# $Hom_{R-Alg}(\prod R, S)$

Given an $R$-algebra $S$, is $Hom_{R-Alg}(\prod_{i=1}^n R, S) = Hom_{S-Alg}(\prod_{i=1}^n S, S)$?

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If $R$ is a commutative ring and $T$ an $R$-algebra, a morphism $\phi:R\times \cdots\times R\to T$ of $R$-algebras is in particular left $R$-linear, so there exist $a_1$, $\dots$, $a_n\in T$ such that $\phi(x_1,\dots,x_n)=x_1a_1+\cdots x_ny_n$. Since $\phi$ is unitary, $\phi(1,\dots,1)=1$, so $a_1+\cdots+a_n=1$. Let $e_i=\phi(0,\dots,0,1,0,\dots,1)$, with the one in the $i$th position. Then multiplicativity implies that $e_i^2=e_i$ for all $i$ and $e_ie_j=0$ if $i\neq j$. This gives a concrete description of the set $\hom_{R\mathrm{-alg}}(R\times \cdots\times R, T)$, and you should be able to use it to answer your question.
Tensor products commute with finite products (of modules, but the same isom. gives it for rings), in particular $(\prod_{i=1}^{n} R) \otimes_R S = \prod_{i=1}^{n} S$.
$Hom_S(A \otimes_R S,T) = Hom_R(A,T)$. – Martin Brandenburg Nov 8 '10 at 23:37
The $\hom$'s in the question are sets of algebra homomorphisms. Your isomorphism involves, in principle, $\hom$'s of modules, so you still need to check that it induces an isomorphism in the subsets of ring homomorphisms. – Mariano Suárez-Alvarez Nov 9 '10 at 3:48
I also meant algebra homomorphisms. It is a well-known fact that the forgetful functor $S-Alg \to R-Alg$ has the tensor product as a left adjoint ... – Martin Brandenburg Nov 10 '10 at 10:22