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Let $F$ be a field and $f(x)\in F[x]$ such that $f(x)$ divides $g(x)$ for every non constant polynomial $g(x)\in F[x].$ Show that $f(x)$ is a constant polynomial.

My solution goes like this: Let $a\in F.$ We assume, $g(x)=x+a\tag 1$ This means that, $f(x)|g(x)=x+a$. Also, if $g(x)=x$ we have $f(x)|x\tag 2$

From $(1)$ and $(2)$ we have, $f(x)|a.$ But $a\in F$ is arbitrary due to which $f(x)|a,\forall a\in F.$ This means $\deg f(x)$ can't be positive and so, $\deg f(x)=0$ which implies $f(x)$ is a constant polynomial.

Can the readability of the solution be improved? I want to know specifically, if I can make some changes in the above solution to make it more readable. Any suggestions on how to improve the solution will be greatly appreciated. Finally, if there are any mistakes in the solution above, please do consider pointing it out.

Thomas Finley
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    No mistake. More readable: $f(x)$ divides $x$ and $x+1$ hence $$f(x)\mid(x+1)-x=1,$$ i.e. $f(x)\in F^*$. – Anne Bauval May 15 '24 at 14:00
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    @AnneBauval Thanks a lot for your suggestions! – Thomas Finley May 15 '24 at 14:59
  • More generally, in any gcd domain: $,f,$ is a unit (invertible) $\iff d,$ divides two coprime elements. In OP it is trivial to find two coprimes among nonconstant polynomials. This viewpoint often proves handy, for example a fraction writable with coprime denominators is an integer (since its least denom $,d,$ dvides these coprime denoms so $,d,$ is a unit). $\ \ $ – Bill Dubuque May 15 '24 at 16:59
  • For a solution-verification question to be on topic you must specify precisely which step in the proof you question, and why so. This site is not meant to be used as a proof checking machine. – Bill Dubuque May 15 '24 at 17:01

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