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 $S$ 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
 A: 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


*

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

*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 $f(e') \cdot f(e) =  1_S$.
A: 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 $S$.
