Homework 5

Problem 1

Show that $C \otimes_{R} C$ and $C \otimes_{C} C$ are not isomorphic as $R$ -modules.

Note: Our notation is intentionally a bit sloppy here, since this is how you'll often see statements like this written. For clarity, we note that:

In the first tensor product, the copy of $C$ on the right is the $R$ -module (i.e., $(R, Z)$ -bimodule) of complex numbers (i.e., the additive group of complex numbers along with scaling by real numbers), while the $C$ on the left is $(C, R)$ -bimodule of complex numbers. The resulting tensor product is therefore a $(C, Z)$ -bimodule, which by "restriction of scalars" we can also view as an $(R, Z)$ -bimodule, i.e., an $R$ -module.
In the second tensor product, the copy of $C$ on the right is the $C$ -module (i.e., $(C, Z)$ -bimodule) of complex numbers (i.e., the ring of complex numbers), while the $C$ on the left is the ring of complex numbers with the standard $(C, C)$ -bimodule structure. The resulting tensor product is therefore a $(C, Z)$ -bimodule, which by "restriction of scalars" we can also view as an $(R, Z)$ -bimodule, i.e., an $R$ -module.
Your argument will prove that the two constructions are also not isomorphic as $C$ -modules.

Hints

You often hear the complex numbers described as "two-dimensional as a real vector space." For us, this means that the $R$ -module $C$ is isomorphic to the free $R$ -module on a two-element set, the latter of which is isomorphic as an $R$ -module to $R \oplus R ≃ R^{2}$ .

With that in mind, show $C \otimes_{R} C$ is isomorphic as an $R$ -module to $R^{4}$ . What can you say about $C \otimes_{C} C$ as a $C$ -module, and as an $R$ -module? (Remember that $R$ -modules are just real vector spaces!)

Solution

We have seen that $C \otimes_{C} C ≃ C$ as $C$ -modules, and hence also as $R$ -modules. We also know that $C ≃ R^{2} ≃ R \oplus R$ as $R$ -modules (from linear algebra, say).

On the other hand, since tensor product commutes with direct sums, we have $R$ -module isomorphisms

C \otimes_{R} C ≃ C \otimes_{R} (R \oplus R) ≃ (C \otimes_{R} R) \oplus (C \otimes_{R} R) ≃ C \oplus C ≃ (R \oplus R) \oplus (R \oplus R) ≃ R^{4} .

In the language of linear algebra, $C \otimes_{R} C$ is isomorphic as a real vector space to $R^{4}$ , while $C \otimes_{C} C$ is isomorphic as a real vector space to $R^{2}$ . Since isomorphisms of vector spaces preserve dimension, it follows that $C \otimes_{R} C$ and $C \otimes_{C} C$ cannot be isomorphic as $R$ -modules.

Problem 2

Suppose $D$ is an integral domain with quotient field^[1] $Q$ and $M$ is a left $D$ -module. Prove that every element of $Q \otimes_{D} M$ can be written as a simple tensor of the form $\frac{1}{d} \otimes m$ for some nonzero $d \in D$ and $m \in M$ .

Hints

Try first showing the result for simple tensors $q \otimes m$ , and then extend the result to handle general tensors $\sum q_{i} \otimes m_{i}$ .

Solution

First note that the zero elements in $Q$ and $D$ coincide, and $0_{Q} \otimes m = (1_{Q} \cdot 0_{D}) \otimes m = 1_{Q} \otimes (0_{D} \cdot m) = 1_{Q} \otimes 0_{m}$ , so the result holds for simple tensors of the form $0_{Q} \otimes m$ .

Now suppose $q \otimes m$ is a simple tensor with $q \in Q$ nonzero and $m \in M$ . We can write $q = \frac{a}{d}$ for some $a, b \in D$ with $b$ nonzero, and we then have

q \otimes m = \frac{a}{b} \otimes m = (\frac{1}{b} \cdot a) \otimes m = \frac{1}{b} \otimes a m,

so the result holds for $q \otimes m$ .

