I was working on a different problem when the following question occurred to me: For a polynomial $f\in K[x]$, is there always a constant $c\in K$ such that $f+c$ is irreducible?
Obviously this is false for some fields, consider $x^2$ over $\mathbb{F}_2$, $x^2$ factors as does $x^2+1$. In general it must be false for finite fields as $x^q+c$ has a root at $c$ for all $c$ in $\mathbb{F}_q$. Or take any algebraically closed field, trivially this cannot be done for a nonlinear polynomial since all nonlinear polynomials are reducible. If $K=\mathbb{R}$, then this is also clearly false for any polynomial of degree greater than 2, but it is true for polynomials of degree 2, since $x^2+ax+b$ has discriminant $a^2-4b$ so for $b>a^2/4$, the polynomial is irreducible.
So my question is actually the following: Are there any fields for which this is true for all polynomials? Also is it true in for all fields in the restricted sense? I.e. as in $\mathbb{R}$, true for all polynomials of at most degree $n$ and only true for these polynomials. Note that $n$ may be 1. Also can we compute this $n$?
I guess what I'm asking is when can I add a constant to make a polynomial irreducible? Also interesting is the case when $K$ is replaced with a ring.
I have no idea how to begin this question, any suggestions would be appreciated. I would prefer a good indication of what direction to go if it would be reasonable to solve this problem with my background (undergraduate algebra/number theory/basic galois theory/very little analysis) rather than a complete solution, but either is appreciated.
I should mention that I did google this and found nothing, but if you find something, that would also be appreciated. Thanks, I hope I'm not missing something obvious.
Take $k = \mathbb{Q}$, and let $f\in k[X]$ be an arbitrary non-constant polynomial. The polynomial $g(X, Y) = f(X) + Y$ is clearly irreducible, so there exists some $y_0\in k$ such that $g(X, y_0) = f(X) + y_0$ is irreducible by Hilbert's irreducibility theorem. The same proof works verbatim for any Hilbertian field $k$ (for example, a finite extension $k/\mathbb{Q}$).