# Restriction of scalars

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

• Any $$\R$$-vector space is a $$\Q$$ vector space.
• Any vector space over $$\Z/p\Z$$ is a module over $$\Z$$.
• There is a ring map $$\C\to M_{2}(R)$$ sending

$i\to\left(\begin{matrix} 0 & 1 \\ -1 & 0\end{matrix}\right)$

and so any left module over $$M_{2}(\R)$$ can be viewed as a complex vector space. There are other elements in $$M_{2}(\R)$$ satisfying $$x^2+1=0$$, and so there are lots of ways to view a left module over $$M_{2}(\R)$$ as a $$\C$$-vector space.

# Extension of scalars

Suppose that $$f:R\to S$$ 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 $$M\to N$$?

Example: Suppose that $$V$$ is a finite dimensional real vector space. Choose a $$v_1,\ldots, v_n$$ for $$V$$. So $$V$$ is isomorphic to $$\R^{n}$$ using this basis. So we can think of $$V$$ as inside $$\C^{n}$$. If we chose a different basis, we’d get a different map $$V\to \C^{n}$$, 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 $$\Z^{n}$$ for some $$n$$, then $$M$$ embeds in $$\Q^{n}$$. On the other hand if $$M$$ is finite, there are no maps from $$M\to \Q^{n}$$ for any $$n$$. If $$M$$ is a mixture of free and torsion parts, we can embed the free part of $$M$$ in $$\Q^{n}$$ but not the torsion part.

Example: If $$M$$ is an abelian group, can we embed $$M$$ into a vector space over $$\Z/p\Z$$ – here the map $$R\to S$$ is the map $$\Z\to \Z/p\Z$$? 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 $$f:R\to S$$ and ask: how can we make an $$S$$-module out of $$M$$ and the map $$f:R\to S$$?

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

$S\times M\to M$

satisfying the axioms

• $$(s,(m_1+m_2))=(s,m_1)+(s,m_2)$$
• $$(s_1+s_2,m)=(s_1,m)+(s_2,m)$$
• $$(sr,m)=(s,rm)$$ for $$r\in R$$.

# 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 $$s\otimes m$$ (or, in cases where we need more context, $$s\otimes_{R} m$$ or even $$s\otimes_{f}m$$). THe group itself is called $$S\otimes_{R} M$$. So the following rules hold:

• $$(s_1+s_2)\otimes m=s_1\otimes m+s_2\otimes m$$.
• $$s\otimes (m_1+m_2)=s\otimes m_1 + s\otimes m_2$$
• $$sr\otimes m=s\otimes rm$$.

A typical element of $$S\otimes M$$ is a sum of the form $$\sum_{i=1}^{n} s_{i}\otimes m_{i}$$. It is an $$S$$-module via multiplication by $$S$$ on the first factor.

We have a map $$M\to S\otimes_{R} M$$ given by $$m\mapsto 1\otimes m$$.

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

# Some examples

Suppose that $$M=\Z^{n}$$ and $$R\to S$$ is $$\Z\to \Q$$. Then $$\Q\otimes M$$ consists of sums of elements $$a\otimes m$$. But $$a=\frac{x}{y}$$ where $$x\in \Z$$, so we can write $$a\otimes m=\frac{1}{y}\otimes xm$$. $$\Q\otimes M$$ is isomorphic to $$\Q^{n}$$.

Suppose that $$M$$ is a finite group of order $$n$$. Then any element of $$\Q\otimes M$$ can be written $$a\otimes m$$ where $$a\in \Q$$. But $$a=n(a/n)$$. Therefore

$a\otimes m = (a/n)n\otimes m = (a/n)\otimes nm=0$

so $$\Q\otimes M$$ is the zero module.

# The universal property

The idea is that $$S\otimes M$$ is the smallest $$S$$ module containing $$M$$, where the action of $$R$$ comes via the map $$R\to S$$. In other words, if $$L$$ is any other $$S$$ module and there is an $$R$$-module map $$f:M\to L$$, then there is a unique $$S$$-module map $$\phi:S\otimes M\to L$$ so that this triangle commutes:

$\begin{xy} \xymatrix@=2cm{ M \ar[r]^{1\mapsto 1\otimes m}\ar[rd]^{f}& S\otimes M\ar[d]^{\phi} \\ & L \\ } \end{xy}$

# More on the universal property

If $$\iota: M\to S\otimes M$$ is the map $$m\mapsto 1\otimes m$$, then $$M/\ker(\iota)$$ maps injectively in $$S\otimes M$$. 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 $$\Z^{n}\oplus T$$ 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 $$\iota$$ is $$T$$. Further, the $$\Q$$-dimension of the vector space $$\Q\otimes G$$ is the rank of the free part of $$G$$.

