Tensor product terminology in category theory?











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Say that I have any homomorphisms of commutative rings, $A rightarrow B, A rightarrow C.$ I recently read that $B otimes_A C$ is the pushout of the morphisms in the category of commutative rings. However, I understood tensor products as defined for modules over a commutative ring. Can we somehow realize $B, C$ as modules over $A$ via the homomorphisms or is $B otimes_A C$ as defined by the pushout a generalization of the 'normal' definition of tensors? It seems unlikely that it is a generalization as it seems to depend on these morphisms whereas the tensors of modules doesn't depend on any sort of morphisms.










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    Say that I have any homomorphisms of commutative rings, $A rightarrow B, A rightarrow C.$ I recently read that $B otimes_A C$ is the pushout of the morphisms in the category of commutative rings. However, I understood tensor products as defined for modules over a commutative ring. Can we somehow realize $B, C$ as modules over $A$ via the homomorphisms or is $B otimes_A C$ as defined by the pushout a generalization of the 'normal' definition of tensors? It seems unlikely that it is a generalization as it seems to depend on these morphisms whereas the tensors of modules doesn't depend on any sort of morphisms.










    share|cite|improve this question


























      up vote
      5
      down vote

      favorite









      up vote
      5
      down vote

      favorite











      Say that I have any homomorphisms of commutative rings, $A rightarrow B, A rightarrow C.$ I recently read that $B otimes_A C$ is the pushout of the morphisms in the category of commutative rings. However, I understood tensor products as defined for modules over a commutative ring. Can we somehow realize $B, C$ as modules over $A$ via the homomorphisms or is $B otimes_A C$ as defined by the pushout a generalization of the 'normal' definition of tensors? It seems unlikely that it is a generalization as it seems to depend on these morphisms whereas the tensors of modules doesn't depend on any sort of morphisms.










      share|cite|improve this question















      Say that I have any homomorphisms of commutative rings, $A rightarrow B, A rightarrow C.$ I recently read that $B otimes_A C$ is the pushout of the morphisms in the category of commutative rings. However, I understood tensor products as defined for modules over a commutative ring. Can we somehow realize $B, C$ as modules over $A$ via the homomorphisms or is $B otimes_A C$ as defined by the pushout a generalization of the 'normal' definition of tensors? It seems unlikely that it is a generalization as it seems to depend on these morphisms whereas the tensors of modules doesn't depend on any sort of morphisms.







      category-theory terminology tensor-products






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      edited Nov 25 at 6:51









      Shaun

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      asked Nov 25 at 6:47









      伽罗瓦

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          Take all rings here to be commutative. A ring homomorphism $f:Ato B$
          makes $B$ into an $A$-module. In detail, the module action is $acdot b=f(a)b$.
          With another ring homomorphism $g:Ato C$ then we have two $A$-modules,
          and can form the tensor product $Botimes_A C$.



          At first $Botimes_A C$ is just a module. But it has a multiplication,
          defined as the composition
          $$(Botimes_A C)times(Botimes_A C)to (Botimes_A C)otimes_A (Botimes_A C)
          to (Botimes_A B)otimes_A (Cotimes_A C)to Botimes_A C.$$

          The middle map is just permuting the factors, and the last map is induced
          by $(botimes b')otimes(cotimes c')mapsto bb'otimes cc'$.
          In terms of elements:
          $$(botimes c)(b'otimes c')=bb'otimes cc'.$$



          Then $Botimes_A C$ is a ring. There are ring homomorphisms from $B$
          and $C$ to it, the first given by $amapsto aotimes 1_C$. Now one sits
          down with a large sheet of paper, and proves that the map $Ato Botimes_A C$
          is the pushout of $Ato B$ and $Ato C$.






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            1 Answer
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            1 Answer
            1






            active

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            active

            oldest

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            active

            oldest

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            up vote
            6
            down vote



            accepted










            Take all rings here to be commutative. A ring homomorphism $f:Ato B$
            makes $B$ into an $A$-module. In detail, the module action is $acdot b=f(a)b$.
            With another ring homomorphism $g:Ato C$ then we have two $A$-modules,
            and can form the tensor product $Botimes_A C$.



