I was given this problem, and the solution I came up with is eerily similar to the classic Euclid proof on infinitude of primes, so I'm feeling like I must be wrong somewhere.
Problem: Given $k$ any field, show that there are infinitely many irreducible monic polynomials in $k[X]$.
My attempt: assume there are finitely many irreducible monic polynomials $f_1, f_2, …, f_n$. Then consider $p = 1 + f_1 f_2 … f_n$, which is monic since all the $f_i$ are monic and $1$ is constant. By assumption, $p$ must be reducible, since it is not equal to any of the $f_i$. So there must exist some $g \in k[X]$ such that $g$ divides $f_1 f_2 … f_n$ — a contradiction since the $f_i$ are assumed to be irreducible.
I think my mistake was in assuming $p$ is not equal to any of the $f_i$ — I suppose in some finite field perhaps there is a case where that could be… I'm trying to work out in my head whether that's possible.
Am I on the right track or just totally off?
TIA!
Jason
EDIT: should be that one of these $f_i$ must divide $p$, since $p$ is reducible, but that's not possible because then they would have to divide 1.
Best Answer
It's easier than that, actually.
If $k$ is infinite, there are an infinite number of monic linear polynomials, and they are all irreducible. If $k$ is finite, $k$ has infinitely many finite extensions (corresponding to adjoining the appropriate roots of unity), and each extension has a generator which has a monic minimal polynomial over $k$ and these are all distinct and all irreducible. QED