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Let $φ:R \to S$ be a homomorphism between two rings with unit element, and let $Φ:R[x] \to S[x]$ be the following homoprhism:

for all $f(x) \in R[x], f(x) = a_nx^n + \cdots +a_0,\ Φ(f(x))=φ(a_n)x^n + \cdots +φ(a_0)$.

Prove that if $Φ(f(x))$ is irreducible, for $f(x) \in R[x]$ with $a_n \notin \kerφ$, then $f(x)$ is also irreducible.

Can you help me? I can't make any use of the $a_n \notin \kerφ$ part.

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    Without any assumptions on $R$ and/or $f$, this does not make any sense. I gave a counterexample as a comment to the 'answer'. – 2017-02-22
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    Thank you. If you make your comment an answer I will accept it. I was stuck because of the no field assumption. – 2017-02-22
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    So you mean homomorphic pull back of an irreducible polynomial is an irreducible polynomial? – 2018-05-12

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It would be attempting to state : If $f$ is reducible then $f = f_1f_2\ldots f_m$ and so $\Phi(f) = \Phi(f_1)\Phi(f_2)\ldots\Phi(f_m)$ is also reducible. In $f$ being reducible it is understood that at least two factors have degree $>0$ (otherwise each polynomial would be reducible with $f = 1\cdot f$). But this is not always true as shown by the example $f(x) = 2x^2+3x+1$ with $\varphi(x) = x \mod 2$. To remediate this situation one could make the statement to be restricted to monic polynomials (polynomials with leading coefficient equal to $1$) only. Indeed if $f$ is monic then its factors can always be rewritten as monic polynomials. Moreover the degree of $\Phi(f_i)$ always equals the degree of $f_i$ because $\varphi(1) = 1$. Then as a corrolary one can make an equivalent statement for $f(x) = a_nx^n + \cdots +a_0$ with $a_n$ invertible in $R$ since $\varphi(a_n)$ will also be invertible in $S$.

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    This would require for $S$ and $R$ to be fields though, right? – 2017-02-24
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    Not necessarily, rings can have invertible elements (called units) too, like in $\Bbb Z_{12}$ $5,7$ and $11$ are invertible. – 2017-02-24
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    Isn't $\Bbb Z_1$ the zero ring? It's finite and has no zero divisors. Doesn't that make it a field? – 2017-02-24
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    That was a typo, has been corrected. – 2017-02-24
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    Alright, I'll check your answer in full later. One more thing. In your example do you mean $f(x)$ in $\Bbb R$ or in $\Bbb Z$? Cause I think it's irreducible in the integers, right? – 2017-02-24
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    Indeed , in $\Bbb Z$. But $f(x) = (2x+1)(x+1)$ – 2017-02-24