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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 $\textbf{C}[x]$.

Pat Muchmore
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    $x^2+2$ is not a unit in $\mathbb{R}[x]$. And he never claims that "neither $2$ nor $x^2+2$ is a unit in $\mathbb{C}[x]$." You're misunderstanding the definition. – Randall May 28 '23 at 13:26
  • The fact that, yes, $2$ is a unit in $\mathbb{C}$ has nearly nothing to say about its reducibility. To be reducible means that there exists one factorization without unit terms, not that ALL factorizations must never have any unit terms. – Randall May 28 '23 at 13:31
  • @Randall, thank you, yes I'm definitely misunderstanding something. The definition says that $f(x)$ is irreducible over $D[x]$, if whenever $f(x)$ is expressed as a product $g(x)h(x)$, then $g(x)$ or $h(x)$ is a unit in $D[x]$ (emphasis added). Is "whenever" just a bad word choice in the text? In their first example, they explicitly state that $f(x)$ is reducible over $Z$ because "neither $2$ nor $x^2+2$ is a unit in $Z[x]$", seeming to imply that the unit test of any $g(x)h(x)$ is enough. – Pat Muchmore May 28 '23 at 13:52
  • No, "whenever" is fine, because it means "in every case." And they are still correct with the last example, because they have exhibited ONE factorization where no term is a unit, which means it IS NOT irreducible. – Randall May 28 '23 at 13:56
  • OK, so if I'm understanding you correctly, this test only works in one direction. When their first example says that $f(x)$ is irreducible over $Q$ but reducible over $Z$ for the reason I already typed, they are apparently not saying that it's irreducible over $Q$ because of the fact that $2$ is a unit? To use the definition as a proof that $f(x)$ is irreducible, I would have to find every possible $g(x)h(x)$ and show that those $g(x)$'s and $h(x)$'s are never units in $Q[x]$? – Pat Muchmore May 28 '23 at 14:23
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    No, to show that $f$ is irreducible you must find every possible factorization $gh$ and show that at least one of $g$ or $h$ is a unit in every one of these factorizations. – Randall May 28 '23 at 14:32

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