# $f(x) \in F[x]$ is irreducible over $F$ if $f(x) = g(x)h(x)$ implies that $g(x) \in F$ or $h(x) \in F$ where $F$ is a field. [duplicate]

$$f(x) \in F[x]$$ is irreducible over $$F$$ if $$f(x) = g(x)h(x)$$ implies that $$g(x) \in F$$ or $$h(x) \in F$$ where $$F$$ is a field.

If $$g(x) \in F$$ or $$h(x) \in F$$, then doesn't that make $$f(x)$$ reducible since it has factors in $$F$$?

• If that was the case, every polynomial in $F[x]$ would be reducible since $f=1\cdot f$. Nov 4 '20 at 12:03

No, because being reducible (if $$f(x)\ne0$$) means that you can factor it as a product of non-invertible elements of $$F[x]$$.