Prove that if $R$ and $S$ are rings, $\theta : R \to S$ is an isomorphism, and $e$ is a unity of $R$, then $\theta (e)$ is a unity of $S$

Question

Prove that if $R$ and $S$ are rings, $\theta : R \to S$ is an isomorphism, and $e$ is a unity of $R$, then $\theta (e)$ is a unity of $S$

So if $\theta : R \to S$ is an isomorphism, that means that they essentially have the same structure. From my understanding, a unity is an element that has a multiplicative inverse in R. So $e$ is the inverse in R and we have to prove the same for S. Not sure where to go from here

Answer 1

Your question is not very clear. What you call a "unity" is usually denoted as a "unit" which is only defined for rings with a multiplicative identity (otherwise speaking about "an element that has a multiplicative inverse" wouldn't make sense).

Nevertheless I guess that your question is based on ring homomorphisms betweeen rings not necessarily having a multiplicative identity (and ring homomorphisms are maps that respect addition and multiplication). Then what you want to show is this:

Let $f : R \to S$ is a ring isomorphism and let $R$ have a multiplicative identity $1_R$. Then

1. $1_S = f(1_R)$ is a multiplicative identity of $S$.

2. If $e \in R$ is a unit, then $f(e)$ is a unit.

To see this, let $s \in S$ and $r = f^{-1}(s) \in R$. We have $1_S \cdot s = f(1_R) \cdot f(r) = f(1_R \cdot r) = f(r) = s$. Similarly $s \cdot 1_S = s$.

Next, let $e \in R$ be a unit. It has an inverse $e'$ characterized by $e \cdot e' = e' \cdot e = 1_R$. Then $f(e) \cdot f(e') = f(e\cdot e') = f(1_R) = 1_S$ and $f(e') \cdot f(e) = 1_S$.

Answer 2

We can get more mileage on this one with virtually no extra work. For suppose

$\theta: R \to S \tag 1$

is merely a ring epimorphism, that is, a surjective homomorphism; we needn't require that

$\ker R = \{0\} \subset R; \tag 2$

for let $1_R$ be the (unique) multiplicative unit of $R$, and

$s \in S, \tag 3$

we have

$\exists r \in R, \; \theta(r) = s; \tag 4$

then

$\theta(1_R) s = \theta(1_R) \theta(r) = \theta(1_R r) = \theta(r) = s, \tag 5$

and

$s \theta(1_R) =\theta(r) \theta(1_R) = \theta(r 1_R) = \theta(r) = s; \tag 6$

(5) and (6) together imply that

$\theta(1_R) = 1_S, \tag 7$

the unique multiplicative unit of $S$.

Now if $e \in R$ is merely a unit (not necessarily the unit $1_R$), then it is by definition a divisor of $1_R$; that is

$\exists f, g \in R, \; fe = eg = 1_R; \tag 8$

note (8) implies

$f = f1_R = f(eg) = (fe)g = 1_Rg = g; \tag 9$

with the existence of $f = g$ such that

$fe = ef = 1_R, \tag{10}$

we see that

$\theta(f) \theta(e) = \theta(fe) = \theta(ef) = \theta(1_R) = 1_S = \theta(e) \theta(f), \tag{11}$

which shows that $\theta(e)$ and $\theta(f)$ are units in $S$.