# polynomials factorization over rings and finite fields

Any nonzero polynomial over a subring $R$ of $\mathbb{C}$ is a product of irreducible polynomials over $R$. And for any subfield $K$ of $\mathbb{C}$, factorization of polynomials over $K$ into irreducible polynomials is unique up to constant factors and the order in which the factors are written.

Is it true or counterexamples that for any subring $R$ of $\mathbb{C}$, factorization of polynomials over $R$ into irreducible polynomials is unique up to constant factors and the order in which the factors are written?

Is it true or counterexamples that Any nonzero polynomial over a commutative ring $R$ with multiplicative identity is a product of irreducible polynomials over $R$?

Is it true or counterexamples that for $\mathbb{F}_p$, $p$ prime, factorization of polynomials over $\mathbb{F}_p$ into irreducible polynomials is unique up to constant factors and the order in which the factors are written?

• General results that come to fore: If $K$ is a field $K[x]$ is a PID, and thus also a UFD. If $R$ is a UFD, so is $R[x]$. Nov 30 '13 at 8:03

Some terms:

• A domain is factorization if every element can be factored into irreducible elements.
• And is unique factorization if the factorizations are unique up to associates and ordering.

If $F$ is a field then $F[x]$ is unique factorization. The standard proof shows that PID $\Rightarrow$ UFD and also shows that $F[x]$ has Euclidean division and hence is a PID (principal ideal domain). You can look this up in standard texts or online notes on abstract algebra and number theory.

There are subrings of $\Bbb C$ that are not factorization, let alone unique factorization. For example if we let $T$ be any transcendental element, if $P:=\{T^r:r\in\Bbb Q,r\ge0\}$, in the ring $R[P]$ the element $T$ cannot be factored into irreducibles (note $T=(\sqrt[n]{T})^n$ for all $n\ge1$). If a ring is not factorization then the formal polynomial ring over it is not factorization. This relies on the fact that "irreducible polynomials" are defined to be polynomials that are irreducible elements, and that if something is irreducible in a domain then it remains irreducible in the polynomial ring.

If we loosen our notion of "associate" to nonunits too (i.e. nonunit nonzero multiples of an element are also associate) our conclusions change. Let's call the corresponding notions "loose factorization" and "loose unique factorization."

Suppose $R\subseteq\Bbb C$ is a subring. Let $F={\rm Frac}(R)$ be the fraction field. Suppose a polynomial of $R[x]$ has two factorizations. These factorizations also exist in $F[x]$, and as $F$ is a field must be associate to each other up to units of $F$ and ordering, which means the two factorizations are equivalent within $R[x]$ by multiplying each factorization by an appropriate element of $R$. (This follows from the fact that all fractions in $F$ can be written as $a/b$ with $a,b\in R$.) Therefore, $R[x]$ has the property of being loose unique factorization.

The same reasoning applies to any domain $R$. For rings with zero divisors I am not sure if the polynomial ring is loose factorization. (One may also change the definition of "associates" again to mean up to nonunit non-zero-divisor multiples.)