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Are coproducts left exact or right exact functors in general?

Let k be a commutative ring (unital assosiative).
Specifically in the category of k-algebras is the tensor exact. (This is not the case in the category of k-modules, but then again the tensor is not a coproduct therein).

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  • $\begingroup$ Colimit functors are right exact. $\endgroup$ – Zhen Lin Feb 9 '14 at 2:13
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Let $A$ be a category with finite colimits. Denote by $\sqcup\colon A\times A\to A$ the functor which sends every pair $(a_1,a_2)$ of objects to their coproduct $a_1\sqcup a_2$. Then $\sqcup$ is left adjoint to the diagonal functor $\Delta\colon A\to A\times A$, then $\sqcup$ preserve all colimits and therefore is right exact.

But they are not left exact in general. For example, functor $\sqcup\colon\mathbf{Set}\times\mathbf{Set}\to\mathbf{Set}$ doesn't preserve terminal objects.

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  • $\begingroup$ Prefect!!! Thanks alot Oskar, I'll have to modify my proof a tiny bit but should be fine :) $\endgroup$ – AIM_BLB Feb 9 '14 at 2:48
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    $\begingroup$ @CSA You are welcome. $\endgroup$ – Oskar Feb 9 '14 at 2:50
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To answer your question about commutative $k$-algebras:

If $A$ is a commutative $k$-algebra, then $A \otimes_k - : \mathsf{CAlg}(k) \to \mathsf{CAlg}(k)$ is right exact, and it is also left exact when $k$ is a field, or more generally when $A$ is flat over $k$ (i.e. as a module). But in general, the functor doesn't even have to preserve monomorphisms.

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