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I am looking for a proof of the result that appears at the end of page 418 here indicating the possibilities for equal characteristic and mixed characteristic.

I could prove the following: Let $(R,m,K=R/m)$ be a local ring.

1) If char $K=0$, then, char $R=0$. So $R$ contains a copy of $\mathbb{Z}$. Now, since $R/m$ has char $0$, none of the images of non-zero integers in $R$ can be in $m$, else their images in $R/m$ will be zero contradicting char $K=0$. Then by the universal property of localization, $R$ contains $\mathbb{Q}$.

2) If char $K=p$, then either char $R$ is $0$ or non-zero (I don't understand why it should be a prime power if it's non-zero. If char $R=p>0$ where $p$ is a prime, then, $R$ contains a copy of $\mathbb{Z}/p\mathbb{Z}$.

So I guess the proof now reduces to proving the question I raise in the paranthesis and why $R$ cannot contain a field in mixed characteristic. Any help will be appreciated.

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It would help if you quoted the question from the book instead of linking to the book. –  Qiaochu Yuan Jun 30 '11 at 4:35

1 Answer 1

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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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

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