Finally, suppose we take a general element of $Q \otimes_{D} M$ , represented as a finite sum of the form $\sum_{i = 1}^{k} q_{i} \otimes m_{i}$ . Using the above argument, we can write each simple tensor in this sum in the form $q_{i} \otimes m_{i} = \frac{1}{b_{i}} \otimes m_{i}^{'}$ for some $b_{i} \in D$ nonzero and $m_{i}^{'} \in M$ . Let $d$ be a least common multiple of the $b_{i}$ in $D$ , so that for each $i$ we can write $\frac{1}{b_{i}} = \frac{c_{i}}{d}$ for some $c_{i} \in D$ . We can then rewrite each simple tensor as

\frac{1}{b_{i}} \otimes m_{i}^{'} = \frac{c_{i}}{d} \otimes m_{i}^{'} = \frac{1}{d} \otimes c_{i} m_{i}^{'} .

Using linearity, we then have

\sum_{i = 1}^{n} (q_{i} \otimes m_{i}) = \sum_{i = 1}^{n} (\frac{1}{b_{i}} \otimes m_{i}^{'}) = \sum_{i = 1}^{n} (\frac{1}{d} \otimes c_{i} m_{i}^{'}) = \frac{1}{d} \otimes \sum_{i = 1}^{n} c_{i} m_{i}^{'} = \frac{1}{d} \otimes m,

as desired.

Problem 3

Let ${e_{1}, e_{2}}$ be a basis for $R^{2}$ as an $R$ -vector space. Show that the element $e_{1} \otimes e_{2} + e_{2} \otimes e_{1}$ in $R^{2} \otimes_{R} R^{2}$ cannot be written as a simple tensor $v \otimes w$ for any $v, w \in R^{2}$ .

Hints

A basis for $R^{2} \otimes_{R} R^{2}$ as an $R$ -module (i.e., $R$ -vector space) is ${e_{1} \otimes e_{1}, e_{1} \otimes e_{2}, e_{2} \otimes e_{1}, e_{2} \otimes e_{2}}$ . Try expressing each simple tensor $v \otimes m$ in $R^{2} \otimes_{R} R^{2}$ in terms of this basis. Use $R$ -linear independence of your basis to eventually arrive at a contradiction.

Solution

We have seen that a basis for $R^{2} \otimes R^{2}$ as an $R$ -module is ${e_{1} \otimes e_{1}, e_{1} \otimes e_{2}, e_{2} \otimes e_{1}, e_{2} \otimes e_{2}}$ . In particular, these four tensors are linearly independent over $R$ . Now suppose, towards a contradiction, that $e_{1} \otimes e_{2} + e_{2} \otimes e_{1} = v \otimes w$ for some $v, w \in R^{2}$ . Let's expand the right-hand side in terms of our basis. We have $v = a_{1} e_{1} + a_{2} e_{2}$ and $w = b_{1} e_{1} + b_{2} e_{2}$ for some unique real numbers $a_{1}, a_{2}, b_{1}, b_{2} \in R$ . We then have

\begin{aligned} v \otimes w & = (a_{1} e_{1} + a_{2} e_{2}) \otimes (b_{1} e_{1} + b_{2} e_{2}) \\ = a_{1} b_{1} (e_{1} \otimes e_{2}) + a_{1} b_{2} (e_{1} \otimes e_{2}) + a_{2} b_{1} (e_{2} \otimes e_{1}) + a_{2} b_{2} (e_{2} \otimes e_{2}) . \end{aligned}

Since the four tensors are a basis for our vector space, the equality $e_{1} \otimes e_{2} + e_{2} \otimes e_{1} = v \otimes w$ forces the following four equalities:

$a_{1} b_{1} = 0$
$a_{2} b_{1} = 1$
$a_{1} b_{2} = 1$
$a_{2} b_{2} = 0$

Equations (2) and (3) imply that all four numbers $a_{1}, a_{2}, b_{1}, b_{2}$ are nonzero (as $R$ is a domain), but then equations (1) and (4) are impossible.

Thus, the tensor $e_{1} \otimes e_{2} + e_{2} \otimes e_{1}$ cannot be written as a simple tensor.

