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Is this proof correct ?

Let $R$ be a PID. Prove $p\in R$ is prime if and only if $p\in R$ is irreducible.

Proof:

$\implies$

Let $p\in R$ and p prime.

Let $ab=p\in\langle p\rangle.$ Thus $p\mid ab$ and $p\mid a$ or $p\mid b$.

w.l.o.g suppose $p\mid a.$ Then $a=kp,$ for some $k\in R$.

$\implies p=ab=kpb \implies p(1-kb)=0 \implies 1-kb=0 \implies b=1 \implies p $ is irreducible.

$\Leftarrow $ (this side of the proof was based on Xam's proof :)

Let D be an euclidean domain and $p\in D$ irreducible.

Let $ab=p\in\langle p\rangle$. Thus $p\mid ab$.

Now if $p$ doesn't divides a, then we have to prove $p\mid b.$

Suppose $u=px+ay,$ for some $x,y\in D.$

$\implies bu=bpx+bay$

$\implies p\mid bay$ and $p\mid bpx$

$\implies p\mid bu$

As $ub\mid b,$ then $p\mid b$. Therefore $p$ is prime.

user441848
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2 Answers2

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As to the first part of your proof, you have that $kb = 1$, which says that $b$ is a unit, not necessarily $1$. However, you only need $b$ to be a unit, so it's a minor fix.

For the second half, what if your PID is not a Euclidean Domain? You have proved the result holds in Euclidean Domains, but not in PIDs.

Alex Ortiz
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As I pointed out in my first comment, the first implication is essentially right, you just need to change $1-kb=0 \implies b=1$ for $1-kb=0 \implies b\in R^\times$ (here $R^\times$ is the set of units of $R$).

For the other implication, the basic idea is right. We need to show that there is some unit, let's say $u$ such that we can write $u=px+ay$. How do we do this? Here we use the hypothesis that $R$ is a PID. Namely, we consider the ideal $(p,a)$. Since $R$ is a PID, the above ideal is principal, therefore there is some $r\in R$, $r\neq 0$, such that $(r)=(p,a)$. As $p\in (p,a)=(r)$, we deduce that $r\mid p$, but since $p$ is irreducible, then $r$ and $p$ are associates or $r$ is an unit.

If $r$ and $p$ are associates, then as $r\mid a$ we conclude that $p\mid a$, but $p$ doesn't divide $a$, contradiction. Hence $r$ is an unit. Now, as $r\in (r)=(p,a)$, we can write $r=px+ay$. (In other words, we can take $u=r$ as our desired unit).

Xam
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  • Why $r|a?$ and why $p$ doesn't divide $a?$ – user441848 Jul 19 '17 at 17:07
  • @AnneliseToft $p$ doesn't divide $a$ is your hypothesis ("Now if $p$ doesn't divides $a$ ... "). On the other hand, $r\mid a$ by the same argument that shows that $r\mid p$. Think about it. – Xam Jul 19 '17 at 17:13
  • oh I wasn't making a relation with the proof above, if $r|a$ then this $a\in (r)$ happens – user441848 Jul 19 '17 at 17:20
  • @AnneliseToft it really doesn't matter. We can start the proof with the assumption that $p\mid ab$ and $p\not\mid a$, in then prove that necessarily $p\mid b$. $r\mid a\iff a\in (r)$. But, why it happens that $a\in (r)$? – Xam Jul 19 '17 at 17:25
  • I don't know. How this ideals $(p,a)$ looks like? I haven't seen it's form. I know how this $(r)$ looks like: $(r)={kr:k\in R}$ – user441848 Jul 19 '17 at 17:30
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    @AnneliseToft $a\in (p,a)=(r)$. That $a\in (p,a)$ should be clear since $a=p\cdot 0+a\cdot 1$. Hence $a\in (r)$. Btw, the general form of the elements of $(p,a)$ is $px+ay$, for some $x,y\in R$. – Xam Jul 19 '17 at 17:32
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    @AnneliseToft You may find helpful this answer, which elaborates on earlier points. – Bill Dubuque Jul 19 '17 at 23:42
  • $ub\mid b $ because $ub\mid b \iff b=\lambda ub$ and $\lambda$ can be taken as $\lambda=u^{-1}$? – user441848 Jul 29 '17 at 20:12
  • @Annet. yes. That's right. – Xam Jul 29 '17 at 21:13