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I understand that for rings R and S, S contains a subring isomorphic to R if and only if there is an injective ring homomorphism from R to S. But I am unsure of how to proceed from this definition?

Also that for rings to be isomorphic you would instead need a bijection.

Any help would be appreciated, thanks!

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  • $\begingroup$ If $f$ is such a homorphism, what $f(1)$ ? And $f(1)+...+f(1)$ ? $\endgroup$ Apr 20, 2018 at 20:00

2 Answers 2

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Hint: the additive group of the intergers modulo $29$ is cyclic of prime order and has no proper non-trivial subgroup.

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Hint: The integers mod $29$ is a field, and the integers mod $8$ has nonzero zero divisors.

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