Problem 4

Give an example to show that tensor product does not commute with direct products.

Hints

Consider the extension of scalars from $Z$ to $Q$ of the direct product of the $Z$ -modules $M_{i} = Z_{2^{i}}$ for $i = 1, 2, \dots$ . Show that $Q \otimes_{Z} M_{i} = 0$ for each $i$ , but that $Q \otimes_{Z} \prod_{i = 1}^{\infty} M_{i}$ is nonzero. For this last part, use the helpful property that relates the torsion of a $D$ -module $M$ to the $Q$ -module $Q \otimes_{D} M$ , whenever $Q$ is the field of fractions of an integral domain $D$ .

Solution

Consider the extension of scalars from $Z$ to $Q$ of the direct product of the $Z$ -modules $M_{i} = Z_{2}$ for $i = 1, 2, \dots$ .

On the one hand, since each $M_{i}$ is a finite abelian group, we know that $Q \otimes_{Z} M_{i} = 0$ for every $i$ . It follows that $\prod_{i = 1}^{\infty} Q \otimes_{Z} M_{i} = 0$ .

On the other hand, we claim that $Q \otimes_{Z} \prod_{i = 1}^{\infty} M_{i}$ is nonzero. To see this, first note that $Q$ is the field of fractions of the integral domain $Z$ , so by our general properties for any $Z$ -module $M$ the $Q$ -module $Q \otimes_{Z} M$ is zero exactly when $M$ is torsion (as a $Z$ -module). However, in our case the $Z$ -module $M = \prod_{i = 1}^{\infty} M_{i}$ has many elements that are not torsion; e.g., the element $m = (1, 1, 1, \dots)$ is not torsion. (For any integer $k \in Z$ , simply note that for all $i$ with $2^{i} > | k |$ the element $k \cdot m = (k, k, k, \dots)$ is nonzero in the $i$ th coordinate.) We thus have that $Q \otimes_{Z} M$ is nonzero, and hence is not isomorphic to the zero module $\prod_{i = 1}^{\infty} Q \otimes_{Z} M_{i}$ .

Problem 5

Suppose $R$ and $S$ are commutative rings (with unity). We can form their tensor product $R \otimes S$ in the category of commutative rings as follows. First, as abelian groups (i.e., $(Z, Z)$ -bimodule) we can form the tensor product $R \otimes_{Z} S$ , which we simply denote $R \otimes S$ . We can then define a multiplication in $R \otimes S$ is "component-wise", i.e., $(r_{1} \otimes s_{1}) \cdot (r_{2} \otimes s_{2}) = (r_{1} r_{2}) \otimes (s_{1} s_{2})$ . This operation gives $R \otimes S$ the structure of a commutative ring (with unity $1_{R} \otimes 1_{S}$ ).

Define $i_{1} : R \to R \otimes S$ by $r \mapsto r \otimes 1_{S}$ , and $i_{2} : S \to R \otimes S$ by $s \mapsto 1_{R} \otimes s$ .

Verify $i_{1}$ and $i_{2}$ are ring morphisms.
Show that the ring $R \otimes S$ together with these ring morphisms is a coproduct of $R$ and $S$ in the category of commutative rings.

Solution

First observe that $$i_1(r_1+r_2)=(r_1+r_2)\otimes 1_S = r_1\otimes 1_S+r_2\otimes 1_S = i_1(r)+i_2(r),$$
and also

i_{2} (r_{1} r_{2}) = r_{1} r_{2} \otimes 1_{S} = r_{1} r_{2} \otimes 1_{S} \cdot 1_{S} = (r_{1} \otimes 1_{S}) \cdot (r_{2} \otimes 1_{S}) = i_{1} (r_{1}) \cdot i_{1} (r_{2}) .

This prove $i_{1}$ is a ring morphism. The same argument, mutatis mutandis, proves that $i_{2}$ is a ring morphism.

