Definition of irreducible polynomial via the units of an integral domain. Let $D$ be an integral domain. A polynomial $f(x)$ from $D[x]$ that is
neither the zero polynomial nor a unit in $D[x]$ is said to be irreducible
over $D$ if, whenever $f(x)$ is expressed as a product $f(x)= g(x)h(x)$, with
$g(x)$ and $h(x)$ from $D[x]$, then $g(x)$ or $h(x)$ is a unit in $D[x]$.
This is the definition of the irreducible polynomial in Gallian's book. What I don't understand is how this definition translate to the fact that the polynomial cannot be factored into a product of polynomials of lower degree when $D$ happens to be a field.
There is also this examples that the polynomial $f(x)=2 x^2+ 4$ is irreducible over $\mathbb R$, but reducible over $\mathbb{C}$.
To me it seems that it is also irreducible over $\mathbb{C}$ since in its factorization $2(x^2+2)$, $2$ is unit over $\mathbb{C}$.
 A: That common definition of an (ir)reducible polynomial works over a field but needs to be modified when you pass to more general coefficient rings, because they may contain constants that are nonzero nonunits. In your example, $\,f = 2x^2+4\,$ is irreducible over $\,\Bbb Q,\,$ but is reducible over $\,\Bbb Z\,$ where $\,f = 2(x^2+2)\,$ and both factors are nonzero nonunits, but both factors are not of lower degree.
Over non-fields, the constants can convey very useful divisibility information so we don't want to ignore them when studying polynomial divisibility. For example, one way to view Gauss's Lemma is that the prime $\,p\in\Bbb Z\,$ extends to a prime in $\,\Bbb Z[x],\,$ i.e. $\,p\mid fg\,\Rightarrow\,p\mid f\,$ or $\,p\mid g.\,$
A: The basic idea is that for any possible combination of factors if none of them is non zero and non unit then the polynomial is reducible over that integral domain.
In ID  z(X)only the constants 1 and -1 are units. So 2(x^2)+2 can be factored as (2)((x^2)+1) both non zero and non unit ( although degree of both is not lower). So it is reducible in z(X).
But in case of fields all constants, non zero are units. So to make combination of non zero and non unit factors the degree of both factors must be lower (any possible case) to show the polynomial is  reducible. If no such combination exists then polynomial is irreducible.
So 2(x^2)+2 if factored as (X+i)(2x-2i) gives terms not belonging to R(X).
And if factored as (2)(x^2 +1) then 2 is a unit in R(X).
In other words it can't be broken into two factors of lower positive degrees. So it is irreducible in R(X).
In C(X) there exists combination (X+i)(2x-2i) both nonzero, non unit and positive lower degrees. So reducible over c(X).