# More examples

Example: Let $$M$$ be an $$R$$ module and let $$f:R\to R/I$$ be the quotient map. Then $$R/I\otimes M$$ is isomorphic to $$M/IM$$. First notice that if $$x\in IM$$, then $$1\otimes x=1\otimes im=i\otimes m=0$$, so $$IM$$ is in the kernel of $$\iota$$. Therefore we have a map $$M/IM\to R/I\otimes M$$. We have a map in the opposite direction $$R/I\otimes M\to M/IM$$ given by $$(r+I)\otimes m\mapsto rm+IM$$. So if $$G$$ is a finite abelian group, then $$\Z/p\Z\otimes G$$ 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 $$F\to E$$ is a field extension, then $$E\otimes V$$ is an $$n$$-dimensional vector space over $$E$$.

# 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

$f: M\times N \to L$

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

• $$(m_1+m_2,n)=(m_1,n)+(m_2,n)$$
• $$(m,n_1+n_2)=(m,n_1)+(m,n_2)$$
• $$(rm, n)=(m,rn)$$

The equivalence class of the pair $$(m,n)$$ is written $$m\otimes n$$. $$M\otimes N$$ is an $$R$$ module via the action $$r(m\otimes n)=(rm\otimes n)=(m\otimes rn)$$ and

$r(\sum m_{i}\otimes n_{i})=\sum r(m_{i}\otimes n_{i}).$

# Universal property

If $$f:M\times N\to L$$ is bilinear, then we can defined $$\overline{f}:M\otimes N\to L$$ by $$\overline{f}(m\otimes n)=f(m,n)$$. This is well defined and it converts a bilinear map into a module homomorphism. To go in the other direction, there is a map $$B:M\times N\to M\otimes N$$ which is bilinear, sending $$(m,n)\to m\otimes n$$. This is the universal" bilinear map. The universal property says that, if $$f:M\times N\to L$$ is any bilinear map, there is a unique module homomorphism $$\overline{f}:M\otimes N\to L$$ such that $$\overline{f}B=f.$$

$\begin{xy} \xymatrix{ M\times N \ar[r]^{B}\ar[dr]^{f} & M\otimes N\ar[d]^{\overline{f}}\\ & L\\ } \end{xy}$

# 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 $$v_1,\ldots, v_n$$ be a basis for $$V$$ and $$w_1,\ldots, w_m$$ a basis for $$W$$.

If $$L$$ is another $$F$$-vector space, then a bilinear map $$f:V\times W\to L$$ is determined by its values on all pairs $$(v_i,w_j)$$.

The tensor product $$V\otimes W$$ is an $$F$$-vector space and is spanned by the tensors $$v_{i}\otimes w_{j}$$.

Now construct a bilinear map $$f_{ij}:V\times W\to F$$ by setting

$f_{ij}(\sum a_{s}v_{s},\sum b_{s}w_{s})=a_{i}b_{j}.$

By the universal property we have $$f_{ij}(v_{r}\otimes w_{s})=0$$ unless $$r=i$$ and $$s=j$$ in which case it is one.

# Tensor product of vector spaces continued

Suppose that

$x=\sum c_{rs}v_{r}\otimes w_{s}=0.$

Then $$f_{ij}(x)=c_{ij}=0$$ for all pairs $$i,j$$ and therefore all $$c_{rs}=0$$; in other words, the $$v_{r}\otimes w_{s}$$ are linearly independent. Thus $$V\otimes W$$ is an $$nm$$ dimensional $$F$$-vector space.

# Endomorphisms

Let $$V$$ be a vector space and let $$V^{\ast}$$ be its dual. Then there is an isomorphism

$V\otimes V^{\ast}\to \End(V)$

where $$(v\otimes f)(w)=f(w)v$$.

# 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 $$M\times N\to L$$ 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 $$f:M\times N\to L$$, where $$L$$ is an abelian group, that are balanced, meaning that $$f(mr,n)=f(m,rn)$$ for $$r\in R$$.

We create an abelian group spanned by $$m\otimes n$$ subject to the relations $$mr\otimes n=m\otimes rn$$ together with the additivity $$(m+m')\otimes n=m\otimes n + m'\otimes n$$ 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

$$M\otimes N$$ still satisfies a universal property. Call a map $$f:M\times N\to L$$, 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 $$r\in R$$.

