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How to prove the following statement?

In $\mathbb Q[x]$, for any irreducible polynomial $g(x)$, there exists an irreducible polynomial $f(x)$ such that $g(f(x))$ is reducible.

(from a discussion. By searching MSE, I found a related question)

Example

Given $g(x) = x^2+1$, there is an irreducible polynomial $f(x) = x^2+3x+3$, and $g(f(x)) = (x^2 + 2x + 2)(x^2 + 4x + 5)$ is reducible.

Example

Given $g(x) = x^2+x+1$, how to find an irreducible polynomial $f(x)$ such that $f(x)^2+f(x)+1$ is reducible?

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    It is a cool fact that if $g(x)$ is any polynomial and $c$ is any constant, then the polynomial $g(x+cg(x))$ is divisible by $g(x)$. (Here is a proof when the roots of $g(x)$ are distinct; the proof is for $c=1$ but immediately generalizes.) So if any of the polynomials $x+cg(x)$ are irreducible ($c\ne0$) then the answer is yes. In particular, if $g(x)=x^2+x+1$, we can choose $f(x)=x+2g(x)=2x^2+3x+2$ which is irreducible, obtaining $g(f(x)) = g(x)(4x^2+8x+7)$. – Greg Martin Oct 17 '23 at 20:19
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    @Greg Simpler: $\bmod \color{#c00}{g(x)}!:\ \color{#c00}{g(x)\equiv 0}\Rightarrow g(x+c\color{#c00}{g(x)})\equiv g(x)\equiv 0,,$ as used here to show nonconstant polynomials cannot have only prime values. – Bill Dubuque Oct 17 '23 at 20:33
  • @BillDubuque I agree, that's the proof that I like—though a lot of undergraduates might be unfamiliar with modulo-a-polynomial. – Greg Martin Oct 17 '23 at 20:42
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    @Greg But as I show in my prior link it can equivalently be proved by the Factor Theorem, i.e. $,g(\color{#0a0}{x+cg(x)})-g(\color{#c00}x),$ is divisible by $\color{#0a0}{x+cg(x)} - \color{#c00}x = cg(x)$ thus $,g(x)\mid g(x+cg(x))\ \ $ – Bill Dubuque Oct 17 '23 at 20:50

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