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 $p$.
['https://math.stackexchange.com/questions/2005378', 'https://math.stackexchange.com', 'https://math.stackexchange.com/users/217745/']
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):

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