If $r$ and $s$ are positive integers, $R$ a commutative ring and $a_0,\dots,a_r$, $b_0,\dots,b_s$ independent variables, we can consider the polynomials $f=\sum_{i=0}^ra_iX^i$ and $g=\sum_{j=0}^sb_iX^j$ in $R[a_0,\dots,a_r,b_0,\dots,b_s][X]$ and then compute their Sylvester resultant $$R_{r,s}=\mathsf{res}_{r,s}(f,g)\in R[a_0,\dots,a_r,b_0,\dots,b_s].$$
It is a result well-known among people that know such things that $R_{r,s}$ is an irreducible element in $R[a_0,\dots,a_r,b_0,\dots,b_s]$ if $R$ is a field of characteristic zero; see, for example, the answer to this MO question and/or, of course, the Gelfand-Kapranov-Zelevinsky book on resultants. In that answer, Vesselin Dimitrov tells us that in general over a field resultants are a power of an irreducible, but I think I can prove in that case that it is actually irreducible (so that maybe he means general resultants?) and my argument seems to work over a ring $R$ with prime nilradical (that is, with irreducible spectrum).
Is $R_{r,s}$ irreducible for what rings $R$?
