Defining irreducible polynomials over polynomial rings

Question

Let R be a ring, and R[x] be a polynomial ring.

Can we define what it means for a polynomial $p(x) \in R[x]$ to be irreducible over R[x]?

Various sources (such as Wikipedia) only provide such definition for a field or a unique factorization domain.

Answer 1

It is not clear what irreducible means in rings that are not domains.

You can define irreducible but it won't have all properties you expect. See this.

In particular, if $R$ is not a domain, it may happen that decomposing a polynomial in $R[x]$ does not simplify it (that is, does not reduce its degree).

For instance: $$ 5x+1=(2x+1)(3x+1) \bmod 6 $$

It is even possible to decompose a linear polynomial as a product of two quadratic polynomials: $$ x+1=(2x^2+x+7)(4x^2+6x+7) \bmod 8 $$

Answer 2

When I taught abstract algebra, I pointed out that any commutative ring with multiplicative unit could be split up into disjoint sets: \begin{align} 1.&\quad\{0\}\\ 2.&\quad\text{nonzero zero-divisors (including nilpotent elements)}\\ 3.&\quad\text{units}\\ 4.&\quad\text{decomposable elements: those writable as}\\ &\quad\text{product two elements not in classes 1, 2, 3}\\ 5.&\quad\text{everything else.} \end{align} The indecomposable elements, also called irreducibles, are those in class 5.

So, for $a$ to be irreducible would mean that $a$ was not writable as product of two non-unit non-zero-divisors, and my “irreducibles” would be the “strong irreducibles” in the categorization mentioned in the link offered by @lhf.