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I am trying to prove without any galois theory that, for a finite field $K$, there exists irreducible polynomials of arbitrarily large degree. There are answers using Galois theory on MSE, e.g. here

My current thought is as follows, although I am uncertain as it seems too simple: Suppose all irreducible polynomials are degree $\leq n$, then as our field is finite, we can enumerate all the irreducible polynomials $P_1, ... P_s$. We then consider $G=\prod_{j=1}^{j=s}P_j +1$. Next, as $K$ is a field, $K[X]$ is a euclidean domain, and hence a PID, and therefore a UFD. In a UFD primes and irreducibles are equivalent. Therefore, if $G$ is not irreducible, we can factor it as a product of powers of $P_1, ..., P_s$.

Without loss of generality, suppose $P_1$ divides $G$. Then we can write $G=P_1f$, and $1=P_1(f-\prod_{j=2}^{j=s}P_j)$. However, as $K$ is a field, and therefore also an integral domain, clearly $P_1$ cannot have an inverse as by definition an irreducible element is not a unit.

(for context, I came across this question in a past paper for our introductory course on Groups Rings, and Modules. We haven't covered Galois theory*, and normally questions for the irreducibility of polynomials we are expected to use eisenstein's criterion, Gauss' lemma and elementary arguments. *for example, terms like 'splitting field' and 'field extension' are not defined in the course.

Also for context: I switched into the second year of a maths degree from economics, and am not particularly confident with abstract algebra compared to areas like real analysis and statistics , so apologies if this is an unintelligent question)

Thank you very much for your help :)

riemann_lebesgue
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    The proof seems correct, but note that you are conflating two questions: whether there are irreducibles of arbitrary degree, and whether there are irreducibles of arbitrarily large degree. The first is harder to show. – Andrew Dudzik Apr 03 '21 at 13:58
  • Ah, thank you!! That would explain why the galois machinery isn't necessary :))) – riemann_lebesgue Apr 03 '21 at 13:59
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    Yes, mimicking Euclid's classical proof of infinitely many primes works in this and related problems, which occurs frequently here, e.g. see the linked dupe. – Bill Dubuque Apr 03 '21 at 14:45
  • perfect thank you – riemann_lebesgue Apr 03 '21 at 15:20
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    Also, note that the accepted answer of the post you linked does not use any Galois theory. It's a counting argument, albeit a somewhat subtle one. The point is that $X^{q^n} - X$ is the product of all irreducibles of degree dividing $n$ (this can be shown in an elementary way), and that the solution to the recurrence $\sum_{d\mid n} a_d = q^n$ is always nonzero. – Andrew Dudzik Apr 04 '21 at 12:27

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