1
$\begingroup$

If $A$ is a PID and $p\in A$, I want to prove that the following are equivalent:
a) $p$ is prime
b) $(p)$ is a non-zero prime ideal
c) $p$ is irreducible
d) $(p)$ is a maximal ideal

I assume the key here is proving that $a)\implies b)$ and $c)\implies d)$ and then proving that either $a) \implies c)$ or $b) \implies d)$

Since $A$ is a PID, I use the definition that it is an integral domain where every ideal is principal to use the proof of $p$ prime $\implies$ $p$ irreducible in an integral domain.
(ie. $p=ab$ with $p$ prime $\implies p|a$ or $p|b$ so wlog let $a=pc \implies p=pcb \implies cb=1 \implies b $ is a unit $\implies p$ irreducible.) There is also a proof that in a PID $p$ irreducible $\implies p$ prime, but it is quite long. Is this necessary also? If so, is there a relatively short proof of it?

Using the definitions of prime and prime ideal (resp. irreducible and maximal ideal) I am finding it hard to come up with a proof of their equivalence. It just seems obvious! Is there a neat way of showing it?

$A$- PID, $p\in A$, $(p)$ ideal of $A$ and $a,b\in A$.

maximal ideal: no ideals $d$ of $A$ exist such that $(p) \subset d \subset A$
irreducible: $p=ab \implies a$ or $b$ is a unit

prime ideal: $ab \in (p) \implies a \in (p)$ or $b \in (p)$
prime: $p|ab \implies p|a$ or $p|b$

$\endgroup$
2
$\begingroup$

$(a)\Rightarrow(b)$:
If p is a prime, then (p) is not empty,since p is in it.
suppose $ab \in (p)$, then p|ab ,since p is prime, so we have p|a or p|b,which means $a \in (p)$ or $b \in (p)$.
(p) is a prime ideal.

$(b)\Rightarrow(c)$:
suppose (p) is a non-zero prime ideal, but p is not irreducible.
$\exists a,b$ non-unit such that p=ab.
$p \in (a)$ and $p \in (b)$.
$(p) \subseteq (a)$ and $(p) \subseteq (b)$
We notice $ab \in (p)$,so $a \in (p)$ or $b \in (p)$.
so $(a) \subseteq (p)$ or $ (b) \subseteq (p)$
$(a)=(p)$ or $(b)=(p)$
If $(a)=(p)$,then there is a unit u,such that p=au.
since $p=ab$ , so $ab=au$ $\Rightarrow a(b-u)=0$.
A is a domain and $a \neq 0$ , so $b-u=0 \Rightarrow b=u$
Contradiction!!!! Since we assume b is not a unit.
similar proof for case $(b)=(p)$.
p is irreducible.

$(c) \Rightarrow (d)$:
suppose P is irreducible and (p) is not maximal.
$\exists$ an proper ideal (t) such that $(p) \subset (t)$.
So $p \in (t) \Rightarrow \exists$ a non-unit $s$ in A such that $p=ts$, otherwise $(p)=(t)$
$t$ is not a unit,otherwise $(t)=A$,and s is not a unit.
so p is not irreducible.Contradiction!!!!
(p) is a maximal ideal.

$(d) \Rightarrow (a)$:
suppose (p) is a maximal ideal ,but p is not a prime.
Since p is not a prime, we can factor it in a non-trivial way.
That is to say, we can find a prime $p'$ such that $p'|p$ and $p=p'N$, where $N$ is not a unit.
That is to say we find a ideal $(p') \supset (p)$.
Contradiction to the maximality of (p).
p is a prime.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.