I'm trying to make sense of two different definitions of an algebra over a ring. The definitions are as follows: If R is a commutative ring, then 1) S is an R-algebra if S is an R-module and has a compatible ring structure (such that addition agrees) 2) If [itex] \alpha:R \to S [/itex] is a ring homomorphism such that [itex] \alpha(R) [/itex] is in the centre of S, then [itex] \alpha [/itex] is an R-algebra. We normally abuse notation in this instance and say that S is the R-algebra when [itex] \alpha [/itex] is understood. I think the second definition is more technically valid, albeit more complicated. So my first question: Via the first definition, are all commutative rings R also R-algebras? It seems like this would be true, since R is an R-module over itself and it certainly has a compatible ring structure. Secondly, I want to make sure we can get from the second definition the first. I think I can do it as follows: First, the ring homomorphism [itex] \alpha:R \to S [/itex] defines an R-module structure on S via the map [itex] \rho: R\times S \to S, \rho(r,s) = \alpha(r)s [/itex]. Next, since [itex] \alpha(r)s = s \alpha(r) [/itex] for all r and s and [itex] \alpha [/itex] is a homomorphism, then [tex] [\alpha(r_1)s_1 ][\alpha(r_2)s_2] = \alpha(r_1r_2)s_1 s_2. [/tex] This last step seems right, but in my head I can't figure out why this gives compatibility with the ring structure of S.