Prove the converse to Euclid's lemma [e.g. for polynomials]

Question

Let $k$ be a field and let $f(x)$ be a nonconstant polynomial. If, whenever $f$ divides a product of two polynomials, it necessarily divides one of the factors, then $f$ is irreducible.

Answer 1

In a UFD (e.g. $\Bbb Z$ or $K[x])$ suppose if $f$ divides a product it divides some factor. If $\,f=gh\,$ then

$f = gh\mid gh\,\Rightarrow\, gh\mid g\ $ or $\ gh\mid h\,\Rightarrow\, h\mid 1\ $ or $\ g\mid 1,\,$ so $f\:\!$ is irreducible (so prime in a UFD)

Remark $ $ Generally, as above, if $\, f\mid gh,\ f\nmid g,\ f\nmid h\,$ then $\,\gcd(f,g)\,$ is a proper factor of $\,f,\,$ i.e. if we have specific witnesses $\,g,h\,$ to $\,f\,$ fails Euclid's Lemma then we can use them split $\,f\,$ into proper factors using an efficient gcd algorithm. In other words, there is a constructive proof of the contrapositive of prime $\,\Rightarrow\,$ irreducible in Euclidean (gcd) domains (the converse of the OP). See this answer for further discussion.