Show that if in a commutative ring $R$, if $p$ is prime, $pu$ is also prime. I want to show that if $p$ is prime, $pu$ is also prime, where $u$ is an invertible element of $R$.
Therefore, knowing that if $p\mid ab \implies$ $p\mid a$ or $p\mid b$ then I want to see that if $pu\mid ab\implies pu\mid a$ or $pu\mid b$. This should be an easy exercise, but I can't solve it. 
I have tried to work with the definitions, from $ab=\alpha pu$ I want to get to $a=\beta pu$ or  $b=\beta pu$ without succeeding.
 A: If $pu\mid ab$ than $p\mid ab$ and so $p\mid a$ or $p\mid b$ but if, for example, $p\mid a$ than $a=\alpha p=\alpha u^{-1} (pu)$ and so $pu\mid a$
A: We recall that $p$ is prime if and only if $p \mid ab$ implies $p \mid a$ or $p \mid b$; another way of saying this is that if $p \mid ab$ and $p \not \mid a$ then $p \mid b$.
We wish to show that if $p$ is prime and $u$ is a unit, that is, there exists $v$ such that $uv = 1$, then $pu$ is prime; that is, if $pu \mid ab$ and $pu \not \mid a$ then $pu \mid b$.
We have, for any $p$ (prime or not) and unit $u$:
$p \mid a \Longleftrightarrow pu \mid a, \tag 1$
for
$p \mid a \Longrightarrow \exists k \; kp = a \Longrightarrow k(vu)p =(kv)(pu) = a \Longrightarrow pu \mid a$
$\Longrightarrow \exists l \; l(pu) = a \Longrightarrow (lu)p = a \Longrightarrow p \mid a;\tag 2$
now for any $p$, prime or not, by (1),
$pu \mid ab \Longrightarrow p \mid ab, \tag 3$
and 
$pu \not \mid a \Longrightarrow p \not \mid a; \tag 4$
for primes $p$,
$p \mid ab, p \not \mid a \Longrightarrow p \mid b, \tag 5$
so again by (1),
$p \mid b \Longrightarrow pu \mid b; \tag 6$
we conclude from (3)-(6) that $pu$ is prime.
We can in fact see from the above that $pu$ prime implies $p = p(uv) = (pu)v$ is prime where $uv = 1$.  So it goes both ways.
