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Let $f:R \rightarrow S$ be an isomorphism of rings.

Let $g:S \rightarrow R$ be the inverse function of $f$.

Show that $g$ is also an isomorphism.

I know I can show and isomorphism by first showing $g$ is injective, surjective, and operation preserving. I am unclear on how to do that exactly. Any help would be greatly appreciated.

Raul Quintanilla Jr.
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  • The only problem here, assuming you already know the basic set theory stuff for inverse functions, is to show the inverse preserves the operations. So suppose $;s,s'\in S;$ . You must prove $;g(ss')=g(s)g(s');,;;g(s+s')=g(s)+g(s');$ ...Use bijectivity of $;f;$ for this – DonAntonio Nov 19 '18 at 01:19
  • Related: https://math.stackexchange.com/questions/2039702/what-is-an-homomorphism-isomorphism-saying/2039715#2039715 – Ethan Bolker Nov 19 '18 at 01:42

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because $f$ is a bijection, there are $r,r' \in R$ such that $f(r)=s$ and $f(r')=s'$. So now ,

$g(ss')=g(f(r)f(r'))=g(f(rr'))=rr'=g(s)g(s')$

also

$g(s+s')=g(f(r)+f(r'))=g(f(r+r'))=r+r'=g(s)+g(s')$.

finally,

$g(1)=g(f(1))=1$

Hence, $g$ is a homomorphism and since inverse of a bijective function $f$ is bijective, $g$ is also a bijective hence an isomorphism.

Q the Platypus
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user614287
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