We need to prove that for every pair of ring morphisms $f : R \to T$ , $g : S \to T$ there is a unique ring morphism $h : R \otimes S \to T$ through which $f$ and $g$ factor. So, suppose $f : R \to T$ , $g : S \to T$ is a pair of ring morphisms. Define a set map $ϕ : R \times S \to T$ by $ϕ (r, s) = f (r) g (s)$ . Observe that this is a bilinear map:
$\begin{aligned} ϕ ((r_{1}, s) + (r_{2}, s)) & = ϕ (r_{1} + r_{2}, s) = f (r_{1} + r_{2}) g (s) \\ = (f (r_{1}) + f (r_{2})) g (s) \\ = f (r_{1}) g (s) + f (r_{2}) g (s) \\ = ϕ (r_{1}, s) + ϕ (r_{2}, s) \end{aligned}$
and similarly $$\phi((r,s_1)+(r,s_2))=\phi(r,s_1)+\phi(r,s_2).$$

By the universal property of the tensor product, we then have a corresponding ring morphism $h : R \otimes S \to T$ defined on simple tensors by $h (r \otimes s) = f (r) g (s)$ . Then note that
$(h \circ i_{1}) (r) = h (r \otimes 1_{S}) = f (r) g (1_{S}) = f (r) 1_{T} = f (r),$
so we indeed have $f = h \circ i_{1}$ . We similarly have $g = h \circ i_{2}$ , since
$(h \circ i_{2}) (s) = h (1_{R} \otimes s) = f (1_{R}) g (s) = 1_{T} g (s) = g (s) .$
So we've shown $f$ and $g$ factor through $h$ .

As for uniqueness, suppose $h^{'} : R \otimes S \to T$ is another ring morphism such that $f$ and $g$ factor through $h^{'}$ . Then observe that
$\begin{aligned} h^{'} (r \otimes s) & = h^{'} ((r \otimes 1_{S}) \cdot (1_{R} \otimes s)) \\ = h^{'} (r \otimes 1_{S}) \cdot h^{'} (1_{R} \otimes s) \\ = h^{'} (i_{1} (r)) \cdot h^{'} (i_{2} (s)) \\ = f (r) g (s) \\ = h (r \otimes s) . \end{aligned}$
This proves uniqueness of $h$ , which establishes that $R \otimes S$ (together with the ring morphisms $i_{1}, i_{2}$ ) satisfy the universal property of the coproduct of $R$ and $S$ (in the category of commutative rings).

Problem 6

Show that for each $(R, S)$ -bimodule $M$ and $(R^{'}, S)$ -bimodule $N$ , the set ${Hom}_{S} (M, N)$ of right $S$ -module morphisms between $M$ and $N$ (viewed as right $S$ -modules) has the structure of a $(R^{'}, R)$ -bimodule.

Notes

You take can as given that the set ${Hom}_{S} (M, N)$ is an abelian group, with operation defined by $(f + g) (n) = f (n) + g (n)$ .

For the bimodule structure, rather than getting buried in all of the tiny details, focus on describing:

The left action of $R^{'}$ on ${Hom}_{S} (M, N)$ :
- What is the definition of $r^{'} \cdot f$ ?
- Does this appear to actually define a left action of $R^{'}$ on ${Hom}_{S} (M, N)$ ? (Verify at least one of the required properties is true; e.g., show $r^{'} \cdot (f + g) = r^{'} \cdot f + r^{'} \cdot g$ .)
The right action of $R$ on ${Hom}_{S} (M, N)$
- What is the definition of $f \cdot r$ ?
- Does this also appear to actually define a right action of $R$ on ${Hom}_{S} (M, N)$ ? (Again, verify at least one of the required properties is true.)
Do these two actions verify the one required compatibility to give ${Hom}_{S} (M, N)$ the structure of an $(R^{'}, R)$ -bimodule?

Solution

For those who might be curious, here we'll present a full solution covering (nearly) every detail one should officially check for this problem.

