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Is homomorphism between two rings a symmetric relation? That is, suppose $\exists \phi:R\rightarrow R'$ a ring homomorphism. Then does there exist a ring homomorphism from $R'$ to $R$?

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  • $\begingroup$ Yes this should be true, but the other ring homomorphism will be trivial as in the cases described below. $\endgroup$
    – jfarchione
    Apr 28, 2017 at 1:18
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    $\begingroup$ @Hayden: Your comment is better than the two answers (as of now). Care to make it an answer? $\endgroup$ Apr 28, 2017 at 1:26

4 Answers 4

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This will depend somewhat on your definition of a ring and a ring homomorphism, as there are a few standard ways of defining these notions.

Firstly, for rings, there is sometimes a distinction between a ring and a unital ring, which additionally has a multiplicative identity $1$ (whereas a ring need not have one).

Secondly, and more importantly, even when ring is taken to mean unital ring, there is sometimes a distinction between a ring homomorphism (which simply preserves addition and multiplication) and a unital ring homomorphism (which much additionally preserve the multiplicative identity).

If a ring homomorphism need not preserve the multiplicative identities (assuming these are present in our rings), then between any two rings $R,S$ there is the zero map $\mathbf{0}: R \to S$ defined by $\mathbf{0}(r) = 0$.

As such, if your ring homomorphisms need not be unital, then you will always have such trivial homomorphisms. Note that a non-trivial homomorphism between unital rings will usually preserve $1$, so whether you instead want to consider "non-trivial" or "unital" ring homomorphisms, the answers given by Hurkyl or Lord Shark give examples of (unital) rings which don't admit unital/non-trivial ring homomorphisms in both directions.

For example, there is the obvious inclusion of $\mathbb{Z}$ into $\mathbb{Q}$ (and this is a unital ring homomorphism), but a non-trivial homomorphism $\varphi: \mathbb{Q} \to \mathbb{Z}$ must have $2\varphi(1/2) = \varphi(1)=1$, so thus $\varphi(1/2)$ is some integer which, when multiplied by $2$, gives $1$. Of course, this can't happen, so we get a contradiction.

(I said above that a non-trivial ring homomorphism between unital rings will usually preserve $1$. A quick and easy example of a non-trivial ring homomorphism not preserving $1$ between unital rings is given by the following: Take $\mathbb{F}_2 = \mathbb{Z}/2\mathbb{Z}$ and consider the inclusion of $\mathbb{F}_2$ into $\mathbb{F}_2^2$ by sending $0\mapsto (0,0)$ and $1\mapsto (1,0)$. Now, $(1,0)$ is not the multiplicative identity of $\mathbb{F}^2$, as $(1,1)$ is. However, it is a multiplicative identity of the image of the inclusion. In general, a ring homomorphism sends a multiplicative identity to a multiplicative identity of the image. In many cases, especially rings which have some sort of multiplicative cancellation property, that will be enough to mean that it actually sends $1$ to $1$. This is why I was able to justify the fact that $\varphi(1)=1$ above.

For more about this idea, see here.)

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  • $\begingroup$ Thank you sir for this wonderful explanation. $\endgroup$
    – mudok
    Apr 28, 2017 at 1:55
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No. $\mathbb{Z} \to \mathbb{F}_2$ and $\mathbb{Z} \to \mathbb{Q}$ are counterexamples.

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  • $\begingroup$ More generally, Z -> F for any non-trivial field F. $\endgroup$ Apr 28, 2017 at 14:17
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Certainly not. Take for instance $\Bbb Z$ and $\Bbb Q$. Or $\Bbb R$ and $\Bbb C$.

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  • $\begingroup$ I dont understand why? Z to Q there is the inclusion map. but why there is not a homomorphism from Q to Z? $\endgroup$
    – mudok
    Apr 28, 2017 at 1:22
  • $\begingroup$ @mudok $1$ must go to $1$ but where would $1/2$ go? $\endgroup$ Apr 28, 2017 at 1:34
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    $\begingroup$ The material you flesh out in the Comments might well be edited into the body of your Answer. $\endgroup$
    – hardmath
    Apr 28, 2017 at 1:50
  • $\begingroup$ @LordSharktheUnknown you sir did not explain why 1 must go to 1? $\endgroup$
    – mudok
    Apr 28, 2017 at 1:57
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    $\begingroup$ @mudok As Hayden explains it is conventional to insist that ring homomorphisms are unital, taking $1$ to $1$. If one does not adopt this convention then the zero map is always a ring homomorphism and your original question becomes idle. $\endgroup$ Apr 28, 2017 at 2:04
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If we add, to the question, the reasonable requirement that the homomorphism is nonzero, the answer is still no.

For instance consider $R=M_2 (\mathbb R) $, $R'=M_3 (\mathbb R) $. There are many nontrivial homomorphisms $R\to R'$: for any invertible $B $, $$A\longmapsto B\,\begin{bmatrix}A&0\\0&0\end{bmatrix}\,B^{-1}, $$ is a ring homomorphism. But there are no nonzero homomorphisms $R'\to R $.

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