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I am asked to prove or disprove that given a finite field $\mathbf{F}_q$, the ring $\mathbf{F}_q[x]$ contains irreducible polynomials of arbitrarily large degree. I couldn't think of a reason why this should be false so I tried to prove it. Inspired by Euclid:

Take arbitrary $\mathbf{F}_{q}$. Consider $X_{n}=\{f\in\mathbf{F}_{q}[x]:f\;\text{irreducible}, 1\leqslant\deg f\leqslant n\}$ (finite). Define: $$h(x)=\prod_{f\in X_n} f(x)+1$$ Then $\deg h>n$ and $h\in \mathbf{F}_q[x]$ irreducible.

Does this work? If not any pointers/should I be proving the opposite?

Mike Pierce
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user329864
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    You can't conclude that $h$ is irreducible, but you can surely say that it has an irreducible factor of degree $> n$. By arbitrarity of $n$, you are done. – Crostul May 25 '16 at 23:34
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    $h$ is not necessarily irreducible, but it is not divisible by any of the $f\in X_n$. I think you can't really avoid an argument by contradiction (if there was a largest degree, then $X_n$ would consist of all irreducible polynomials for some $n$), but the idea is very nice indeed. – Luiz Cordeiro May 25 '16 at 23:36
  • More can be said. There exists an irreducible polynomial of every positive integer degree over a finite field. – hardmath May 26 '16 at 00:14
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    See for example Counting Irreducible Polynomials over a finite field. – hardmath May 26 '16 at 02:35
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    Thanks all, I can see now that $h$ is not necessarily irreducible (much like $\prod p + 1$ is not necessarily prime in Euclid's proof) but that it is sufficient to guarantee the existence of an irreducible of higher degree. – user329864 May 26 '16 at 07:20

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