The abelian group structure

We first show ${Hom}_{S} (M, N)$ is an abelian group. The addition in ${Hom}_{S} (M, N)$ is defined by addition of outputs; i.e., for every pair of right $S$ -module morphisms $f, g : M \to N$ , we define $f + g : M \to N$ by $(f + g) (m) = f (m) + g (m)$ . Of course, we need to verify this operation is well defined; i.e., that $f + g$ is still a right $S$ -module morphism. It is additive because $f$ and $g$ are additive.

\begin{aligned} (f + g) (m_{1} + m_{2}) & = f (m_{1} + m_{2}) + g (m_{1} + m_{2}) \\ = f (m_{1}) + f (m_{2}) + g (m_{1}) + g (m_{2}) \\ = (f (m_{1}) + g (m_{1})) + (f (m_{2}) + g (m_{2})) \\ = (f + g) (m_{1}) + (f + g) (m_{2}) . \end{aligned}

It is compatible with the right $S$ -action because $f$ and $g$ are compatible with the right $S$ -action:

\begin{aligned} (f + g) (m) \cdot s & = (f (m) + g (m)) \cdot s \\ = f (m) \cdot s + g (m) \cdot s \\ = f (m s) + g (m s) \\ = (f + g) (m s) . \end{aligned}

So we have a well-defined binary operation on ${Hom}_{S} (M, N)$ . It is evidently commutative and associative, since addition in $N$ is commutative and associative. The zero map from $M$ to $N$ is a right $S$ -module morphism, and it plays the role of the additive identity under this operation. Additive inverses are given via additive inverses of outputs; i.e., $(- f) (n) = - f (n)$ . It is straightforward to verify $- f$ is indeed another right $S$ -module morphism; let's skip checking that, at least.

The left $R^{'}$ -module structure

We've thus established that ${Hom}_{S} (M, N)$ has the structure of an abelian group. We next consider the left and right actions.

We define the left $R^{'}$ -action using the left $R^{'}$ -action on $N$ , namely by defining $r^{'} \cdot f$ as the function given by $(r^{'} \cdot f) (m) = r^{'} \cdot f (m)$ . Once again, we need to verify this is well defined; i.e., the map $r^{'} \cdot f$ is another right $S$ -module morphism. The map $r^{'} \cdot f$ is certainly additive:

\begin{aligned} (r^{'} \cdot f) (m_{1} + m_{2}) & = r^{'} \cdot f (m_{1} + m_{2}) \\ = r^{'} \cdot (f (m_{1}) + f (m_{2})) \\ = r^{'} \cdot f (m_{1}) + r^{'} \cdot f (m_{2}) \\ = (r^{'} \cdot f) (m_{1}) + (r^{'} \cdot f) (m_{2}) . \end{aligned}

It is still compatible with the right $S$ -action:

\begin{aligned} (r^{'} \cdot f) (m) \cdot s & = r^{'} \cdot f (m) \cdot s \\ = r^{'} \cdot f (m s) \\ = (r^{'} \cdot f) (m s) . \end{aligned}

So, $r^{'} \cdot f$ is indeed another right $S$ -module morphism from $M$ to $N$ .

To verify this defines a left action of $R^{'}$ on ${Hom}_{S} (M, N)$ , observe that we certainly have $1_{R^{'}} \cdot f = f$ , since for all $m \in M$ we have

(1_{R^{'}} \cdot f) (m) = 1_{R^{'}} \cdot f (m) = f (m) .

Similarly, we have $(r_{1}^{'} + r_{2}^{'}) \cdot f = r_{1}^{'} \cdot f + r_{2}^{'} \cdot f$ , as

\begin{aligned} ((r_{1}^{'} + r_{2}^{'}) \cdot f) (m) & = (r_{1}^{'} + r_{2}^{'}) \cdot f (m) \\ = r_{1}^{'} \cdot f (m) + r_{2}^{'} \cdot f (m) \\ = (r_{1}^{'} \cdot f) (m) + (r_{2}^{'} \cdot f) (m) . \end{aligned}

It is also compatible with our additive operation in ${Hom}_{S} (M, N)$ :

\begin{aligned} (r^{'} \cdot (f_{1} + f_{2})) (m) & = r^{'} \cdot (f_{1} + f_{2}) (m) \\ = r^{'} \cdot (f_{1} (m) + f_{2} (m)) \\ = r^{'} \cdot f_{1} (m) + r^{'} \cdot f_{2} (m) \\ = (r^{'} \cdot f_{1}) (m) + (r^{'} \cdot f_{2}) (m) \\ = (r^{'} \cdot f_{1} + r^{'} \cdot f_{2}) (m), \end{aligned}

thus $r^{'} \cdot (f_{1} + f_{2}) = r^{'} \cdot f_{1} + r^{'} \cdot f_{2}$ .

