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$f(x) \in F[x]$ 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 $F$?

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  • $\begingroup$ If that was the case, every polynomial in $F[x]$ would be reducible since $f=1\cdot f$. $\endgroup$ Nov 4 '20 at 12:03
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No, because being reducible (if $f(x)\ne0$) means that you can factor it as a product of non-invertible elements of $F[x]$.

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    $\begingroup$ This is just a comment (which says quite a bit about the question, if you ask me). $\endgroup$ Nov 4 '20 at 12:01

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