            At first $Botimes_A C$ is just a module. But it has a multiplication,
            defined as the composition
            $$(Botimes_A C)times(Botimes_A C)to (Botimes_A C)otimes_A (Botimes_A C)
            to (Botimes_A B)otimes_A (Cotimes_A C)to Botimes_A C.$$

            The middle map is just permuting the factors, and the last map is induced
            by $(botimes b')otimes(cotimes c')mapsto bb'otimes cc'$.
            In terms of elements:
            $$(botimes c)(b'otimes c')=bb'otimes cc'.$$



            Then $Botimes_A C$ is a ring. There are ring homomorphisms from $B$
            and $C$ to it, the first given by $amapsto aotimes 1_C$. Now one sits
            down with a large sheet of paper, and proves that the map $Ato Botimes_A C$
            is the pushout of $Ato B$ and $Ato C$.






            share|cite|improve this answer

























              up vote
              6
              down vote



              accepted










              Take all rings here to be commutative. A ring homomorphism $f:Ato B$
              makes $B$ into an $A$-module. In detail, the module action is $acdot b=f(a)b$.
              With another ring homomorphism $g:Ato C$ then we have two $A$-modules,
              and can form the tensor product $Botimes_A C$.



              At first $Botimes_A C$ is just a module. But it has a multiplication,
              defined as the composition
              $$(Botimes_A C)times(Botimes_A C)to (Botimes_A C)otimes_A (Botimes_A C)
              to (Botimes_A B)otimes_A (Cotimes_A C)to Botimes_A C.$$

              The middle map is just permuting the factors, and the last map is induced
              by $(botimes b')otimes(cotimes c')mapsto bb'otimes cc'$.
              In terms of elements:
              $$(botimes c)(b'otimes c')=bb'otimes cc'.$$



              Then $Botimes_A C$ is a ring. There are ring homomorphisms from $B$
              and $C$ to it, the first given by $amapsto aotimes 1_C$. Now one sits
              down with a large sheet of paper, and proves that the map $Ato Botimes_A C$
              is the pushout of $Ato B$ and $Ato C$.






              share|cite|improve this answer























                up vote
                6
                down vote



                accepted







                up vote
                6
                down vote



                accepted






                Take all rings here to be commutative. A ring homomorphism $f:Ato B$
                makes $B$ into an $A$-module. In detail, the module action is $acdot b=f(a)b$.
                With another ring homomorphism $g:Ato C$ then we have two $A$-modules,
                and can form the tensor product $Botimes_A C$.



                At first $Botimes_A C$ is just a module. But it has a multiplication,
                defined as the composition
                $$(Botimes_A C)times(Botimes_A C)to (Botimes_A C)otimes_A (Botimes_A C)
                to (Botimes_A B)otimes_A (Cotimes_A C)to Botimes_A C.$$

                The middle map is just permuting the factors, and the last map is induced
                by $(botimes b')otimes(cotimes c')mapsto bb'otimes cc'$.
                In terms of elements:
                $$(botimes c)(b'otimes c')=bb'otimes cc'.$$



                Then $Botimes_A C$ is a ring. There are ring homomorphisms from $B$
                and $C$ to it, the first given by $amapsto aotimes 1_C$. Now one sits
                down with a large sheet of paper, and proves that the map $Ato Botimes_A C$
                is the pushout of $Ato B$ and $Ato C$.






                share|cite|improve this answer












                Take all rings here to be commutative. A ring homomorphism $f:Ato B$
                makes $B$ into an $A$-module. In detail, the module action is $acdot b=f(a)b$.
                With another ring homomorphism $g:Ato C$ then we have two $A$-modules,
                and can form the tensor product $Botimes_A C$.



                At first $Botimes_A C$ is just a module. But it has a multiplication,
                defined as the composition
                $$(Botimes_A C)times(Botimes_A C)to (Botimes_A C)otimes_A (Botimes_A C)
                to (Botimes_A B)otimes_A (Cotimes_A C)to Botimes_A C.$$

                The middle map is just permuting the factors, and the last map is induced
                by $(botimes b')otimes(cotimes c')mapsto bb'otimes cc'$.
                In terms of elements:
                $$(botimes c)(b'otimes c')=bb'otimes cc'.$$



                Then $Botimes_A C$ is a ring. There are ring homomorphisms from $B$
                and $C$ to it, the first given by $amapsto aotimes 1_C$. Now one sits
                down with a large sheet of paper, and proves that the map $Ato Botimes_A C$
                is the pushout of $Ato B$ and $Ato C$.







                share|cite|improve this answer












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                answered Nov 25 at 7:00









                Lord Shark the Unknown

                98.4k958131




                98.4k958131






























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