Lastly, we should verify that $r_{1}^{'} \cdot (r_{2}^{'} \cdot f) = (r_{1}^{'} r_{2}^{'}) \cdot f$ for all $r_{1}^{'}, r_{2}^{'} \in R^{'}$ . To see this is true, observe that for every $m \in M$ we have

(r_{1}^{'} \cdot (r_{2}^{'} \cdot f)) (m) = r_{1}^{'} \cdot ((r_{2}^{'} \cdot f) (m)) = r_{1}^{'} \cdot (r_{2}^{'} \cdot f (m)) = (r_{1}^{'} r_{2}^{'}) \cdot f (m) = (r_{1}^{'} r_{2}^{'} \cdot f) (m),

where the second-to-last equality holds by the fact that we have a left $R^{'}$ -action on $N$ .

The right $R$ -module structure

We define the right $R$ -module structure on ${Hom}_{S} (M, N)$ using the left $R$ -action on $M$ , namely by defining $f \cdot r$ as the function given by $(f \cdot r) (m) = f (r m)$ . We now repeat the same analysis as above to verify this is a well-defined right $R$ -action on ${Hom}_{S} (M, N)$ .

The map $f \cdot r$ is additive:

\begin{aligned} (f \cdot r) (m_{1} + m_{2}) & = f (r (m_{1} + m_{2})) \\ = f (r m_{1} + r m_{2})) \\ = f (r m_{1}) + f (r m_{2}) \\ = (f \cdot r) (m_{1}) + (f \cdot r) (m_{2}) . \end{aligned}

It is still compatible with the right $S$ -action:

\begin{aligned} (f \cdot r) (m) \cdot s & = f (r m) \cdot s \\ = f (r m s) \\ = (f \cdot r) (m s) . \end{aligned}

So, $f \cdot r$ is indeed another right $S$ -module morphism from $M$ to $N$ .

To verify this defines a right action of $R$ on ${Hom}_{S} (M, N)$ , observe that we certainly have $f \cdot 1_{R} = f$ , since for all $m \in M$ we have

(f \cdot 1_{R}) (m) = f (1_{R} \cdot m) = f (m),

and similarly $f \cdot (r_{1} + r_{2}) = f \cdot r_{1} + f \cdot r_{2}$ , as

\begin{aligned} (f \cdot (r_{1} + r_{2})) (m) & = f ((r_{1} + r_{2}) m) \\ = f (r_{1} m + r_{2} m) \\ = f (r_{1} m) + f (r_{2} m) \\ = (f \cdot r_{1}) (m) + (f \cdot r_{2}) (m_{2}) \\ = (f \cdot r_{1} + f \cdot r_{2}) (m) . \end{aligned}

It is also compatible with our additive operation in ${Hom}_{S} (M, N)$ :

\begin{aligned} ((f_{1} + f_{2}) \cdot r) (m) & = (f_{1} + f_{2}) (r m) \\ = f_{1} (r m) + f_{2} (r m) \\ = (f_{1} \cdot r) (m) + (f_{2} \cdot r) (m) \\ = (f_{1} \cdot r + f_{2} \cdot r) (m), \end{aligned}

thus $(f_{1} + f_{2}) \cdot r = f_{1} \cdot r + f_{2} \cdot r$ .

Finally, we should verify that $(f \cdot r_{1}) \cdot r_{2} = f \cdot (r_{1} r_{2})$ for all $r_{1}, r_{2} \in r$ . To see this, observe that for every $m \in M$ we have

((f \cdot r_{1}) \cdot r_{2}) (m) = (f \cdot r_{1}) (r_{2} m) = f (r_{1} (r_{2} m)) = f ((r_{1} r_{2}) m) = (f \cdot (r_{1} r_{2})) (m),

where the second-to-last equality holds by the fact that we have a left $R$ -action on $M$ .