Then given such a balanced map from $$M\times N$$ to $$L$$, there is a unique map of abelian groups $$\phi : M\otimes N\to L$$ such that the following diagram commutes.

$\begin{xy} \xymatrix{ M\times N \ar[r]^{\iota}\ar[dr]^{f} & M\otimes N\ar[d]^{\phi}\\ & L\\ } \end{xy}$

Here the map $$M\times N\to M\otimes N$$ is the expected one: $$(m,n)\mapsto m\otimes n$$.

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 $$R=M_{2}(F)$$, $$S=F$$, and $$M$$ is the space $$F^{2}$$ 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 $$M\otimes_{R}N$$ which is an abelian group; but we can furthermore let $$S$$ act by $$s(m\otimes n)=(sm\otimes n)$$.

This makes $$M\otimes_{R}N$$ 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 $$M\otimes_{R}N$$ 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 $$M\otimes_{R}N$$ is automatically an $$R$$-module if $$R$$ is commutative.

# Tensor product of maps

If $$f:M\to M'$$ and $$g:N\to N'$$ are maps of right/left $$R$$-modules, then $$f\otimes g:M\otimes N\to M'\otimes N'$$ defined by $$(f\otimes g)(m\otimes n)=f(m)\otimes g(n)$$ is a well defined group homomorphism.

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

Further, provided everything makes sense, $$(f\otimes g)\circ (f'\otimes g')=(f\circ f')\otimes (g\circ g').$$

# The Kronecker Product

If $$L$$ and $$M$$ are matrices giving linear maps from $$\R^{n}$$ to $$\R^{n'}$$ and $$\R^{m}$$ to $$\R^{m'}$$.

Then the tensor product of these maps is a linear map from $$\R^{n}\otimes \R^{m}$$ to $$\R^{n'}\otimes \R^{m'}$$.

The standard bases of $$\R^{k}$$ give us bases $$e_{i}\otimes e_{j}$$ of $$\R^{n}\otimes R^{m}$$ and $$\R^{n'}\otimes\R^{m'}$$. Thus the tensor product is represented as an $$nm\times n'm'$$ matrix. This is called the Kronecker Product of the matrices $$L$$ and $$M$$.

# Associativity

The tensor product is associative in the sense that $$(M\otimes_{R} N)\otimes_{T} L$$ is isomorphic to $$M\otimes_{R} (N\otimes_{T} L)$$ 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 $$N\otimes_{T}L$$ is a left $$R$$-module and $$M\otimes_{R}N$$ is a right $$T$$-module.

For fixed $$l\in L$$, the map $$(m,n)\mapsto m\otimes (n\otimes l)$$ is $$R$$-balanced, so there is a well-defined map

$M\otimes_{R}N\to M\otimes_{R}(N\otimes_{T}L)$

This gives a well-defined map

$M\otimes_{R}N\times L \to M\otimes_{R}(N\otimes_{T}L)$

and by the universal property this translates to a map

$(M\otimes_{R}N)\otimes_{T}L\to M\otimes_{R}(N\otimes_{T}L).$

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 $$M\otimes_{R}N$$ is an $$(S,T)$$ bimodule and $$(M\otimes_{R}N)\otimes_{T}L$$ is a left $$S$$-module.

# Commutativity

If $$R$$ is commutative, the tensor product is commutative in the sense that $$M\otimes N$$ is isomorphic to $$N\otimes M$$.

# Distributive law

$$(M\oplus M')\otimes N$$ is isomorphic to $$(M\otimes N)\oplus (M'\otimes N)$$ 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

$F: (M\oplus M')\times N \to (M\otimes N)\oplus (M'\otimes N)$

defined by $$F((m,m'),n)=(m\otimes n,m'\otimes n)$$

and so we have a map

$(M\oplus M')\otimes N\to (M\otimes N)\oplus (M'\otimes N).$

On the other hand we have balanced maps $$M\times N\to (M\oplus M')\otimes N$$ sending $$m\otimes n$$ to $$(m,0)\otimes n$$ and similarly for $$M'\times N$$. These give a map from $$M\otimes N\oplus M'\otimes N\to (M\oplus M')\otimes N$$ which is inverse to the map above.

# Tensor product of algebras

If $$A$$ and $$B$$ are $$R$$ algebras where $$R$$ is commutative, then $$A\otimes_{R}B$$ is an $$R$$ algebra with multiplication $$(a\otimes b)(a'\otimes b')=(aa')\otimes (bb')$$. (remember that an $$R$$ algebra is a ring in which $$R$$ is mapped into the center of the ring).