\mathfrak{m})$ contains a field if and only if $\mathrm{char}(R)$ and $\mathrm{char} (R/\mathfrak{m})$ are equal. To this we must relate the characteristic of $R$ to the characteristic of it's residue field $K=R/\mathfrak{m}$. If $\mathrm{char}(K)=0$, then, $\mathrm{char}(R)=0$. So $R$ contains a copy of $\mathbb{Z}$. Now, since $R/\mathfrak{m}$ has characteristic $0$, none of the images of non-zero integers in $R$ can be in $\mathfrak{m}$, else their images in $R/\mathfrak{m}$ will be zero, contradicting $\mathrm{char}(K)=0$. Then by the universal property of localization, $R$ contains $\mathbb{Q}$.

Now for the converse, I know that if $\mathrm{char}(K)=p$, then $\mathrm{char}(R)$ is $0$ a power of $p$, but I don't know how to completely prove this part. Why can't the characteristic be some non-prime integer? And furthermore, why can't $R$ contain a field in any case besides when $\mathrm{char}(R)= p$?

  • 0
    Am I missing something here? The statement you want to prove seems to be false... What about $\Bbb F_p\subseteq \Bbb F_p[\![x]\!]$?2018-09-22
  • 0
    @Stahl It might be false. The original thing that the OP wanted to prove some statement behind a link, but the link was dead. See the [original post](https://math.stackexchange.com/revisions/48595/1). I tried to "reverse engineer" what the actual question was based on OPs writing and the answer below. Do you have any idea what the actual claim might have been?2018-09-22
  • 0
    @MikePierce I didn't realize how old this question was, I'm not sure why it popped up in my feed just now! In any case, I think the correct statement is "$R$ contains a field iff the characteristic of $R$ is the same as the characteristic of the residue field," and I've posted a proof of this below.2018-09-22
  • 0
    @Stahl It popped up because I edited the post. Editing a questions will (sometimes unfortunately) bump the post to the front page. In this case its a good thing that it got bumped and you saw it since I got it wrong, and Arturo's answer wasn't quite right. Thank you for posting an answer! :)2018-09-22

2 Answers 2