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?


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    $\begingroup$ It's a good thing that the "last step" feels strange. It's incorrect. What you need to show is that if $a$ and $b$ are any elements of $R$, and $ab\in (p)$, then either $a\in(p)$ or $b\in (p)$. What you argued was just that $p\in (p)$, which is true but irrelevant here. $\endgroup$ – Arturo Magidin Jun 26 '11 at 3:07
  • $\begingroup$ I don't see how you conclude that $(p)$ is a prime ideal. If "hence" should be "because", then note that in order to show that an ideal is prime, you must consider every product of two elements that lands in that ideal. For example, the ideal $(4)$ in $\mathbf{Z}$ has elements of the form $4a$ for $a \in \mathbf{Z}$, and certainly $4 \in (4)$, but this does not show that $(4)$ is prime. $\endgroup$ – Dylan Moreland Jun 26 '11 at 3:13

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 $p\mid b$.
  • $\begingroup$ Thanks for the explanation. I'm still wondering though if it is correct to say that (a) is prime ideal if (a) != R. Hence a is contained within (a) we know that for every element ra $\in (a)$ a is an element in (a). $\endgroup$ – sxd Jun 26 '11 at 2:26
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    $\begingroup$ @Dimitri: No, that's not correct. Consider the ideal $(6)\subset\mathbb{Z}$. Obviously $(6)\neq\mathbb{Z}$, but $(6)$ is not a prime ideal, because we have that $2\cdot 3\in (6)$, but $2\notin (6)$ and $3\notin (6)$. $\endgroup$ – Zev Chonoles Jun 26 '11 at 2:26
  • $\begingroup$ In order for $(p)$ to be a prime ideal, we need $ab\in (p)$ to imply that either $a\in (p)$ or $b\in (p)$, for any $a,b\in R$. $\endgroup$ – Zev Chonoles Jun 26 '11 at 2:29
  • $\begingroup$ Oh, yes I think i forgot that there could exists bc = ap , where b != c and a != p, right? $\endgroup$ – sxd Jun 26 '11 at 2:34
  • $\begingroup$ @Dmitri: I think you meant to say, "$ab=cp$ where $a\neq c$ and $b\neq p$". But this can be the case even when $p$ is prime: for example, $2$ is a prime element in $\mathbb{Z}$, and $$4\cdot 4 = 8\cdot 2.$$ $\endgroup$ – Zev Chonoles Jun 26 '11 at 2:46

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...

  • $\begingroup$ Yes, i clearly see where my mistake is now, i made a wrong assumption that there is no cb = ap, so that c != a and b != p, as clearly they could exists and thus either b or c have to be in (p). Thanks for your response! $\endgroup$ – sxd Jun 26 '11 at 3:11
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    $\begingroup$ @Dimitri: Ah, the too common error of thinking that just because every element can be written $rp$, concluding (incorrectly) that every instance of the element must be of the form $rp$. $\endgroup$ – Arturo Magidin Jun 26 '11 at 3:12

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


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