Theorem 9.3 from 'Integers, Polynomials, and Rings' by Ronald S. Irving is stated as follows:
Let K be a field and let n be a positive integer. Suppose f(x) is a polynomial in K[x] of degree n. If f(x) is not irreducible, then f(x) has a divisor of degree less than or equal to n/2.
My question is, are units in a ring of polynomials over a field considered as divisors?, and more generally, are units in a commutative ring divisors? the definitions from ProofWiki
https://proofwiki.org/wiki/Definition:Divisor_of_Polynomial
https://proofwiki.org/wiki/Definition:Divisor_(Algebra)
don't discard units as divisors. Consequently, Theorem 9.3 is trivially true since all the units are divisors of all polynomials. For example, let $f(x)\in K[x],f(x)\neq 0_{K[x]}$ & $p(x)=\frac{1}{2}$, then $\exists 2f(x)\in K[x],f(x)=p(x)\cdot 2f(x)$ where $p(x)\cdot 2f(x)=\frac{1}{2}\cdot 2f(x)=f(x)$, consequently $p(x)$ is a divisor of $f(x)$.
Best Answer
Units are divisors, but they are uninteresting: units divide $1$, and $1$ divides everything, so units divide everything: if $uv = 1$ then $a = a \cdot 1 = u(av)$. Every number in a commutative ring is divisible by (i) all units and (ii) all unit multiples of itself. Therefore these automatic factors, which you can never avoid, are considered trivial. When people speak about trying to factor an integer or a polynomial, they are for the most part interested in factorizations other than than those where one of the factors is a unit.
In particular, when saying a polynomial of (positive) degree $n$ in $K[x]$ for a field $K$ that is not irreducible must have a factor of degree at most $n/2$, the intended meaning here is nontrivial factor. Nonzero constant factors are automatic and therefore of no interest. It would be more accurate to say a nonconstant polynomial in $K[x]$ that is reducible has a nonconstant factor of degree at most $n/2$. The reader is expected to realize that is the intended meaning, I guess. Nobody fundamentally cares about trivial factorizations of the form $c((1/c)f(x))$ for $c \in K^\times$. There is no knowledge gained from such a factorization because you can do it for all polynomials.
Do you know what an integral domain is? If so, then each nonzero element of an integral domain $R$ is exactly one of the following types:
(i) unit (a factor of $1$),
(ii) irreducible (no matter how you write it as a product of two terms, exactly one of them is a unit)
(iii) reducible (everything else: it has some factorization into two numbers that are not both units).
There is a fourth type, called a prime, which is a number that divides a product of two terms only by dividing one of them. Every prime is irreducible, and for many $R$ that don't have unique factorization there will be some irreducibles in $R$ that aren't prime. Since primes are always irreducible, it's convenient to focus on (i), (ii), and (iii) when talking about factorization numbers as far as possible: "into irreducible parts" more or less.
Some $R$ without unique factorization have no irreducibles at all: the ring $\overline{\mathbf Z}$ of all* algebraic integers has lots of numbers that aren't units (such as integers greater than $1$) but it has no irreducible elements since if $\alpha$ is an algebraic integer that is not a unit then we can write $\alpha = \sqrt{\alpha}\sqrt{\alpha}$. Every nonzero algebraic integer that is not a unit is reducible in $\overline{\mathbf Z}$.
When $R = K[x]$ for a field $K$, the units are the nonzero elements of $K$ and that is why nonzero constant polynomials are considered trivial for the purpose of factoring (their appearance in a factorization conveys no knowledge). But when $R = \mathbf Z[x]$, the units are just $\pm 1$ and other nonzero integers are not units in $\mathbf Z[x]$: $2x + 4$ is irreducible in $\mathbf Q[x]$ (all linear polynomials in $K[x]$ are irreducible when $K$ is a field), but $2x+4$ is reducible in $\mathbf Z[x]$ since it is $2(x+2)$ and neither factor is a unit in $\mathbf Z[x]$. A polynomial in $\mathbf Z[x]$ is called primitive when its coefficients collectively are relatively prime, such as $6x^2 + 10x + 15$: the coefficients are not pairwise relatively prime but they are relatively prime as a triple and this polynomial is primitive. For primitive polynomials in $\mathbf Z[x]$, irreducibility in $\mathbf Z[x]$ and in $\mathbf Q[x]$ are equivalent conditions, but for non-primitive polynomials the situation is different, as with $2x+4$. By factoring out the gcd $a$ of the coefficients of a nonzero polynomial $f(x) \in \mathbf Z[x]$, you can write $f(x) = aF(x)$ where $a$ is a nonzero integer and $F(x)$ is primitive in $\mathbf Z[x]$. This breaks up the task of factoring $f(x)$ into two parts: factoring $a$ into prime numbers and factoring $F(x)$ into irreducible polynomials (necessarily primitive) in $\mathbf Z[x]$, with the latter task being essentially the same as factoring $F(x)$ into irreducible factors in $\mathbf Q[x]$ (any such factorization of $F(x)$ can be rescaled in each part to be a factorization into irreducibles in $\mathbf Q[x]$ that are primitive in $\mathbf Z[x]$, all automatically irreducible in $\mathbf Z[x]$).