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nLab > Latest Changes: tensor product of algebras

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    • CommentRowNumber1.
    • CommentAuthorUrs
    • CommentTimeSep 24th 2012
    • CommentRowNumber2.
    • CommentAuthorTobyBartels
    • CommentTimeSep 27th 2012

    Tensor product is coproduct only for commutative algebras.

    • CommentRowNumber3.
    • CommentAuthorUrs
    • CommentTimeAug 20th 2020

    After the statement that pushout = tensor product in CRingCRing, i have added pointer to category of monoids – pushouts – of commutative monoids

    diff, v6, current

    • CommentRowNumber4.
    • CommentAuthorRodMcGuire
    • CommentTimeAug 23rd 2020

    I’m not sure why Urs used this strange link syntax that doesn’t work. I tried to fix it and make it more standard changing

    See at _[pushouts of commutative monoids](category+of+monoids#PushoutOfCommutativeMonoids)_

    to

    See at _[[category of monoids#PushoutOfCommutativeMonoids|pushouts of commutative monoids]]_

    diff, v7, current

    • CommentRowNumber5.
    • CommentAuthorUrs
    • CommentTimeAug 24th 2020

    Thanks for noticing that my link didn’t work. I wonder why: I use the syntax all the time (maybe it dates from a time when the other syntax wasn’t available yet), and it seems I didn’t have a typo in it.

    Checking what happens when I paste my code into the Sandbox…

    Ah, there it works. (!?)

    • CommentRowNumber6.
    • CommentAuthorGeoffVooys
    • CommentTimeDec 10th 2024

    I’m a little worried that the definition of R↓RigR \downarrow \mathbf{Rig} or R↓RingR \downarrow \mathbf{Ring} does not necessarily capture the correct category of RR-algebras (in the sense that for a commutative object AA and an object RR with a map f:A→Rf:A \to R, the map determines an AA-algebra structure if and only if it factors through the centre Z(R)Z(R) of RR).

    Here’s an explicit example of what can go wrong: if you let A=ℂA = \mathbb{C} and σ:ℂ→ℂ\sigma:\mathbb{C} \to \mathbb{C} denote complex conjugation then the ring of twisted polynomials ℂ[x,σ]\mathbb{C}[x,\sigma] is a ring with a map ℂ→ℂ[x,σ]\mathbb{C} \to \mathbb{C}[x,\sigma] but which does not factor through Z(ℂ[x,σ])Z(\mathbb{C}[x,\sigma]) because Z(ℂ[x,σ])=ℝ[x 2]Z(\mathbb{C}[x,\sigma]) = \mathbb{R}[x^2]. Alternatively and more explicitly, i⋅x=i⋅xi\cdot x = i\cdot x but x⋅i=ı¯x=−i⋅x x\cdot i= \overline{\imath}x = -i\cdot x. This suggests that in this case ℂ↓Ring≠ℂAlg\mathbb{C} \downarrow \mathbf{Ring} \ne \mathbb{C}\mathbf{Alg}, or at least not in the “usual” sense.

    Did I misinterpret the sense in which algebra is meant on this page? Certainly taking the coslice category of the commutative object over the base is a nice generalization.

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