Why is $E \otimes_B B^n$ isomorphic to $E^n$ (for finite dimensional commutative $B$-algebras)? $B$ field, $E$ field and finite dimensional (as a $B$-vector space) $B$-algebra. In categorical terms, why is $E \otimes_B B^n \cong E^n$?
In $(B \downarrow \mathbf{CRing}) \cong B\text{-Alg}$, we have the initial object $(B, 1_B)$. Hence under the left adjoint $E \otimes_B (-): (B \downarrow \mathbf{CRing}) \to (E \downarrow \mathbf{CRing})$, $E \otimes_B B \cong E$. But products of initial are not obvious to me.
I am thinking the following:


*

*The following red diagram is commutative.

*$B\text{-Vect}$ (and $E\text{-Vect}$) are linear categories hence finite products and finite coproducts coincide.

*The bottom functor $E \otimes_B(-)$ is a left adjoint, hence preserves coproducts.



Then the forgetful functor $U_E: (E \downarrow \mathbf{CRing}) \to E\text{-Vect}$ creates and preserves limits, so one should have $E \otimes_B B^n \cong E^n$.
But this cannot be the correct argument because then for any $B$-algebra $A$ which is finite dimensional as a $B$-vector space, $E \otimes_B A \cong E^n$, which is not true unless $B \subseteq E$ is a Galois extension and $A$ is split by the extension. So what has gone wrong?
 A: This is really a statement about the fact that the tensor product of algebras commutes with finite products, and not necessarily one about. If we write $E \cong B(\alpha_1, \cdots, \alpha_m)$ (by virtue of the fact that $E$ is a finite field extension of $B$) and if we write
$$
B^n = \prod_{i=1}^{n} B \cong \bigoplus_{i=1}^{n} Be_i
$$
for $\lbrace e_i \; | 1 \leq i \leq n \rbrace$ a system of orthogonal idempotents spanning $B^n$, it follows that 
$$
B(\alpha_1, \cdots, \alpha_m) \otimes_B B^n \cong B(\alpha_1, \cdots, \alpha_m) \otimes_B\left( \bigoplus_{i=1}^{n}Be_i \right) \cong \bigoplus_{i=1}^{n}(B(\alpha_1, \cdots, \alpha_m) \otimes_B Be_i) \cong \bigoplus_{i=1}^{n} B(\alpha_1, \cdots, \alpha_m)e_i \cong E^n.
$$
Note that I never had to leave the category of $B$-algebras; I only had to consider a specific endofunctor (namely $(-)\otimes_B B^n)$ and then use the fact that the tensor product commutes with finite products. 
This should make sense categorically and geometrically for the following reason: In $\mathbf{ASch}_{/\operatorname{Spec} B}$, the category of affine schemes over $B$, 
$$
\operatorname{Spec} B^n \cong \coprod_{i=1}^{n} \operatorname{Spec} B.
$$
Then, because pullbacks commute with finite coproducts in the category of schemes over $B$ (and hence, in particular, in the category of affine schemes over $B$)
$$
\operatorname{Spec}(E \otimes_{B} B^n) \cong \operatorname{Spec}E \times_B \operatorname{Spec}B^n \cong \operatorname{Spec} E \times_B \left(\coprod_{i=1}^{n}  \operatorname{Spec} B\right) \cong \coprod_{i=1}^{n}\left( \operatorname{Spec} E \times_B \operatorname{Spec}B\right) \\
\cong \coprod_{i=1}^{n} \operatorname{Spec} E \cong \operatorname{Spec} E^n.
$$
Note that this argument above follows from the opposite equivalence of categories $\mathbf{ASch}_{/\operatorname{Spec B}} \simeq (B/\mathbf{Cring})^{op} = (B-\mathbf{Alg})^{op}$.
