If $I\lhd A$ and $P$ is a prime ideal in $I$, prove $P\lhd A$.

Question

Q: Let $I$ be an ideal in $A$. If $P$ is a prime ideal in $I$, prove that $P$ is an ideal in $A$.

First of all, if $p \in P$ and $q \in P$, then $p - q \in P$, because $P$ is an ideal in $I$.

So, i need to prove that if $a \in A$ and $p \in P$, then $a*p \in P$

I don't know how to prove this. I know that if $a*p \in P$, then $a \in P$ or $p \in P$, but the reciprocal afirmattion isn't necessarily true. Also know that $a*p \in I$, cauz $a \in A$ and $p \in I$.

Any ideas on how to proceede?

Answer 1

Let $a\in A$ and $p\in P$. Then at least we have $ap\in I$ since $I$ is an ideal. If $P=I$ then we are done. Otherwise, choose $x\in I\setminus P$. Then $apx=p(ax)\in P$ since $ax\in I$, and since $apx=(ap)(x)$ and $x\notin P$ it follows that $ap\in P$ since $P$ is prime in $I$.

This works assuming $A$ is a commutative ring, not necessarily with unity.

Answer 2

From where you left off, you already had $ap\in I$.

$(ap)^2=(apa)p\in P$, since $apa\in I$. By primeness, $ap\in P$.