Written by dustbringer on 14 March 2024 . View source.

Tensor-Hom adjunctions for modules

To make the module structures explicit, we may write AMB_A M_B for an (A,B)(A,B)-bimodule MM. Also we write HomR(,)\Hom_{R_\ell}(-,-) for morphisms of left RR-modules, and HomRr(,)\Hom_{R_r}(-,-) for morphisms of right RR-modules.

Hom\Hom as a (bi)module

Let A,B,CA,B,C be rings. Then let

  • MM be an (A,B)(A,B)-bimodule,
  • NN be a (B,C)(B,C)-bimodule,
  • and KK be an (A,C)(A,C)-bimodule.

BHomA(AMB,AKC)C_B \Hom_{A_\ell}(_A M_B,_A K_C) _C

The additive group HomA(AMB,AKC)\Hom_{A_\ell}(_A M_B,_A K_C) can be endowed with the structure of a (B,C)(B,C)-bimodule. That is, for some left AA-module homomorphism f:MKf: M \to K, define

  • the left BB-action to be (bf)(m)=f(mb)(b \cdot f)(m) = f(m \cdot b)
  • and the right CC-action to be (fc)(m)=f(m)c(f \cdot c)(m) = f(m) \cdot c.

The left action is well-defined because (bf)(am)=f(amb)=af(mb)=a(bf)(m)(b \cdot f)(a \cdot m) = f(a \cdot m \cdot b) = a \cdot f(m \cdot b) = a \cdot (b \cdot f)(m), so bfb \cdot f is a left AA-module homomorphism. Furthermore it forms a module because

(b1b2f)(m)=f(mb1b2)=(b2f)(mb1)=(b1(b2f))(m),(b_1 b_2 \cdot f)(m) = f(m \cdot b_1 \cdot b_2) = (b_2 \cdot f)(m \cdot b_1) = (b_1 \cdot (b_2 \cdot f))(m),

where the other properties follow trivially.

The right action is well-defined because (fc)(am)=f(am)c=af(m)c=a(fc)(m)(f \cdot c)(a \cdot m) = f(a \cdot m) \cdot c = a \cdot f(m) \cdot c = a \cdot (f \cdot c)(m), so faf \cdot a is a left AA-module homomorphism. It also forms a module since

(fc1c2)(m)=f(m)c1c2=(fc1)(m)c2=((fc1)c2)(m).(f \cdot c_1 c_2)(m) = f(m) \cdot c_1 \cdot c_2 = (f \cdot c_1)(m) \cdot c_2 = ((f \cdot c_1) \cdot c_2)(m).

This is a bimodule because ((bf)c)(m)=(bf)(m)c=f(mb)c=(fc)(mb)=(b(fc))(m)((b \cdot f) \cdot c)(m) = (b \cdot f)(m) \cdot c = f(m \cdot b) \cdot c = (f \cdot c)(m \cdot b) = (b \cdot (f \cdot c))(m).

Notice that the bimodule structure was inherited from the "unused" right module structures of MM and KK.

BHomCr(AKC,BNC)A_B \Hom_{C_r}(_A K_C,_B N_C) _A

Similarly on the right, we can endow the additive group HomCr(AKC,BNC)\Hom_{C_r}(_A K_C,_B N_C) with the structure of a (B,A)(B,A)-bimodule. That is, for some right CC-module homomorphism f:KNf: K \to N, define

  • the left BB-action to be (bf)(k)=bf(k)(b \cdot f)(k) = b \cdot f(k)
  • and the right AA-action to be (fa)(k)=f(ak)(f \cdot a)(k) = f(a \cdot k).

The left action is well-defined because (bf)(kc)=bf(kc)=bf(k)c=(bf)(k)c(b \cdot f)(k \cdot c) = b \cdot f(k \cdot c) = b \cdot f(k) \cdot c = (b \cdot f)(k) \cdot c, so bfb \cdot f is a right CC-module homomorphism. This has a module structure since

(b1b2f)(k)=b1b2f(k)=b1(b2f)(k)=(b1(b2f))(k). (b_1 b_2 \cdot f)(k) = b_1 \cdot b_2 \cdot f(k) = b_1 \cdot (b_2 \cdot f)(k) = (b_1 \cdot (b_2 \cdot f))(k).

The right action is well-defined because (fa)(kc)=f(akc)=f(ak)c=(fa)(k)c(f \cdot a)(k \cdot c) = f(a \cdot k \cdot c) = f(a \cdot k) \cdot c = (f \cdot a)(k) \cdot c, so faf \cdot a is a right CC-module homomorphism. This gives a module structure with

(fa1a2)(k)=f(a1a2k)=(fa1)(a2k)=((fa1)a2)(k). (f \cdot a_1 a_2)(k) = f(a_1 \cdot a_2 \cdot k) = (f \cdot a_1)(a_2 \cdot k) = ((f \cdot a_1)\cdot a_2)(k).

