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Suppose that $M$ is an $A$-module, and $N$ is a $B$-module. The coproduct of $A$ and $B$ is $A\otimes_{\mathbb{Z}}B$, and the coproduct of $M$ and $N$ is $M\oplus N$. I was wondering if $M\oplus N$ could be a module over $A\otimes_{\mathbb{Z}}B$, so that it is a "coproduct of two modules." But it appears that there is no $A\otimes_{\mathbb{Z}}B$-module structure on $M\oplus N$. Is it true that "the category of modules" doesn't have coproducts?

Edit: All rings are assumed to be commutative. I'm envisaging a "category of modules," where objects are pairs $(A,M)$, where $A$ is a ring and $M$ is an abelian group, together with $A$-scalar multiplication structure. A morphism $(A,M)→(B,N)$ is a pair $(f,ϕ)$, where $f:A→B$ is a ring homomorphism, $ϕ:M→N$ is an abelian group homomorphism, such that $ϕ(am)=f(a)ϕ(m)$. When I said "coproduct of $M$ and $N$", I'm treating them as abelian groups.

The question is whether this category has coproducts. My initial guess was that the coproduct of $(A,M)$ and $(B,N)$ is $(A\otimes_{\mathbb{Z}}B,M\oplus N)$. I think this is wrong, but I think a more complicated coproduct exists for any family of modules $(A_\alpha,M_\alpha)_\alpha$. My current guess looks something like $(\varinjlim\otimes A_\alpha,\varinjlim\otimes_{\beta\neq\alpha}A_\beta\otimes M_\alpha)$.

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You said "and the coproduct of $M$ and $N$ is $M\oplus N$." while you asked "Is it true that "the category of modules" doesn't have coproducts?" I don't think you make sense. –  Makoto Kato Sep 17 '12 at 21:46
    
I'm envisaging the "category of modules," where the objects are pairs $(A,M)$, where $A$ is a ring and $M$ is an abelian group, together with $A$-scalar multiplication structure. When I said "coproduct of $M$ and $N$", I'm treating them as abelian groups. –  ashpool Sep 17 '12 at 21:48
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I suggest you rewrite your question. Nobody can read your mind. –  Makoto Kato Sep 17 '12 at 21:52
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Well, edit the question to remove any possibility of confusion! (You seem to be assuming that the rings are commutative: being explicit about that too never hurts, either) –  Mariano Suárez-Alvarez Sep 17 '12 at 22:01
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The total category of modules is a Grothendieck opfibration over the category of rings, and there is a procedure to construct colimits in such things out of colimits in the base and colimits in the fibres. I think the coproduct you're thinking of is actually $(B \otimes_\mathbb{Z} M) \oplus (A \otimes_\mathbb{Z} N)$. –  Zhen Lin Sep 18 '12 at 0:55
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2 Answers 2

The natural way to write down an $A \otimes_{\mathbb{Z}} B$-module from this data is to actually take the tensor product $M \otimes_{\mathbb{Z}} N$. This follows from a careful inspection of the universal property of the tensor product.

$M \oplus N$ instead inherits an action of $A \oplus B$. The motivating example here is when $M, N$ are two finite-dimensional vector spaces $V, W$ and $A = \text{End}(V), \text{End}(W)$.

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But I can't think of an abelian group homomorphism $M\rightarrow M\otimes_{\mathbb{Z}}N$. I'm thinking of maybe $(M\otimes_{\mathbb{Z}}B)\oplus(A\otimes_{\mathbb{Z}}N)$. If $(f,\phi):(A,M)\rightarrow(C,L)$, $(g,\psi):(B,N)\rightarrow(C,L)$ are morphisms of modules, then the induced morphism will be $(A\otimes_{\mathbb{Z}}B,(M\otimes_{\mathbb{Z}}B)\oplus(A\otimes_{\mathbb{Z}}N))‌​\rightarrow(C,L),\quad(a\otimes b,m\otimes n)\mapsto g(b)\phi(m)+f(a)\psi(n)$. –  ashpool Sep 17 '12 at 22:25
    
Why do you want an abelian group homomorphism $M \to M \otimes_{\mathbb{Z}} N$? –  Qiaochu Yuan Sep 17 '12 at 22:34
    
