Rings | Homomorphisms | Units

Question

Question

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

Attempt

Let $f:R\rightarrow S$ be a homomorphism. Then

$$ f(a)=f(a\cdot 1_R)=f(a)f(1_R),$$

so

$$ f(1_R)=\frac{f(a)}{f(a)}=1_S, $$

which is me trying to show why $(6)$ holds:

Consequently, we then have that,

$$ f(a\cdot a^{-1})=f(a)f(a^{-1})=f(1_R)=1_S, $$

so it is then easy to see that because $f(a)f(a^{-1})=1_S$, then $f(a)$ is a unit in $S$. Furthermore, by the same equality

$$ f(a^{-1})=\frac{1_S}{f(a)}=f(a)^{-1}, $$

as needed.

Answer 1

Hint: if $a\in R$ is a unity then you can write $1_R=a\cdot b$. Use the fact that $f(1_R)=1_S$ and compute $f(a\cdot b)$.

Answer 2

You've got something of the right idea, here, but are off in a few details. I assume that your rings are abelian groups under addition, monoids under multiplication, with multiplication being left- and right-distributive over addition. I further assume your definition of a ring homomorphism $R\to S$ requires $1_R\mapsto 1_S.$ Let me know if my assumptions are correct.

Now, given those assumptions, we know for any unit $a\in R$ that $$1_S=f(1_R)=f(aa^{-1})=f(a)f(a^{-1}),$$ and likewise, $$1_S=f(a^{-1})f(a).$$ Thus, indeed, we have that $f(a)$ is a unit of $S$ and $f(a)^{-1}=f(a^{-1}).$

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Added: The assumptions that I made above are actually necessary. Without them, the proposition is false.

On the one hand, if we only require a ring homomorphism $f:R\to S$ to satisfy $$f(x+y)=f(x)+f(y)\quad\text{and}\quad f(xy)=f(x)f(y)\tag{$\clubsuit$}$$ for all $x,y\in R$, then let's consider $R=S=\Bbb Z$, and let $f:R\to S$ be given by $f(n)=0$ for all $n.$ Then $-1$ is a unit of $R$, but $f(-1)=0$ is not a unit of $S$, but $f$ is a ring homomorphism by this definition. This shows that without requiring $1_R\mapsto 1_S$, the proposition fails.

Now, suppose we don't require that a ring be a monoid under multiplication. (The even integers are then a ring, for example.) The most we can require in general, then, is that ring homomorphisms $f$ on $R$ satisfy $(\clubsuit)$, which we already saw wasn't enough.