For a polynomial $f(x)$ in Integral Domain $D[x]$ to be irreducible, it has to be
neither the zero polynomial nor a unit
if $f(x)$ can be expressed as $f(x) = g(x)h(x)$ with $g(x), h(x) \in D[x]$, then $g(x)$ or $h(x)$ is a unit in $D[x]$
I assume unit here means an element which has a multiplicative inverse. I am unable to understand the significance of having (or not having) a multiplicative inverse in both the conditions.
Why isn't the condition for being an irreducible polynomial not just "one which cannot be factored into lower degree polynomials"? What extra does having these extra conditions add?
Also, in a field, these extra conditions don't exist - I can understand why the 2nd condition isn't required in a field - it's because every element of a field has a multiplicative inverse. However, if we do add the first condition, then there can never be a irreducible polynomial in a field - is that the sole reason for not having the first condition in a field or is there some other reason why that condition loses it's significance in a field?
This relates to the criteria that are used to define a unique factorization domain (UFD). In a ring of polynomials, for example, we would like each polynomial to have a unique factorization into irreducible elements, up to multiplication by a unit. If you do not include this last condition ("up to multiplication by a unit"), then you end up having multiple factorizations of the same polynomial, which creates unnecessary confusion. Consider a polynomial $p$ in the integral domain $F[X]$, where $F$ is a field. Then $p(X) = a_0 + a_1X + ... + a_nX^n$, for some $a_0, a_1,..., a_n \in F$. If $p$ appears in the factorization of some other polynomial $q(X) \in F[X]$, then we could define a "new" factorization of $q(X)$ by simply "factoring" $p(X)$ as $a_0(1 + a_0^{-1}a_1X + ... + a_0^{-1}a_nX^n)$. This doesn't really tell us anything new about either $p(X)$ or $q(X)$, so we tend to treat such a distinction as in some sense irrelevant.