If $(A\otimes_{\mathbb{Z}}B,M\otimes_{\mathbb{Z}}N)$ is a coproduct of $(A,M)$ and $(B,N)$, we need morphisms $(f,\phi):(A,M)\rightarrow(A\otimes_{\mathbb{Z}}B,M\otimes_{\mathbb{Z}}N)$ and $(g,\psi):(B,N)\rightarrow(A\otimes_{\mathbb{Z}}B,M\otimes_{\mathbb{Z}}N)$ in the category of modules. –  ashpool Sep 17 '12 at 22:46
    
@ashpool: in the first version of your question it was not clear that you wanted to construct coproducts in the "category of modules" (this is a confusing name). –  Qiaochu Yuan Sep 17 '12 at 22:47
    
Writing latex without preview is a nightmare! Here is a correction to my first comment: $(a\otimes b,m\otimes n)\mapsto(f(a)g(b),g(b)\phi(m)+f(a)\psi(n))$. –  ashpool Sep 17 '12 at 22:57
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$\newcommand{\dlim}{\varinjlim} \newcommand{\sst}{\subset} \newcommand{\sbe}{_\beta} \newcommand{\sal}{_\alpha} \newcommand{\iso}{isomorphism} \newcommand{\al}{\alpha} \newcommand{\be}{\beta} \newcommand{\fall}{\ \ \forall} \newcommand{\dra}{\dashrightarrow} \newcommand{\mt}{\mapsto} \newcommand{\ra}{\rightarrow} \newcommand{\cc}{\circ} \newcommand{\id}{\operatorname{id}} \newcommand{\itk}[1]{{\it #1}} \newcommand{\ras}{\overset{\sim}{\ra}} \newcommand{\amod}{$A$-module} \newcommand{\Hom}{\operatorname{Hom}} \newcommand{\ten}{\otimes} \newcommand{\opl}{\oplus} \newcommand{\rf}[1]{\overset{#1}{\rightarrow}} \newcommand{\ifof}{if and only if } \newcommand{\la}{\leftarrow} \newcommand{\Ra}{\Rightarrow} \newcommand{\set}[1]{\left\{ #1 \right\}} \newcommand{\setst}[2]{\left\{ #1 \ | \ #2 \right\}} \newcommand{\lf}[1]{\overset{#1}{\leftarrow}} \newcommand{\lff}[1]{\overset{#1}{\longleftarrow}} \newcommand{\im}{\operatorname{im}} \newcommand{\sur}{\twoheadrightarrow} \newcommand{\krn}{\operatorname{ker}} \newcommand{\hra}{\hookrightarrow} \newcommand{\las}{\overset{\sim}{\la}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\bb}{\backslash} \newcommand{\os}[2]{\overset{#1}{#2}} \newcommand{\us}[2]{\underset{#1}{#2}} \newcommand{\amods}{$A$-modules} \newcommand{\sm}{\smallskip} \newcommand{\beq}[1]{\begin{equation}#1\end{equation}} \newcommand{\ffall}{\qquad\forall} \newcommand{\del}{\delta} \newcommand{\isos}{isomorphisms} \newcommand{\Tcd}{There is a commutative diagram} \newcommand{\cd}{commutative diagram} \newcommand{\tcd}{there is a \cd} \newcommand{\tis}{there is an \iso} \newcommand{\Tis}{There is an \iso} \newcommand{\gam}{\gamma} \newcommand{\hmms}{homomorphisms} \newcommand{\Del}{\Delta} \newcommand{\fgas}{f.g.~\amods} \newcommand{\lam}{\lambda} \newcommand{\Tesq}{There is an exact sequence} \newcommand{\isoc}{isomorphic} \newcommand{\bopl}{\bigoplus} \newcommand{\wrt}{with respect to} \newcommand{\esqs}{exact sequences} \newcommand{\qqq}{\qquad\quad} \newcommand{\ds}{_*} \newcommand{\dla}{\dashleftarrow} \newcommand{\sab}{_{\al\be}} \newcommand{\sx}{_x} \newcommand{\Calg}{\mathsf{Alg}} \newcommand{\Cab}{\mathsf{Ab}} \newcommand{\ang}[1]{\langle #1\rangle} \newcommand{\hpsi}{\hat{\psi}} \newcommand{\hphi}{\hat{\phi}} \newcommand{\resp}{respectively} \newcommand{\cds}{commutative diagrams} \newcommand{\qq}{\qquad} \newcommand{\mf}{\mapsfrom} \newcommand{\aalg}{$A$-algebra} \newcommand{\aalgs}{$A$-algebras} \newcommand{\Cring}{\mathsf{Ring}} \newcommand{\Cmod}{\mathsf{Mod}} \newcommand{\alg}{algebra} \newcommand{\algs}{algebras} \newcommand{\hmm}{homomorphism}$

Proposition. The category $\Cmod$ has coproducts.

