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Suppose for a UFD $R,$ we have $F \in R[X]$ factor into $F = GH$ for $G \in R$ a non-unit. By definition, does this mean $F$ is reducible?

Furthermore, why do we only care about irreducibles in UFDs? The definition of irreducible polynomials seems to only be stated in fields or UFDs but not in any other context. I know $R[X]$ is a UFD iff $R$ is a UFD. But what is so important about uniqueness that we define them only in these rings?

green frog
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  • If $G$ and $H$ are non-units, then $F$ is not irreducible. Not really sure what your second paragraph is asking. Factorization is as important in rings as it is for the integers. – Dzoooks Jul 01 '18 at 02:34
  • You can actually define irreducible elements in any integral domain, see the end of the wikipedia page on irreducible polynomial https://en.wikipedia.org/wiki/Irreducible_polynomial – Aaron Johnson Jul 01 '18 at 03:06
  • As for why $R$ has to be an integral domain, see the question https://math.stackexchange.com/questions/1838066/irreducible-elements-for-a-commutative-ring-that-is-not-an-integral-domain/1838593#1838593 – Aaron Johnson Jul 01 '18 at 03:13

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By definition, $F\in R[X]$ is irreducible in $R[X]$ if and only if whenever we write $F=GH$, for $G,H\in R[X]$, then either $G$ or $H$ is a unit. So $F$ is reducible if and only if there exists a factorization of $F$ as $F=GH$ with both $G$ and $H$ non units. (As an aside: The units of $R[X]$ are precisely the units of $R$, i.e. $(R[X])^\times = R^\times$.)

So, assuming $F=GH$ and $G\in R$ is a nonunit, this does not necessarily mean that $F$ is reducible. As a silly example, in $\Bbb Z[X]$, let $F=G=2$, and $H=1$. Then $2=2\cdot 1$, with $G=2$ a non unit. But $2$ is not reducible in $\Bbb Z[X]$.

As to your question about why we care about irreducibles in UFDs: Well, "UFD" = unique factorization domain. And this unique factorization of each element is, by definition, a factorization into irreducibles.

So, when you ask "why do we care about irreducibles in UFDs", you're essentially asking, "Why do we care about irreducibles in a ring which we care about irreducibles?"

Nicholas Camacho
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