How can I prove using integral extensions that, if $R$ is a commutative ring with $1$, and $f \in R[x]$, then there is a ring extension $R \subseteq S$ such that $S$ contains all the roots of $f$? Trying to mimic the proof for splitting field extensions, the first step would be finding an extension containing one root, and the rest would follow by induction. Problem is, the obvious injection $\iota: R \longrightarrow R[x]/(f)$ sending $r \in R$ to $r+(f) \in R[x]/(f)$ doesn't work anymore, because $R$ is not necessarily a field, and thus I can't take $R[x]/(f)$ as an extension of $R$ having a root.
I guess the proof would involve integral extensions, as this is a part of a problem involving them.