Homomorphisms

**dwsmith** · July 12th 2011, 10:14 AM

R and S are rings.
Show that if $f:R\to S$ is a homomorphism, and if a is a unit of S. Show, in fact, that $f(a^{-1})=f(a)^{-1}$ for any unit of a of R.

$f(a^{-1})-f(a)^{-1}=f((a^{-1})+(-a)^{-1})=f(0)=0$

Is this how I show this?

**hatsoff** · July 12th 2011, 10:43 AM

Originally Posted by dwsmith

R and S are rings.
Show that if $f:R\to S$ is a homomorphism, and if a is a unit of S. Show, in fact, that $f(a^{-1})=f(a)^{-1}$ for any unit of a of R.

$f(a^{-1})-f(a)^{-1}=f(a^{-1})+(-a)^{-1})=f(0)=0$

Is this how I show this?

No. When you use the expression $f(a)^{-1}$ you have assumed it exists, but you have to first show that it exists. Also, in the first equality it looks like (trying to read around your type) you have assumed that $-f(a)^{-1}=f((-a)^{-1})$ , which is neither given.

Instead, just notice that $f(a^{-1})f(a)=f(a^{-1}a)=f(1)=1$ . The conclusion follows immediately.

**dwsmith** · July 12th 2011, 11:08 AM

Originally Posted by hatsoff

you have assumed that $-f(a)^{-1}=f((-a)^{-1})$ , which is neither given.

My book states the following:

f is a homomorphism if f satisfies the following properties:
(1) $f(r+r')=f(r)+f(r')$
(2) $f(r\cdot r')=f(r)f(r')$
(3) $f(1)=1$

If f satifies 1-3, then
(4) $f(0)=0$
(5) $f(-r)=-f(r)$

So I didn't assume anything. I used what was given. It is stated that f is a homomorphism so it satisfies 1-3 which means 4 and 5 are also satisfied.

**Drexel28** · July 12th 2011, 11:50 AM

Originally Posted by dwsmith

My book states the following:

f is a homomorphism if f satisfies the following properties:
(1) $f(r+r')=f(r)+f(r')$
(2) $f(r\cdot r')=f(r)f(r')$
(3) $f(1)=1$

If f satifies 1-3, then
(4) $f(0)=0$
(5) $f(-r)=-f(r)$

So I didn't assume anything. I used what was given. It is stated that f is a homomorphism so it satisfies 1-3 which means 4 and 5 are also satisfied.

Well the first mistake you made was assuming that $f(a)\in S^\times$ .

**dwsmith** · July 12th 2011, 12:18 PM

Originally Posted by Drexel28

Well the first mistake you made was assuming that $f(a)\in S^\times$ .

I don't know what you mean by that so it doesn't make any sense to me.

**Drexel28** · July 12th 2011, 12:24 PM

Originally Posted by dwsmith

I don't know what you mean by that so it doesn't make any sense to me.

$S^\times$ is the set of units of $S$ --I was saying that in your proof you are assuming $f(a)$ is a unit, which is part of what you are trying to prove.

**Deveno** · July 13th 2011, 06:09 AM

since $a$ is a unit, $a^{-1}$ exists. since f is a homomorphism:

$1_S = f(1_R) = f(aa^{-1}) = f(a)f(a^{-1})$ , so....

**Drexel28** · July 13th 2011, 11:12 AM

Originally Posted by Deveno

since $a$ is a unit, $a^{-1}$ exists. since f is a homomorphism:

$1_S = f(1_R) = f(aa^{-1}) = f(a)f(a^{-1})$ , so....

A previous user had said the exact same thing.

**Deveno** · July 13th 2011, 11:40 AM

no, but it was isomorphic.

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