In Chapter 17 of Gallian's Contemporary Abstract Algebra, 8th Edition, irreducible polynomials are defined as: in an integral domain $D$, whenever $f(x)$ from $D[x]$ 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 $D[x]$. Their first example is $f(x)=2x^2+4$, which is irreducible over $Q$ (presumably because $2$ is a unit, since it's inverse $\frac{1}{2}$ is also in $Q[x]$?), but is reducible over $Z$, because neither $2$ nor $x^2+2$ is a unit in $Z[x]$.
This makes sense to me, but the very next example says that the same $f(x)$ is irreducible over $\textbf{R}$ but reducible over $\textbf{C}$, and I don't understand. By the same definition, I think that either (or both) $2$ or $x^2+2$ is a unit in $\textbf{R}[x]$, but that neither is a unit in $\textbf{C}[x]$, but I don't understand why $2^{-1}=\frac{1}{2}$ isn't part of $\textbf{C}[x]$. To be clear, I do understand that $f(x)=2x^2+4$ must be reducible over $\textbf{C}$ by later theorems (because it has a zero in the field $\textbf{C}$), I just don't understand the implication that neither $2$ nor $x^2+2$ is a unit in $\textbf{C}[x]$.