# The local ring $(R,\mathfrak{m})$ contains a field if and only if $\mathrm{char}(R) = \mathrm{char} (R/\mathfrak{m})$.

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

• Am I missing something here? The statement you want to prove seems to be false... What about $\Bbb F_p\subseteq \Bbb F_p[\![x]\!]$? – Stahl Sep 22 '18 at 21:00
• @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? – Mike Pierce Sep 22 '18 at 21:23
• @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. – Stahl Sep 22 '18 at 21:50
• @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! :) – Mike Pierce Sep 22 '18 at 22:01

## 2 Answers

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 $1$).

• 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. – Brittany Murphy Jun 30 '11 at 11:59
• @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 $Z_3\times Z_2\cong Z_6$. – Arturo Magidin Jun 30 '11 at 14:37
• @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 $\mathbb{Z}/p\mathbb{Z}$. – Lao-tzu Jan 29 '17 at 10:48
• @Lao-tzu Or even worse, if you think fields don't count as a counterexample: $R$ could be something like $\Bbb F_p[\![x]\!].$ – Stahl Sep 22 '18 at 21:02

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.)