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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 $\mathrm{char}(R)= p$?

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  • $\begingroup$ Am I missing something here? The statement you want to prove seems to be false... What about $\Bbb F_p\subseteq \Bbb F_p[\![x]\!]$? $\endgroup$ – Stahl Sep 22 '18 at 21:00
  • $\begingroup$ @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. 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? $\endgroup$ – Mike Pierce Sep 22 '18 at 21:23
  • $\begingroup$ @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. $\endgroup$ – Stahl Sep 22 '18 at 21:50
  • $\begingroup$ @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! :) $\endgroup$ – Mike Pierce Sep 22 '18 at 22:01
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The reason the characteristic of the local ring $R$ will be a prime power if it is not equal to $0$ is that the maximal ideal must contain all zero divisors (consider the image of a zero divisor in $R/\mathfrak{m}$ since the maximal ideal in a local ring contains all nonunits). If the characteristic is $d=ab$, with $\gcd(a,b)=1$, then both $a$ and $b$ must lie in $\mathfrak{m}$, hence $1\in\mathfrak{m}$, which is impossible. Thus, the characteristic must be either $0$ or a prime power.

If $\mathrm{char}(R) = 0$ and $\mathrm{char}(K)=p\gt 0$, then $R$ cannot contain a field: if it contained one, the field would contain $1$, hence $\mathbb{Z}$, hence $\mathbb{Q}$, but then $p$ would be a unit lying in $\mathfrak{m}$, which is impossible.

If $\mathrm{char}(R) = p^m$, $m\gt 0$, and $\mathrm{char}(K)=p$, then $R$ cannot contain a field: the field would necessarily contain $1$ and be of characteristic $p$ (since the characteristic of a subring must divide the characteristic of the ring), but then the subring generated by $1$ would be $\mathbb{Z}/p\mathbb{Z}$; however, the condition that $\mathrm{char}(R)=p^m$ means that the additive order of $1$ in $R$ is $p^m$, which shows $R$ cannot contain a field.

Added. As pointed out in comments, in fact the field wold contain $1$, and hence be of the same characteristic as $R$ (since the characteristic equals the additive order of $1$).

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  • $\begingroup$ Thank you, Arturo. I understand all of your answer except the paranthetical comment in the last paragraph - since the characteristic of a subring must divide the characteristic of the ring. Shouldn't the characteristic of any subring of a ring be the same as that of the ring. This would still give us the contradiction we are seeking, since if $R$ contained a field, it will have a characteristic which is $p^m$ with $m>1$ but fields can only have a characteristic which is $0$ or a prime. $\endgroup$ – Brittany Murphy Jun 30 '11 at 11:59
  • $\begingroup$ @Brittany: There is actually a specious argument before (the reason $\mathfrak{m}$ contains all zero divisors is that it contains all non-units). As to your comment: if your subrings must include $1$, then you are correct, as the characteristic of the ring is the additive order of $1$. But if subrings need not contain $1$, the characteristic is the smallest $n$ such that $na=0$ for all $a\in R$, so if you quantify over a subring, the set of possible $n$ is potentially larger, so the characteristic could be smaller (but must divide it anyway); think $Z_3$ in $Z_3\times Z_2\cong Z_6$. $\endgroup$ – Arturo Magidin Jun 30 '11 at 14:37
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    $\begingroup$ @Arturo Magidin In your second case, we should say that if $m>1$, then $R$ cannot contain a field; but for $m=1$ it's possible, for instance, $R$ could be $\mathbb{Z}/p\mathbb{Z}$. $\endgroup$ – Lao-tzu Jan 29 '17 at 10:48
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    $\begingroup$ @Lao-tzu Or even worse, if you think fields don't count as a counterexample: $R$ could be something like $\Bbb F_p[\![x]\!].$ $\endgroup$ – Stahl Sep 22 '18 at 21:02
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The result you want is that $R$ contains a field $K$ iff $\operatorname{char}(R) = \operatorname{char}(R/\mathfrak{m}).$

$\implies$ If $K\subseteq R,$ then clearly $K$ and $R$ must have the same characteristic, as $1 + 1 + \dots + 1 = 0$ in $K$ if and only if $1 + 1 + \dots + 1 = 0$ in $R.$ Moreover, the composition $K\to R\to R/\mathfrak{m}$ gives a homomorphism of fields $K\to R/\mathfrak{m},$ so that $\operatorname{char}(R/\mathfrak{m}) = \operatorname{char}(K) = \operatorname{char}(R).$

$\impliedby$ Conversely, suppose $\operatorname{char}(R) = \operatorname{char}(R/\mathfrak{m}).$ There are two cases to consider.

  1. $\operatorname{char}(R) = \operatorname{char}(R/\mathfrak{m}) = 0:$ If the characteristic is $0,$ then every integer $n$ is invertible in $R$. Indeed, $n$ is invertible if and only if $n\not\in\mathfrak{m}.$ If $n\in\mathfrak{m},$ then $n = 0$ in $R/\mathfrak{m}.$ But this implies that $\operatorname{char}(R/\mathfrak{m})\mid n.$ Thus, there exists a map $\phi : \Bbb Q = (\mathbb{Z}\setminus\{0\})^{-1}\Bbb Z\to R$ by the universal property of localization, which is necessarily an injection.
  2. $\operatorname{char}(R) = \operatorname{char}(R/\mathfrak{m}) = p:$ In this case, $R$ contains a copy of $\Bbb F_p.$ (Just map $\Bbb F_p\to R$ by $1\mapsto 1,$ and verify that this is a well-defined ring homomorphism and an injection.)
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