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When I want to construct some ring $R$ of positive characteristic, say $n \ge 2$, I start by saying "let's consider the ring $\Bbb{Z}/n \Bbb{Z}$ and its polynomial ring etc...".

I realize that very often (*) such construction can be made with $\Bbb{Z}$ as starting point, so that the new resulting ring $\widetilde{R}$ would have characteristic $0$, and $R$ would be simply $\widetilde{R} / \langle n \rangle$.

As a consequence, this question popped into my mind:

Does every ring of positive characteristic arise as a quotient of a ring of zero characteristic?

(*): Actually this is not always the case: for example I could start with the algebraic closure of $\Bbb{F}_p$ (where $p$ is a prime), or with a Boolean ring $(\mathcal{P}(X) , \Delta , \cap)$ (which has characteristic $2$); these do not look like quotients of some ring of characteristic $0$ at first sight, but I suspect they are.

Until now, I have found no counterexamples, nor reasons to believe this is true: can you help me?

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Yes; simply take the ring $\mathbb Z\langle R \rangle$, i.e. the ring freely generated by the elements of $R$. Then the map $\mathbb Z \langle R \rangle \to R$ which takes a variable with name $r$ to the element $r \in R$ provides a surjection. Edit: if we are working with commutative rings, as your commutative-algebra tag seems to indicate, use the polynomial ring instead (which is just the commutative ring freely generated by elements of $R$).

Edit: for something even more trivial, use $\mathbb Z \times R$ for a ring with the projection $\mathbb Z \times R \to R$ as a quotient mapping.

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  • $\begingroup$ I'd like to add that if we take $A / \Bbb F _ p$ we can find a isomorphisms $\Bbb F _p [x_i] / (f_i) \to A$ this we can lift to $\Bbb Z [x_i] / (f_i) $, if we take any representatives for the $f_i$. Then we have $\Bbb Z [x_i] / (f_i) \otimes _{\Bbb Z} \Bbb F _p \cong \Bbb F _p [x_i] /(f_i) \cong A$ which in a way shows that $_ \otimes _{\Bbb Z } \Bbb F _p $ is surjective. $\endgroup$ – user369397 Sep 21 '16 at 14:46

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