If $f : R \to S$ is a surjective ring morphism and if $R,S$ are commutative and finite, how do I show in a simple way that the induced group morphism $f' : R^{\times} \to S^{\times}$ is surjective?

According to this MO post the result is true (even more general results are true). But the proofs given there are too complicated for me. If someone has another idea (maybe assuming other conditions on the rings), I would be interested. Thank you very much!

  • $\begingroup$ Theorem 3.8 here might be of interest : if $f$ is surjective and has finite kernel, then so is $f^{\times}$. $\endgroup$ – Watson Dec 4 '18 at 19:32

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