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Let $R$ be a UFD if $f\neq 0$ and $\langle f\rangle$ is a prime ideal then $f$ is irreducible. Here is a short and easy proof for this claim but I do not understand the last bit.

Assume for a contradiction that $f$ is not irreducible then $f=uv$ where $u$ and $v$ are not units. By $\langle f\rangle$ a prime ideal we have either $f|u$ or $f|v.$ WLOG, assume former, then $uv=f|u$, this contradicts irreducibility of $u$.

My question:

Where in this proof have we assumed $u$ is irreducible?

{I can see that we can alter this proof a bit by using UFD property to say $f=r_1...r_k$ say where $r_i$ are irreducible then the same can follow. Alternatively $uv|u$ also contradicts $v$ is not a unit. But my question is where on this proof did we assume $u$ is irreducible?}

Many thanks in advance!

UnsinkableSam
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  • It should say this contradicts $u$ is a nonunit, not thet $u$ is irreducible. Double check that you did not miscopy the proof. – Bill Dubuque Dec 10 '20 at 00:09
  • @BillDubuque Hi Bill, thank you for helping! I double checked and I believe my book is incorrect for this one. However, wouldnt it contradict that $v$ is a non-unit instead of contradicting $u$ is a non-unit (where the latter was stated in your comment)? Sorry if I am missing something obvious, many thanks in advance! – UnsinkableSam Dec 10 '20 at 10:17
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    Yes, I meant $,v,,$ since $,uv\mid u\iff v\mid 1,$ by cancelling $,u.,$ Likely it is a typo (or "braino") in the book. – Bill Dubuque Dec 10 '20 at 10:20
  • @BillDubuque Thank you Bill, it has cleared up my confusion :) – UnsinkableSam Dec 10 '20 at 16:57

1 Answers1

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This only contradicts the assumption that $v$ is not a unit, not the irreducibility of $u$.

Assume $f\mid u$. Then $u=af$ for some $a\in R$. Consequently, $$u=af=auv=av\cdot u.$$Since $R$ is a UFD (also an integral domain), we must have $av=1$. That is, $v$ is a unit, contradicting the assumption.

In fact, it has nothing to do with the irreducibity of $u$. You can see the proof goes through as long as $R$ is an integral domain.

Bernard Pan
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