Skip to main content Link Menu Expand (external link) Copy Copied

Tensor Products I: Extension of Scalars

Restriction of scalars

Suppose R and S are rings with unity (but not necessarily commutative), that we have ring homomorphism f:RS and that M is an S module. Then M is an R module by “restricting scalars” so that rm=f(r)m.

  • Any R-vector space is a Q vector space.
  • Any vector space over Z/pZ is a module over Z.
  • There is a ring map CM2(R) sending

    i(0110)

    and so any left module over M2(R) can be viewed as a complex vector space. There are other elements in M2(R) satisfying x2+1=0, and so there are lots of ways to view a left module over M2(R) as a C-vector space.

Extension of scalars

Suppose that f:RS is a map of rings with unity, and M is an R-module. Is there a way to make M into an S-module? Maybe a better way to say it is: can we find a “smallest” S-module N together with a map MN?

Example: Suppose that V is a finite dimensional real vector space. Choose a v1,,vn for V. So V is isomorphic to Rn using this basis. So we can think of V as inside Cn. If we chose a different basis, we’d get a different map VCn, but the two maps would be related by a change of basis transformation so in some sense these are “isomomorphic”.

Extension of scalars (More examples)

Example: Suppose that M is an abelian group (hence a Z-module). Can we embed M in a Q-module? Sometimes, yes: if M is Zn for some n, then M embeds in Qn. On the other hand if M is finite, there are no maps from MQn for any n. If M is a mixture of free and torsion parts, we can embed the free part of M in Qn but not the torsion part.

Example: If M is an abelian group, can we embed M into a vector space over Z/pZ – here the map RS is the map ZZ/pZ? Sometimes yes – if M is p-torsion, we can do it, but for general M no.

Universal approach

To study this in general we take a (left) R-module M and a ring map f:RS and ask: how can we make an S-module out of M and the map f:RS?

An S-module structure on M means we need a map

S×MM

satisfying the axioms

  • (s,(m1+m2))=(s,m1)+(s,m2)
  • (s1+s2,m)=(s1,m)+(s2,m)
  • (sr,m)=(s,rm) for rR.

Extension of scalars via tensor product

Our strategy is to make an abelian group whose elements are pairs (s,m) modulo the relations above. The equivalence classes of this abelian group are written sm (or, in cases where we need more context, sRm or even sfm). THe group itself is called SRM. So the following rules hold:

  • (s1+s2)m=s1m+s2m.
  • s(m1+m2)=sm1+sm2
  • srm=srm.

A typical element of SM is a sum of the form i=1nsimi. It is an S-module via multiplication by S on the first factor.

We have a map MSRM given by m1m.

Important: The elements sm belong to a quotient group and so the representation of an element as a sum of “simple tensors” sm need not be unique! In fact it’s quite possible for sm to be zero.

Some examples

Suppose that M=Zn and RS is ZQ. Then QM consists of sums of elements am. But a=xy where xZ, so we can write am=1yxm. QM is isomorphic to Qn.

Suppose that M is a finite group of order n. Then any element of QM can be written am where aQ. But a=n(a/n). Therefore

am=(a/n)nm=(a/n)nm=0

so QM is the zero module.

The universal property

The idea is that SM is the smallest S module containing M, where the action of R comes via the map RS. In other words, if L is any other S module and there is an R-module map f:ML, then there is a unique S-module map ϕ:SML so that this triangle commutes:

M11mfSMϕL

More on the universal property

If ι:MSM is the map m1m, then M/ker(ι) maps injectively in SM. This is the “largest” quotient of M which embeds into an S-module.

Example: Let G be a finitely generated abelian group. Then G is isomoprhic to ZnT where T is a finite group. If we want to map G to a Q-vector space, the kernel has to include T. And in fact the kernel of ι is T. Further, the Q-dimension of the vector space QG is the rank of the free part of G.

More examples

Example: Let M be an R module and let f:RR/I be the quotient map. Then R/IM is isomorphic to M/IM. First notice that if xIM, then 1x=1im=im=0, so IM is in the kernel of ι. Therefore we have a map M/IMR/IM. We have a map in the opposite direction R/IMM/IM given by (r+I)mrm+IM. So if G is a finite abelian group, then Z/pZG is G/pG which is zero if G has no p-torsion.

Example: If V is a vector space over F of dimension n, and FE is a field extension, then EV is an n-dimensional vector space over E.

Tensor products of modules

The commutative case

Assume for the moment that R is a commutative ring with unity. Suppose that M and N are R-modules. If L is yet another R-module, a bilinear map

f:M×NL

is a map that is linear in each variable separately and also satisfies f(rm,n)=f(m,rn) for rR. The tensor product MRN of M and N is the free abelian group on pairs (m,n) modulo the relations:

  • (m1+m2,n)=(m1,n)+(m2,n)
  • (m,n1+n2)=(m,n1)+(m,n2)
  • (rm,n)=(m,rn)

The equivalence class of the pair (m,n) is written mn. MN is an R module via the action r(mn)=(rmn)=(mrn) and

r(mini)=r(mini).

Universal property

If f:M×NL is bilinear, then we can defined f:MNL by f(mn)=f(m,n). This is well defined and it converts a bilinear map into a module homomorphism. To go in the other direction, there is a map B:M×NMN which is bilinear, sending (m,n)mn. This is the ``universal” bilinear map. The universal property says that, if f:M×NL is any bilinear map, there is a unique module homomorphism f:MNL such that fB=f.

M×NBfMNfL

Tensor product of vector spaces

Suppose that R is a field and V, W, are vector spaces over R of dimensions n and m respectively. Let v1,,vn be a basis for V and w1,,wm a basis for W.

