1

Let $R$ be a ring with $1$ and $I$ an ideal of $R$. Let $P$ be an ideal of $I$. In this question we see counterexemples that shows that $P$ is not necessarily an ideal of $R$.

My question is: If $P$ is a prime ideal of $I$, then $P$ is an ideal of $R$?

P.G
  • 662

1 Answers1

1

Yes actually.

Let $r\in R$ and $p\in P$. We need to show that $rp, pr\in P$. We'll make use of an equivalent definition of "prime ideal"

$P$ is prime in $I$ iff for all $x,y\in I$, $xIy\subseteq P$ implies $x\in P$ or $y\in P$.

Now, given that we know $rp\in I$ at least (since $I\lhd R$) we can compute that $rpIrp= (rpIr)p \subseteq Ip\subseteq P$. Therefore the definition of primeness of $P$ kicks in and says $rp\in P$. Similarly $pr\in P$.

rschwieb
  • 153,510