This is a bimodule because ((bf)a)(k)=(bf)(ak)=bf(ak)=b(fa)(k)=(b(fa))(k)((b \cdot f) \cdot a)(k) = (b \cdot f) (a \cdot k) = b \cdot f (a \cdot k) = b \cdot (f \cdot a) (k) = (b \cdot (f \cdot a)) (k).

Now notice that this is flipped (compared to above): the left BB-module structure was inherited from BNC_B N_C, and the right AA-module structure was inherited from AKC_A K_C. If we had it the same way around here as before, we could try and define an (A,B)(A,B)-bimodule structure by (af)(k)=f(am)(a \cdot f)(k) = f(a \cdot m) and (fb)(k)=bf(k)(f \cdot b)(k) = b \cdot f(k). Again we can prove that afa \cdot f and fbf \cdot b are right CC-module homomorphisms. However we run into problems when trying to prove it is a valid module structure. For the left action, we see that (a1a2f)(k)=f(a1a2k)=(a1f)(a2k)=(a2(a1f))(k)(a_1 a_2 \cdot f)(k) = f(a_1 \cdot a_2 \cdot k) = (a_1 \cdot f)(a_2 \cdot k) = (a_2 \cdot (a_1 \cdot f))(k) is in the wrong order. For the right action, we see that (fb1b2)(k)=b1b2f(k)=b1(fb2)(m)=((fb2)b1)(m)(f \cdot b_1 b_2)(k) = b_1 \cdot b_2 \cdot f(k) = b_1 \cdot (f \cdot b_2)(m) = ((f \cdot b_2) \cdot b_1)(m) also has the wrong order.

Tensor-Hom Adjunctions

Let A,B,C,DA,B,C,D be rings. Then let

  • MM be an (A,B)(A,B)-bimodule,
  • NN be a (B,C)(B,C)-bimodule,
  • KK be a (D,C)(D,C)-bimodule,
  • and LL be a (A,D)(A,D)-bimodule

BNCHomCr(BNC,)- \otimes_B N_C \dashv \Hom_{C_r}(_B N_C, -)

We prove that DHomCr(AMBNC,DKC)ADHomBr(AMB,DHomCr(BNC,DKC)B)A_D \Hom_{C_r}(_A M \otimes_B N_C, _D K_C)_A \simeq {_D\Hom_{B_r}(_A M_B, _D \Hom_{C_r}(_B N_C,_D K_C)_B)_A} as (D,A)(D,A)-bimodules.

Define ϕ:HomCr(MBN,K)HomBr(M,HomCr(N,K))\phi: \Hom_{C_r}(M \otimes_B N, K) \to \Hom_{B_r}(M, \Hom_{C_r}(N,K)) to map

ϕ:f(m(nf(mn))).\phi: f \mapsto (m \mapsto (n \mapsto f(m \otimes n))).

This is an (D,A)(D,A)-bimodule homomorphism because

  • ϕ(f+g)=(mnf(mn)+g(mn))=ϕ(f)+ϕ(g)\phi(f + g) = (m \mapsto n \mapsto f(m \otimes n) + g(m \otimes n)) = \phi(f) + \phi(g)
  • and
ϕ(dfa)(m)(n)=(dfa)(mn)=df(amn)=dϕ(f)(am)(n)=(d(ϕ(f)a)(m))(n)=(dϕ(f)a)(m)(n). \begin{align*} \phi(d \cdot f \cdot a)(m)(n) &= (d \cdot f \cdot a)(m \otimes n) \\ &= d \cdot f(a \cdot m \otimes n) \\ &= d \cdot \phi(f)(a \cdot m)(n) \\ &= (d \cdot (\phi(f) \cdot a)(m))(n) \\ &= (d \cdot \phi(f) \cdot a)(m)(n). \end{align*}

Conversely, define ψ:HomBr(M,HomCr(N,K))HomCr(MBN,K)\psi: \Hom_{B_r}(M, \Hom_{C_r}(N,K)) \to \Hom_{C_r}(M \otimes_B N, K) to map

ψ:g(mng(m)(n)).\psi: g \mapsto (m \otimes n \mapsto g(m)(n)).

