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The following question is from Pinter's Book of Abstract Algebra:

List all the ring homomorphisms from $\mathbb Z_2$ to $\mathbb Z_4$; and from $\mathbb Z_3$ to $\mathbb Z_6$.

I was wondering, is this a trick question? I can't really think of any homomorphisms that go in this direction. There are some that go the other way, for instance

$$ \ f = \bigl(\begin{smallmatrix} 0 & 1 & 2 & 3 & 4 & 5 \\ 0 & 1 & 2 & 0 & 1 & 2 \end{smallmatrix}\bigr), \ $$ which is from $\mathbb Z_6$ to $\mathbb Z_3$. The problem I seem to have is that if the function maps one element to two or more different elements, then it isn't a function. If my intuition is correct, could someone outline the best way to show/prove that no such homomorphisms exist? Thanks very much.

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  • $\begingroup$ Does Pinter insist that homomorphisms be unital; that is send $1$ to $1$. $\endgroup$ – Lord Shark the Unknown Dec 18 '17 at 8:04
  • $\begingroup$ @LordSharktheUnknown No he doesn't insist on that, as far as I can see. $\endgroup$ – K.Reeves Dec 18 '17 at 8:14
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The short answer: the only ring homomorphisms $\ \mathbb Z_2\rightarrow Z_4\ $ is the ZERO homomorphism (the "For example" example by @wsj84 was false). On the other hand, $\ \mathbb Z_6\ $ is ring-isomorphic to $\ \mathbb Z_2\times\mathbb Z_3.\ $ For this reason, in addition to the ZERO homomorphism $\ \mathbb Z_3\rightarrow\mathbb Z_6,\ $ there is also the homomorphism which sends $\ 1\in \mathbb Z_3\ $ to $\ 4\in\mathbb Z_6\ $ (neat).

An extra comment: a ring element $\ x\in R\ $ is called an idempotent $\ \Leftarrow:\Rightarrow\ x\cdot x =x.\ $ If $\ f :R\rightarrow S\ $ is a ring homomorphism then it sends every idempotent $\ x \in R\ $ into an idempotent $\ f(x)\in S,\ $ indeed:

$$ f(x)\cdot f(x) = f(x\cdot x) = f(x) $$

Obviously, $\ 0\ $ and $\ 1\ $ (if there is any element $\ 1\ $ which would be THE unit) is an idempotent. If $\ p\ $ is a prime then in the field $\ \mathbb Z_p,\ $ just like in an arbitrary field, elements $\ 0\,\ 1\ $ are the only idempotents. The same is true for $\ \mathbb Z_{p^n}\ $ (for any prime $\ p).\ $ But $4\in \mathbb Z_6\ $ is an idempotent; this helps to find an explicit decomposition of $\ \mathbb Z_6.\ $ In general, the direct products of rings with unit have a respective induced power set of the impotents induced by these units.

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    $\begingroup$ I do not understand your first sentence: why do you think the identity map is not a homomorphism? Perhaps you have a typo? $\endgroup$ – ancientmathematician Dec 18 '17 at 9:44
  • $\begingroup$ It was a typo. I have fixed it by now, thank you. $\endgroup$ – Wlod AA Dec 18 '17 at 9:46
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    $\begingroup$ It is @wsj84 (not asj84) who has made a slip. $\endgroup$ – ancientmathematician Dec 18 '17 at 9:58
  • $\begingroup$ @ancientmathematician, thank you again, I have fixed this typo too (you're very good! :) ). $\endgroup$ – Wlod AA Dec 18 '17 at 10:03

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