Not irreducible over F means reduccible over F (you just phrased it in a very strange way).

Now, F[x]/<f(x)> is an integral domain if and only if <f(x)> is a prime ideal.

To complete the proof show that <f(x)> isnota prime ideal (of course you use the fact that f(x) is reducible). We show this like this. <f(x)> is a principal ideal by definition. And f(x)=p(x)q(x) where deg {p(x)} and deg {q(x)} are both strictly less then deg{f(x)} because it is reduccible. Then since f(x) in <f(x)> it means p(x)q(x) in <f(x)> by our assumption that this principal ideal is a prime ideal would imply that p(x) in <f(x)> or q(x) in <f(x)> which is not possible because the degree of any element in <f(x)> is at least as large as f(x) because <f(x)> = {r(x)f(x) | r(x) in F[x]}, and hence cannot be in this ideal. Thus, the assumption that <f(x)> was a prime ideal was a contradiction.

Q.E.D.

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What does this have to do with a ring homomorphism?