A question about prime elements in integral domains I have to show the following:
Let $p \in R\setminus\{0\}$ then: $p$ is prime element in $R$ if and only if $(p)$ is a prime ideal in $R$.
I have real problems doing so. I tried the following:
$\Rightarrow$ let $p$ be a prime element in $R$, then we know that if $p \mid ab$ then $p\mid a$ or $p\mid b$. Also, we know that $$(p) = Rp = \{ rp \mid r \in R\},$$ then we know that $(p)$ is prime ideal because $rp \in (p) \Rightarrow p \in (p)$.
Now how do i show that $(p)\neq R$. Also, the last step feels strange, as this seems to imply that $(a)$ is prime ideal for any $a\in R$ if $(a) \neq R$? 
with the $\Leftarrow$ direction I do not know how to start, could I get any hints?  
Thanks!
 A: Your argument in the $\Rightarrow$ direction is incorrect. 
You assume that $p$ is a prime element. That means that $p$ is not a unit, and if $p|ab$ in $R$, then $p|a$ or $p|b$.
What you need to show is that $(p)$ is a prime ideal; that is, that if $ab\in (p)$, then either $a\in (p)$ or $b\in (p)$.
Now, if $ab\in (p) = \{ rp\mid r\in R\}$, then there exists $r\in R$ such that $rp = ab$. That means that $p|ab$. Since $p$ is a prime element, then...
And if $(p)=R$, then $1\in (p)$. Therefore...
For $\Leftarrow$: Suppose that $(p)$ is a prime ideal; then $(p)\neq R$, and if $ab\in (p)$, then $a\in (p)$ or $b\in (p)$. Then $p$ is not a unit, because if $p$ is a unit, then $(p)$ will...
And finally, suppose $p|ab$. Then there exists $r\in R$ such that $pr=ab$. Therefore, $ab\in (p)$. Since $(p)$ is a prime ideal, then...
A: Here are some facts to keep in mind.


*

*$a\in (p)$ if and only if $p\mid a$.  (Try to prove this on your own, as it is the key fact for this problem, and also not too hard.)

*$(a)=R$ if and only if $a$ is a unit in $R$. Try to prove this on your own too, but below is an explanation if you can't get it.

 This is because, if $a$ is a unit, say with inverse $v$, then
 $$(a)=\{ra\mid r\in R\}\supseteq\{rva\mid r\in R\}=\{r\mid r\in R\}=R$$ 
 and hence $(a)=R$; conversely, if $(a)=R$, then $1\in (a)=\{ra\mid r\in R\}$, so  that there is some $r\in R$ such that $1=ra$, hence $a$ is a unit.



*

*An ideal $I\subseteq R$ is a prime ideal, by definition, if


*

*$I\neq R$, and

*$ab\in I$ implies either $a\in I$ or $b\in I$.


*An element $p\in R$ is a prime element, by definition, if


*

*$p\neq0$,

*$p$ is not a unit, 

*$p\mid ab$ implies either $p\mid a$ or $p\mid b$.


A: HINT $\ $ Utilize the fact that contains is equivalent to divides for principal ideals. Hence
$\rm\qquad (p)\supseteq (ab)\iff (p)\supseteq (a)\:\ or\:\ (p)\supseteq (b)\ $ is equivalent to $\rm\ p\ |\ ab\iff p\ |\ a\:\ or\:\ p\ |\ b$
Alternatively, if you know that $\rm\:P\:$ is prime $\rm\iff R/pR\:$ is a domain, then you may use
$\: $ nonunit $\rm\ p\:$ prime $\rm\iff [\:p\ |\ a\:b\:\Rightarrow p\:|\:a\ or\ p\:|\:b\:]$ $\rm\iff \rm R/p\:R\:$ a domain $\iff$ $\rm\:p\:R\:$ prime
