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The point of a UFD is that any element can be rewritten as a product of irreducible factors, where any other product of irreducible factors is just a rearrangement of the exact same terms, is this correct?

I.e. $f(x)=a_1a_2\cdots a_n$ where $a_i$ are irreducible for all $1\leq i \leq n$.

And if we also have $f(x)=b_1b_2\cdots b_m$ and $f(x)$ is in a UFD, we know that:

1) $m=n$, and

2) These $b_l$ are just a relabelling of the $a_i$

Is that right?

glS
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Galois in the Field
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2 Answers2

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Actually, you should think in this way. UFD means the factorization is unique, that is, there is only a unique way to factor it. For example, in $\mathbb{Z}[\sqrt5]$ we have $4 =2\times 2 = (\sqrt5 -1)(\sqrt5 +1)$. Here the factorization is not unique.

Esther Jacob
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  • @ThomasAndrews I think the answerer here is giving an example of a failure to be a unique factorisation, so a unique factorisation is indeed what I wrote(other than the delicate unit issue) – Galois in the Field Oct 16 '15 at 03:29
  • @GaloisintheField The word "actually" here implies, to most English speakers, that answerer is contradicting and/or correcting. But this answer doesn't really do that, so it might just be a wrong choice of words. – Thomas Andrews Oct 16 '15 at 03:32
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Here's one way to be pedantic about the question of units.

In an integral domain, $R$, we can define the multiplicative group of invertible elements, $U$.

We can then define an equivalence relationship on $R$ by:

$$a\sim b\iff\exists u\in U(au=b)$$

Then if $a\sim b$ and $c\sim d$ then $ac\sim bd$.

So we get a monoid $(R/\sim, \times)$.

An element $a\in(R/\sim,\times)$ is irreducible if $a=bc$ implies that $b=1$ or $c=1$.

Then a unique factorization domain is one where your statement is true in $R/\sim$ (excluding $0$.)

Thomas Andrews
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