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I have been studying the basics of ring theory and ring homomorphisms. I know that for 2 groups, there always exists a homomorphism between them, namely the trivial homomorphism.

Does a corollary exist for rings? i.e., given 2 rings, does there always exist a ring homomorphism between them? If not, what extra conditions are required for a ring homomorphism to exist? Thanks.

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    $\begingroup$ Whether the zero map is a ring homomorphism depends on the conventions, see this question. $\endgroup$ Commented Apr 18, 2017 at 15:05
  • $\begingroup$ The zero ring screws things up. It's the only ring where 0 = 1. $\endgroup$
    – Tac-Tics
    Commented Apr 18, 2017 at 15:07

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No, in general there need not be a ring homomorphism between two given rings, e.g., there is no (unital) ring homomorphism between $M_2(R)$ and $R$ for the integers, $R=\mathbb{Z}$, see this question.

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For $p\not = q$ prime, any map $\mathbb{Z}_p \to \mathbb{Z}_q$ must vanish.

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Quite a lot of nontrivial problems can be expressed in the form "does a ring homomorphism exist between two given rings". As an example, let $f(X,Y)$ be a nontrivial polynomial in two variables with integer coefficients. Then $C=\{(x,y)\in\Bbb R^2:f(x,y)=0\}$ is a curve. The question "does $C$ have any point with rational coordinates" is equivalent to "is the a ring homomorphism from $\Bbb Q[X,Y]/\langle f(X,Y)\rangle$" to $\Bbb Q$.

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There is no (unital) ring homomorphism between $\mathbb{Q}$ and $\mathbb{Z}$. One of the description of the case when such map exists is when the codomain ring has a subring which is isomorphic to a nontrivial quotient of the domain ring. See the First Isomorphism theorem.

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