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 $p$.


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.

  • $\begingroup$ I feel bad! That was obvious! $\endgroup$ – Abellan Nov 8 '16 at 17:46

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.

  • $\begingroup$ There is one more residue field - $\mathbb{Q}$. $\endgroup$ – Pavel Čoupek Nov 8 '16 at 17:34
  • $\begingroup$ Which is the maximal ideal that gives you $\mathbb{Q}$ as quotient ring? $\endgroup$ – Murphy Nov 8 '16 at 17:37
  • $\begingroup$ Not maximal, but the prime ideal $(0)$. : ) $\endgroup$ – Pavel Čoupek Nov 8 '16 at 17:38
  • $\begingroup$ (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. $\endgroup$ – Murphy Nov 8 '16 at 17:39
  • $\begingroup$ 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. $\endgroup$ – Pavel Čoupek Nov 8 '16 at 17:42

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