This is an (D,A)(D,A)-bimodule homomorphism because

  • ψ(f+g)=(mnf(m)(n)+g(m)(n))=ψ(f)+ψ(g)\psi(f + g) = (m \otimes n \mapsto f(m)(n) + g(m)(n)) = \psi(f) + \psi(g)
  • and
ψ(dfa)(mn)=(dfa)(m)(n)=(df(am))(n)=df(am)(n)=dψ(f)(amn)=(dψ(f)a)(mn). \begin{align*} \psi(d \cdot f \cdot a)(m \otimes n) &= (d \cdot f \cdot a)(m)(n) \\ &= (d \cdot f(a \cdot m))(n) \\ &= d \cdot f(a \cdot m)(n) \\ &= d \cdot \psi(f)(a \cdot m \otimes n) \\ &= (d \cdot \psi(f) \cdot a)(m \otimes n). \end{align*}

These functions are clearly inverses (by construction). Explicitly, this looks like

ψϕ(f)(mn)=ψ(mnf(mn))(mn)=(mn(mnf(mn)))(mn)=f(mn) \begin{align*} \psi \circ \phi (f)(m \otimes n) &= \psi(m \mapsto n \mapsto f(m \otimes n))(m \otimes n) \\ &= (m \otimes n \mapsto (m \mapsto n \mapsto f(m \otimes n)))(m \otimes n) \\ &= f(m \otimes n) \end{align*}

and

ϕψ(g)(m)(n)=ϕ(mng(m)(n))(m)(n)=(mn(mng(m)(n)))(m)(n)=g(m)(n). \begin{align*} \phi \circ \psi (g)(m)(n) &= \phi(m \otimes n \mapsto g(m)(n))(m)(n) \\ &= (m \mapsto n \mapsto (m \otimes n \mapsto g(m)(n)))(m)(n) \\ &= g(m)(n). \end{align*}

Therefore ϕ\phi and ψ\psi define the desired bijection.

AMBHomA(AMB,)_A M \otimes_B - \dashv \Hom_{A_\ell}(_A M_B, -)

We prove that CHomA(AMBNC,ALD)DCHomB(BNC,BHomA(AMB,ALD)D)D_C \Hom_{A_\ell}(_A M \otimes_B N_C, _A L_D)_D \simeq {_C \Hom_{B_\ell}(_B N_C, _B\Hom_{A_\ell}(_A M_B, _A L_D)_D)_D} as (C,D)(C,D)-bimodules.

Define ϕ:HomA(MBN,L)HomB(N,HomA(M,L))\phi: \Hom_{A_\ell}(M \otimes_B N, L) \to \Hom_{B_\ell}(N, \Hom_{A_\ell}(M, L)) to map

ϕ:f(n(mf(mn))).\phi: f \mapsto (n \mapsto (m \mapsto f(m \otimes n))).

This is an (C,D)(C,D)-bimodule homomorphism because

  • ϕ(f+g)=(nmf(mn)+g(mn))=ϕ(f)+ϕ(g)\phi(f + g) = (n \mapsto m \mapsto f(m \otimes n) + g(m \otimes n)) = \phi(f) + \phi(g)
  • and
ϕ(cfd)(n)(m)=(cfd)(mn)=f(mnc)d=ϕ(f)(nc)(m)d=(ϕ(f)(nc)d)(m)=(cϕ(f)d)(n)(m). \begin{align*} \phi(c \cdot f \cdot d)(n)(m) &= (c \cdot f \cdot d)(m \otimes n) \\ &= f(m \otimes n \cdot c) \cdot d \\ &= \phi(f)(n \cdot c)(m) \cdot d \\ &= (\phi(f)(n \cdot c) \cdot d)(m) \\ &= (c \cdot \phi(f) \cdot d)(n)(m). \end{align*}

Conversely, define ψ:HomB(N,HomA(M,L))HomA(MBN,L)\psi: \Hom_{B_\ell}(N, \Hom_{A_\ell}(M, L)) \to \Hom_{A_\ell}(M \otimes_B N, L) to map

ψ:g(mng(n)(m)).\psi: g \mapsto (m \otimes n \mapsto g(n)(m)).

This is an (C,D)(C,D)-bimodule homomorphism because

  • ψ(f+g)=(mnf(n)(m)+g(n)(m))=ψ(f)+ψ(g)\psi(f + g) = (m \otimes n \mapsto f(n)(m) + g(n)(m)) = \psi(f) + \psi(g)
  • and
ψ(cfd)(mn)=(cfd)(n)(m)=(f(nc)d)(m)=f(nc)(m)d=ψ(f)(mnc)d=(cψ(f)d)(mn). \begin{align*} \psi(c \cdot f \cdot d)(m \otimes n) &= (c \cdot f \cdot d)(n)(m) \\ &= (f(n \cdot c) \cdot d)(m) \\ &= f(n \cdot c)(m) \cdot d \\ &= \psi(f)(m \otimes n \cdot c) \cdot d \\ &= (c \cdot \psi(f) \cdot d)(m \otimes n). \end{align*}

These functions are clearly inverses (by construction). The calculation looks very similar to the previous adjunction.

Remark

All the above isomorphisms also hold for left modules and right modules separately.