# Irreducibles are Primes in a UFD

Let $U$ be a Unique Factorization Domain. Let $a \in U$ be irreducible. Consider the proof that $a$ must be prime.

1. Suppose $a \mid bc$ so that $ar = bc$ for some $r \in U$.

2. Let $b = f_1 \cdot \ldots \cdot f_n$ s.t. each of the $f_i$'s are irreducible.

3. Let $c = f_{n+1} \cdot \ldots \cdot f_{m}$ s.t. each of the $f_i$'s are irreducible.

4. Finally, let $r = d_1 \cdot \ldots \cdot d_k$ s.t. each of the $d_i$'s are irreducible.

5. Then $ar = bc = (f_1 \cdot \ldots \cdot f_n)(f_{n+1} \cdot \ldots f_{m})$.

6. Now since $a(d_1 \cdot \ldots \cdot d_k)$ and $(f_1 \cdot \ldots \cdot f_n)(f_{n+1} \cdot \ldots f_{m})$ both represent irreducible factorizations of $ar$, then we have that $a$ is associates with at least one of the $f_i$'s. If $1 \le i \le n$, then $a \mid b$. Otherwise, $a \mid c$.

7. Then $a$ is prime as desired.

EDIT: As was pointed out in the comments, I wrote the proof wrong above. The question below now doesn't make any sense. My writing down the proof wrong is what spawned it. In fact, this post should just be deleted.

Now my question is as follows: if $a$ is associates with some $f_i$, does that mean that the remaining elements of $\{f_j\}$ s.t. $1 \le j \ne i \le n$ are all units?

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Don't you want to show that : If $a\mid bc$ then either $a\mid b$ or $a\mid c$? –  Prahlad Vaidyanathan Dec 3 '13 at 7:32
I thought that was shown? My question is not about the proof's validity, but just about whether I am understanding a tangential implication in one of the steps. –  Techn1cal Dec 3 '13 at 7:34
No. I meant that (1) should read "Suppose $a\mid bc$", not $a=bc$. If $a=bc$, then by definition either $b$ or $c$ must be a unit. –  Prahlad Vaidyanathan Dec 3 '13 at 7:35
Yup -- me having the proof down wrong is what caused my own confusion. Thanks for pointing that out. It's now edited correctly above and in fact this post should just be deleted. –  Techn1cal Dec 3 '13 at 7:45