The bimodule condition

The last property we need to verify is that our two actions on ${Hom}_{S} (M, N)$ are compatible. To that end, suppose $r^{'} \in R$ , $r \in R$ , and $f \in {Hom}_{S} (M, N)$ . For every $m \in M$ we have

(r^{'} \cdot (f \cdot r)) (m) = r^{'} \cdot (f \cdot r) (m) = r^{'} \cdot f (r m),

while

((r^{'} \cdot f) \cdot r) (m) = (r^{'} \cdot f) (r m) = r^{'} \cdot f (r m) .

Thus, we do indeed have $r^{'} \cdot (f \cdot r) = (r^{'} \cdot f) \cdot r$ .

We've at last established ${Hom}_{S} (M, N)$ does indeed have the structure of an $(R^{'}, R)$ -bimodule.

Problem 7

Suppose $R, S$ , and $T$ are rings (with unity), $M$ is an $(R, S)$ -bimodule and $N$ is an $(S, T)$ -bimodule.

Define functors $F, G : (R, T) - Bimod \to S e t$ such that on objects
$\begin{aligned} F (P) & = {Hom}_{(R, T) - Bimod} (M \otimes_{S} N, P) \\ G (P) & = {Hom}_{(R, S) - Bimod} (M, {Hom}_{Mod - T} (N, P)) . \end{aligned}$
In other words, what are the maps on arrows?
For every $(R, T)$ -bimodule $P$ there is a set bijection
$τ_{P} : {Hom}_{(R, T) - Bimod} (M \otimes_{S} N, P) \tilde{\to} {Hom}_{(R, S) - Bimod} (M, {Hom}_{Mod - T} (N, P)) .$
See these notes for the explicit description of the set map $τ_{P}$ , as well as the verification that $τ_{P}$ is a bijection. In short, for each $(R, T)$ -bimodule morphism $f : M \otimes_{S} N \to P$ , $τ_{P} (f)$ is the $(R, S)$ -bimodule morphism $τ_{P} (f) : M \to {Hom}_{Mod - T} (N, P)$ that assigns to each $m \in M$ the right $T$ -module morphism $n \mapsto f (m \otimes n)$ .

Show that these bijections $τ_{p}$ define a natural transformation $τ : F \Rightarrow G$ . Since every $τ_{P}$ is a bijection, we call $τ$ a natural isomorphism between the functors $F$ and $G$ .

Solution

Suppose $f : P_{1} \to P_{2}$ is an $(R, T)$ -bimodule morphism. For each $(R, T)$ -bimodule morphism $g : M \otimes_{S} N \to P_{1}$ , composition with $f$ yields an $(R, T)$ -bimodule morphism $f \circ g : M \otimes_{S} N \to P_{2}$ . This defines a set map from $F (P_{1}) = {Hom}_{(R, T)} (M \otimes_{S} N, P_{1})$ to $F (P_{2}) = {Hom}_{(R, T)} (M \otimes_{S} N, P_{2}) .$

Optional functoriality check

We can verify this map on arrows (together with the given map on objects) does indeed define a functor $F : (R, T) - Mod \to Set$ . First note that for each identity arrow $1_{P} : P \to P$ we have $F (1_{P}) = 1_{P} \circ -$ , which sends each $(R, T)$ -morphism $g : M \otimes_{S} N \to P$ to $1_{P} \circ g = g$ ; i.e., is the identity map on the set ${Hom}_{(R, T)} (M \otimes_{S} N, P) = F (P)$ .

Also, for each pair of composable $(R, T)$ -bimodule morphisms $f_{1} : P_{1} \to P_{2}$ and $f_{2} : P_{2} \to P_{3}$ , for every $(R, T)$ -module morphism $g : M \otimes_{S} N : \to P$ we have $F (f_{2} \circ f_{1}) (g) = (f_{2} \circ f_{1}) \circ g$ and $(F (f_{2}) \circ F (f_{1})) (g) = f_{2} \circ (f_{1} \circ g)$ . By associativity of composition in $(R, T) - Mod$ these compositions are always equal, hence $F (f_{2} \circ f_{1}) = F (f_{2}) \circ F (f_{2})$ . Thus, $F$ does indeed define a functor.