If L is another F-vector space, then a bilinear map f:V×WL is determined by its values on all pairs (vi,wj).

The tensor product VW is an F-vector space and is spanned by the tensors viwj.

Now construct a bilinear map fij:V×WF by setting

fij(asvs,bsws)=aibj.

By the universal property we have fij(vrws)=0 unless r=i and s=j in which case it is one.

Tensor product of vector spaces continued

Suppose that

x=crsvrws=0.

Then fij(x)=cij=0 for all pairs i,j and therefore all crs=0; in other words, the vrws are linearly independent. Thus VW is an nm dimensional F-vector space.

Endomorphisms

Let V be a vector space and let V be its dual. Then there is an isomorphism

VVEnd(V)

where (vf)(w)=f(w)v.

The noncommutative case

Now suppose that R is a noncommutative ring. If M and N are left modules, then we have a problem defining a bilinear map M×NL where L is also a left module. Namely, on the one hand, we would need:

f(rsm,n)=rf(sm,n)=f(sm,rn)=sf(m,rn)=f(m,srn)

but on the other hand

f((rs)m,n)=(rs)f(m,n)=f(m,(rs)n)

and since sr and rs are different we can’t define this consistently.

The noncommutative case continued

In the non-commutative case (with unity) we have to make some compromises:

  • First, we assume M is a right module and N is a left module.
  • Next, we are only going to construct an abelian group, not a module, from M and N.
  • Finally, we are going to consider maps f:M×NL, where L is an abelian group, that are balanced, meaning that f(mr,n)=f(m,rn) for rR.

We create an abelian group spanned by mn subject to the relations mrn=mrn together with the additivity (m+m)n=mn+mn and similarly for N.

This is the tensor product of the modules M and N – remember it is an abelian group, NOT an R-module in general.

Universal property

MN still satisfies a universal property. Call a map f:M×NL, where M is a right R-module, N is a left R-module, and L is an abelian group, balanced if f is additive in M and N separately and satisfies f(mr,n)=f(m,rn) for all rR.

Then given such a balanced map from M×N to L, there is a unique map of abelian groups ϕ:MNL such that the following diagram commutes.

M×NιfMNϕL

Here the map M×NMN is the expected one: (m,n)mn.

As is always the case, the universal property characterizes the tensor product up to isomorphism.

Bimodules

Now suppose that S and R are rings with unity and that M is simultaneously a left S-module and a right R module, so that (sm)r=s(mr). Such an object M is called an (S,R)-bimodule.

For example, suppose R=M2(F), S=F, and M is the space F2 viewed as row vectors with R acting on the right as matrix multiplication and S on the left as scalar multiplication.

Tensor product of bimodules

If N is a left R module, we can form the tensor product MRN which is an abelian group; but we can furthermore let S act by s(mn)=(smn).

This makes MRN into a left S-module. (If N is an (R,S)-bimodule so that R acts on the left and S on the right, then MRN is a right S-module.)

If R is commutative, and M is a left R module, it is also a right R-module via (mr)=rm. So it is automatically an (R,R)-bimodule. This is how MRN is automatically an R-module if R is commutative.

General Properties

Tensor product of maps

If f:MM and g:NN are maps of right/left R-modules, then fg:MNMN defined by (fg)(mn)=f(m)g(n) is a well defined group homomorphism.

If M and M are (S,R) bimodules and f and g are S-module homomorphismsm then fg is an S-module homomorphism. (If R is commutative all this is automatic).

Further, provided everything makes sense, (fg)(fg)=(ff)(gg).

The Kronecker Product

If L and M are matrices giving linear maps from Rn to Rn and Rm to Rm.

Then the tensor product of these maps is a linear map from RnRm to RnRm.

The standard bases of Rk give us bases eiej of RnRm and RnRm. Thus the tensor product is represented as an nm×nm matrix. This is called the Kronecker Product of the matrices L and M.

Associativity

The tensor product is associative in the sense that (MRN)TL is isomorphic to MR(NTL) provided that M is a right R module, N is an (R,T) bimodule, and L is a left T-module. If R and T are commutative this is automatic.

First check that both versions of the tensor product make sense.

Notice that NTL is a left R-module and MRN is a right T-module.

For fixed lL, the map (m,n)m(nl) is R-balanced, so there is a well-defined map

MRNMR(NTL)

This gives a well-defined map

MRN×LMR(NTL)

and by the universal property this translates to a map

(MRN)TLMR(NTL).

You can also reverse this construction to create the inverse map.

If M is an (S,R) bimodule then both constructions yield left S-modules. Then MRN is an (S,T) bimodule and (MRN)TL is a left S-module.

Commutativity

If R is commutative, the tensor product is commutative in the sense that MN is isomorphic to NM.

Distributive law

(MM)N is isomorphic to (MN)(MN) and similarly if N is a direct sum. By induction this extends to finite direct sums. With care it holds for infinite direct sums.

The proof of this uses the fact that there is a well-defined balanced map

F:(MM)×N(MN)(MN)

defined by F((m,m),n)=(mn,mn)

and so we have a map

(MM)N(MN)(MN).

On the other hand we have balanced maps M×N(MM)N sending mn to (m,0)n and similarly for M×N. These give a map from MNMN(MM)N which is inverse to the map above.

Tensor product of algebras

If A and B are R algebras where R is commutative, then ARB is an R algebra with multiplication (ab)(ab)=(aa)(bb). (remember that an R algebra is a ring in which R is mapped into the center of the ring).

View as slides