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Let $p$ be a prime number and $A$ be a commutative ring with unity. We say that $A$ has characteristic $p$ if $p\cdot 1_A=0$. I would like to know if you could have a ring $A$ with all residue fields (= $\operatorname{Frac}(A/\mathfrak{p}$) with $\mathfrak{p}$ a prime ideal) of characteristic $p$ but $A$ itself not being of characteristic $p$.

Abellan
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2 Answers2

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You can take $\mathbb{Z}/4\mathbb{Z}.$ This has characteristic $4$, the only prime is $(2)$ and the residue field $\mathbb{F}_2$ is of characteristic $2$.

EDIT: You may also say something positive (but not really surprising either):

If $A$ is an integral domain such that all the residue fields are of equal characteristic $p$, then $\mathrm{char}(A)=p$.

This is because of the prime ideal $(0)$: in this case, we have $A \subseteq \mathrm{Frac}(A)=\mathrm{Frac}(A/(0)),$ and the claim follows.

Pavel Čoupek
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Have you think in $\mathbb{Z}_{(p)}=\{\frac{a}{b}\mid p\nmid b\}$?. This ring has charateristic $0$. But it is a local ring, with unique maximal ideal $p\mathbb{Z}_{(p)}$. So its residue field $\mathbb{Z}_{(p)}/p\mathbb{Z}_{(p)}$ is isomorphic to $\mathbb{Z}_p$ the integeres modulo p.

Math.mx
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  • There is one more residue field - $\mathbb{Q}$. – Pavel Čoupek Nov 08 '16 at 17:34
  • Which is the maximal ideal that gives you $\mathbb{Q}$ as quotient ring? – Math.mx Nov 08 '16 at 17:37
  • Not maximal, but the prime ideal $(0)$. : ) – Pavel Čoupek Nov 08 '16 at 17:38
  • (0) gives you the original ring, which is a subring of the rationals, morover, this ring is local, meaning that it has an unique maximal ideal, so $(0)$ cannot be a maximal ideal. – Math.mx Nov 08 '16 at 17:39
  • No one is saying that the ideal is maximal. Read the OP's description of what he means by a residue field, if it does not coincide with your definition (the OP's definition seems standard, though). And $(0)$ cannot give you $\mathbb{Z}{(p)}$ as the residue _field - $\mathbb{Z}_{(p)}$ is not a field. – Pavel Čoupek Nov 08 '16 at 17:42
  • Your are saying that in $\mathbb{Z}{(p)}$ there is another residue field $\mathbb{Q}$. First, there is no way to get $\mathbb{Q}$ as quotient of $\mathbb{Z}{(p)}$. You said that is using the ideal $(0)$, and you're claiming that is prime but not maximal, but as the quotient is a field, it must be maximal. It seems I lost some information. I have noticed that for residue field the question says prime ideal, but I assuming that a residue field is a field? Is that correct? =P – Math.mx Nov 08 '16 at 17:51
  • By definition, a residue field of a commutative ring $A$ at a prime ideal $\mathfrak{p}$ is the field of fractions of the integral domain $A/\mathfrak{p}$. Taking $A= \mathbb{Z}{(p)}$ and $\mathfrak{p}=(0),$ this yields $A/\mathfrak{p}=\mathbb{Z}{(p)}$, but the fraction field of this is $\mathbb{Q}$. – Pavel Čoupek Nov 08 '16 at 17:55
  • Thanks, that was claryfing, – Math.mx Nov 08 '16 at 17:56