Let $\phi:R\rightarrow R'$ be a (not necessarily unital) ring homomorphism. If $S'$ is a subring of $R'$, it is easy to show that $\phi^{-1}(S')$ is a subring of $R$. But if $S'$ is also unital, will $\phi^{-1}(S')$ be unital as well? In other words, can we always find a multiplicative identity in $\phi^{-1}(S')$?

PS: By a subring $S$ of $R$ I mean a subset $S\subseteq R$ such that it is a ring itself under the same operations of $R$.


1 Answer 1


Consider the embedding $2\mathbb{Z}\to\mathbb{Z}$. The inverse image of the codomain, which is unital, is the whole domain, which is not unital.

If you want a surjective homomorphism, consider the trivial homomorphism $R\to\{0\}$. The codomain is unital, but the domain may not be.

  • $\begingroup$ @Easy It’s simply $f(x)=x$ $\endgroup$
    – egreg
    Aug 15, 2018 at 9:00
  • $\begingroup$ @Easy You're interested in inverse images: $f^{-1}(\mathbb{Z})=2\mathbb{Z}$. $\endgroup$
    – egreg
    Aug 15, 2018 at 9:22
  • $\begingroup$ @Easy That's irrelevant and you didn't ask for a surjective map. If $f\colon A\to B$ is a map and $C\subseteq B$, then $f^{-1}(C)=\{a\in A:f(a)\in C\}$. There's no surjectivity requirement. $\endgroup$
    – egreg
    Aug 15, 2018 at 9:28
  • $\begingroup$ @Easy A counterexample suffices. $\endgroup$
    – egreg
    Aug 15, 2018 at 9:37

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