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.
I don't understand why the proof being similar to the classic Euclid proof on infinitude of primes would cause you to wonder if it is wrong. I do not find this similarity to be surprising since the statements being proved are similar.
I think your proof is close. Once you get to the step where $1+f_1\cdots f_n$ is reducible, then you can say that one of the $f_i$ (which is of degree $>0$) divides it.
Also, add that there exist an irreducible polynomial since $x$ is irreducible.