# Irreducible Polynomial definition

I was reading Gallian's Contemporary Abstract Algebra where I came across the following :

Let $D$ be an integral domain. A polynomial $f(x)\in D[x]$ that is neither the zero polynomial nor a unit is said to be irreducible over $D$ if whenever $f(x)$ is expressed as a product $f(x)=g(x)h(x)$, then $g(x)$ or $h(x)$ is a unit.

In the case that an integral domain is a field $F$, it is equivalent to define a nonconstant $f(x)\in F[x]$ to be irreducible if $f(x)$ cannot be expressed as a product of two polynomials of lower degree.

Two things changed when the integral domain is a field :

1. He used the term "nonconstant" instead of "neither zero nor unit".

2. He replaced "$f(x)=g(x)h(x)\implies g(x)$ or $h(x)$ is a unit" by "$f(x)$ cannot be expressed as a product of two polynomials of lower degree".

My questions regarding 1 and 2:

1. Aren't the nonzero non-invertible elements of $D[x]$ all the nonconstant polynomials anyway? I know that $2x+1$ in $\Bbb Z_4[x]$ is a non constant invertible polynomial, but $\Bbb Z_4$ isn't an integral domain. Is there such an example?

2. Again, could someone give an example showing that the two statements aren't the same in any $D[x]$?

Consider $4\in\mathbb{Z}[x]$. It cannot be expressed as the product of two polynomials of lower degree. But it is not irreducible, since $4=2\times2$.
• You are right. Sorry about that. It's like the integers. We divided into three classes: $0$, those which are invertible ($\pm1$) and the remaing ones. And the remaing ones can alwyas be written as a product of irreducible integers, that is, of prime numbers. We want to have such a classification in $D[x]$. – José Carlos Santos Sep 19 '18 at 14:40