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this is the third part of a question I've been working on from Hungerford's Algebra. It is exercise 15 in the first section of Chapter III.

$(c)$ If $f\colon R\to S$ is a homomorphism of rings with identity and $u$ is a unit in $R$ such that $f(u)$ is a unit in $S$, then $f(1_R)=1_S$ and $f(u^{-1})=f(u)^{-1}$.

I see how $f(u^{-1})=f(u)^{-1}$ follows from $f(1_R)=1_S$, for if that is so, then $$f(1_R)=f(uu^{-1})=f(u^{-1}u)=1_S\implies f(u)f(u^{-1})=f(u^{-1})f(u)=1_S. $$ It seems easy, but I can't manage to show $f(1_R)=1_S$. I was hoping that someone could perhaps give me a hint on how to proceed on showing the first part? Thank you.

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You're almost there already. The multiplicative inverse in a ring, if it exists, is unique, so $f(u^{-1})f(u) = 1_S$ demonstrates $f(u^{-1}) = f(u)^{-1}$. – Zhen Lin May 22 '11 at 7:29
@Zhen, I see that $f(u^{-1})=f(u)^{-1}$ under the assumption that $f(1_R)=1_S$, but I don't know how to prove that $f(1_R)=1_S$. – yunone May 22 '11 at 7:31
@Pete: Are you sure that we need a rngs tag? Is this actually a word? – Rasmus Jun 30 '11 at 15:53
Well, maybe I can answer the first question myself. There have already been five questions concerning exactly what this tag is about and a tag might be helpful to avoid duplicates or refer to already given answers. – Rasmus Jun 30 '11 at 16:10
@Rasmus: right. As for whether it's a word: yes in the sense that mathematicians use it repeatedly. See e.g. – Pete L. Clark Jun 30 '11 at 19:16
up vote 6 down vote accepted

You can cancel $f(u)$ in the equation $f(u)=f(1_R)f(u)$.

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Dang, so simple. Must be getting too late, thanks Jonas. – yunone May 22 '11 at 7:47
You're welcome, yunone. – Jonas Meyer May 22 '11 at 7:51

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