Proof. Let $(A\sal,M\sal)_{\al\in\Lambda}$ be a nonempty family in $\Cmod$. Let $A'$ be the direct limit of finite tensor products of $(A\sal)\sal$, i.e., $A'=\dlim\ten_{\al\in I}A\sal$, where $I\sst\Lambda$ is a finite subset. Given $\al\in\Lambda$, let $A'\sal:=\dlim\ten_{\be\in I\sal}A\sbe$, where $I\sal\sst\Lambda\bb\set{\al}$ is a finite subset. We claim that $A'\simeq A\sal\ten A'\sal\fall\al\in\Lambda$. To see this, first note that for every $\al\in\Lambda$, there is an obvious ring \hmm\ $\tau'\sal:A'\sal\dra A'$. This induces a ring \hmm $$\phi:A\sal\ten A'\sal\dra A',\qqq a\sal\ten x\mt\tau_{\set{\al}}(a\sal)\tau'\sal(x),$$ where $(\tau_I)_{I\sst\Lambda}$ is the direct limit for $A'$.

On the other hand, let $(\sigma_{I\sal})_{I\sal}$ be the direct limit for $A'\sal$. Given a finite subset $I\sst\Lambda$, define a ring \hmm $$\psi_I:A\sal\ten A'\sal\la \ten_{\al\in I}A\sal$$ as follows: if $\al\not\in I$, then $\psi_I(x):=1_{A\sal}\ten\sigma_I(x)$. If $\al\in I$, then $$\psi_I(\ten_{\be\in I}a\sbe)=a\sal\ten\sigma_{I'\sal}(\ten_{\be\in I'\sal}a\sbe),$$ where $I'\sal:=I\bb\set{\al}$. It is easy to see that $\psi_I$ is compatible with inclusions of the finite subsets $I$. Hence, there is an induced ring \hmm $$\hpsi:A\sal\ten A'\sal\dla A'.$$ It is easy to see that $\phi$ and $\hpsi$ are inverses of each other.

Now let $M'\sal:=M\sal\ten A'\sal$. It is then an $A'$-module, and we will show that $(A',\opl\sal M'\sal)$ is the coproduct of $(A\sal,M\sal)\sal$.

There is an abelian group \hmm $$j\sal:M\sal\ra M'\sal,\qqq m\sal\mt m\sal\ten1'\sal.$$ Let $(i\sal)\sal$ be the coproduct for $\opl\sal M'\sal$. It is easy to see that $$(\tau_{\set{\al}},i\sal\cc j\sal):(A\sal,M\sal)\dra(A',\opl\sal M'\sal)$$ is a morphism in $\Cmod$.

Let $(\phi\sal,\psi\sal)\sal$ be a family of morphisms in $\Cmod$, where $$(\phi\sal,\psi\sal):(A\sal,M\sal)\ra(B,N).$$ Then there are induced ring \hmms\ $\hphi:A'\dra B$, $\hphi\sal:A'\sal\dra B$, and a unique abelian group \hmm $$\hpsi\sal:M'\sal\dra N$$ \st $m\sal\ten x\mt\hphi\sal(x)\cdot\psi\sal(m\sal)$. Note that $\hpsi\sal\cc j\sal=\psi\sal$. It is easy to see that $$(\hphi,\hpsi\sal):(A',M'\sal)\ra(B,N)$$ is a morphism of modules, so there is an induced morphism of modules $$(\hphi,\hpsi):(A',\opl\sal M'\sal)\dra(B,N).$$ It is easy to see that $\hpsi\sal$ is the unique abelian group \hmm\ \st for every $\al\in\Lambda$, $\hpsi\sal\cc j\sal=\psi\sal$ and $(\hphi,\hpsi\sal)$ is a morphism of modules. Hence, the uniqueness of $(\hphi,\hpsi)$ follows.

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What \hmm\ is supposed to mean? And \st? –  user26857 Feb 26 '13 at 8:48
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