Given the category of commutative R- or k-Algebras, it is often mentioned that the coproduct is the same as the tensor product. I'm interested in the proof of this statement.

One idea would be to proof some unique property, but I'm not sure how to do this (or if this is actually the right way). An elaborative explanation would be much appreciated.


Proving the tensor product of commutative $R$-algebras satisfies the universal property of the coproduct is straightforward, but it might be more illuminating to figure out the more general universal property of the tensor product of non-necessarily commutative $R$-algebras.

Let $A$, $B$ and $C$ be $R$-algebras, not necessarily commutative. Let's compute $\mathrm{Hom}(A \otimes_R B, C)$, where $\mathrm{Hom}$ means morphisms of $R$-algebras. An $R$-algebra homomorphism $f : A \otimes_R B \to C$ is an $R$-linear map such that $$f(a_1 a_2 \otimes b_1 b_2) = f(a_1 \otimes b_1) f(a_2 \otimes b_2).$$

Let $g : A \to C$ and $h:B \to C$ be given by $g(a) = f(a \otimes 1)$ and $h(b) = f(1 \otimes b)$. Plugging in $b_1 = b_2 = 1$ to the displayed equation, we see that $g$ is an $R$-algebra homomorphism. Similarly $h$ is also an $R$-algebra homomorphism and $f$ is determined by $f(a \otimes b) = f((a \otimes 1)(1 \otimes b)) = g(a) h(b)$. But we can't fully go backwards: if you start with $R$-algebras maps $g$ and $h$ and try to use this last formula to define an $f$, you find it is only well defined if $g(a)$ and $h(b)$ commute for any $a \in A$ and $b\in B$. Indeed, that is necessary because $(a \otimes 1)(1 \otimes b) = a \otimes b = (1 \otimes b) (a \otimes 1)$, so, applying $f$, we'd find $g(a) h(b) = f(a \otimes b) = h(b) g(a)$.

You can easily check that that commutativity is enough to make $f$ well defined when starting from $g$ and $h$ and thus we get the universal property of the tensor product:

$$ \mathrm{Hom}(A \otimes_R B, C) = \{ (g,h) \in \mathrm{Hom}(A,C) \times \mathrm{Hom}(B,C) : \forall a \in A, b \in B, g(a) \text{ and } h(b) \text{ commute}\}.$$

Clearly, if we restrict to the subcategory of commutative $R$-algebras, the commutation condition is automatic and this reduces to the universal property of the coproduct.

Notice that the situation in the category of groups is quite similar: the direct product of groups has a universal property analogous to this one for the tensor product, i.e., group homorphisms from $G \times H$ into $K$ are given by pairs of homomorphisms $G \to K$ and $H \to K$ whose images commute in $K$. The direct product of groups is also not the coproduct (that would be the free product of groups), but is the coproduct when you restrict to Abelian groups. (One place where this analogy breaks down is that the direct product of groups is the product of groups, but the tensor product of $R$-algebras is not the product.)

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    $\begingroup$ Actually this is more than an analogy. Take any symmetric monoidal category, then the tensor product of algebra objects makes sense and satisfies the universal property that it classifies "commuting" morphisms of algebras. For $R$-modules this gives the tensor product of $R$-algebras, for sets this gives the (tensor) product of monoids (for example, groups). Your final comment does not say that this analogy doesn't work; it merely observes that the monoidal category of sets is cartesian, but the monoidal category of $R$-modules is not. $\endgroup$ – Martin Brandenburg Oct 21 '14 at 18:00
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    $\begingroup$ Thanks alot, I'll have to meditate a while over this! $\endgroup$ – cbb Oct 23 '14 at 15:32

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