We next consider the functor $G$ . Suppose $f : P_{1} \to P_{2}$ is an $(R, T)$ -bimodule morphism. For each $(R, S)$ -bimodule morphism $g : M \to {Hom}_{T} (N, P_{1})$ , let's introduce the temporary notation $g_{m}$ for the right $T$ -module morphism $g (m) : N \to P_{1}$ . In particular, note that composition with $f$ then yields a right $T$ -module morphism $f \circ g_{m} : N \to P_{2}$ . In other words, $f \circ g_{m} \in {Hom}_{T} (N, P_{2}) .$ It therefore makes sense to define a set map $M \to {Hom}_{T} (N, P_{2})$ by $m \mapsto f \circ g_{m}$ . Again, in order to avoid getting overwhelmed by parentheses, let's denote this map by $(G f)_{g}$ . Of course, we need to verify that $(G f)_{g}$ is actually an $(R, S)$ -bimodule morphism. We first note that $g$ is a module morphism and hence additive:
$g_{m_{1} + m_{2}} = g (m_{1} + m_{2}) = g (m_{1}) + g (m_{2}) = g_{m_{1}} + g_{m_{2}}$
Then we can observe that
$\begin{aligned} (G f)_{g} (m_{1} + m_{2}) & = f \circ g_{m_{1} + m_{2}} \\ = f \circ (g_{m_{1}} + g_{m_{2}}) \\ = f \circ g_{m_{1}} + f \circ g_{m_{2}} \\ = (G f)_{g} (m_{1}) + (G f)_{g} (m_{2}) . \end{aligned}$
We next verify $(G f)_{g}$ is compatible with the left $R$ -action. Observe that since both $f$ and $g$ are compatible with the left $R$ -action we have
$\begin{aligned} (G f)_{g} (r m) & = f \circ g_{r m} \\ = f \circ (g (r m)) \\ = f \circ (r \cdot g (m)) \\ = f \circ (r \cdot g_{m}) \\ = r \cdot (f \circ g_{m}) \\ = r \cdot (G f)_{g} (m) . \end{aligned}$
It is also compatible with the right $S$ -action, once we recall that the right $S$ action on ${Hom}_{T} (N, P)$ is by left multiplication on the inputs:
$\begin{aligned} (G f)_{g} (m s) & = f \circ g_{m s} \\ = f \circ (g (m s)) \\ = f \circ (g (m) \cdot s) \\ = f \circ (g_{m} \cdot s) \\ = f \circ (g_{m} (s \cdot -)) \\ = (f \circ g_{m}) (s \cdot -) \\ = (f \circ g_{m}) \cdot s \\ = (G f)_{g} (m) \cdot s . \end{aligned}$
So, we've verified $(G f)_{g}$ is indeed an $(R, S)$ -bimodule morphism and hence we have defined a set map
$G (f) : {Hom}_{(R, S)} (M, {Hom}_{T} (N, P_{1})) \to {Hom}_{(R, S)} (M, {Hom}_{T} (N, P_{2}))$
that given by $g \mapsto (G f)_{g}$ .

Optional Functoriality Check

Lastly, we should verify this map on arrows (together with the given map on objects) does indeed define a functor $G : (R, T) - Mod \to Set$ . First note that for each identity arrow $1_{P} : P \to P$ we have $G (1_{P})$ is the set map $g \mapsto (G 1_{P})_{g}$ . But for every $m \in M$ we have $(G 1_{P})_{g} (m) = 1_{P} \circ g_{m} = g_{m} = g (m)$ , which is the identity on $G (P)$ .

Also, for each pair of composable $(R, T)$ -bimodule morphisms $f_{1} : P_{1} \to P_{2}$ and $f_{2} : P_{2} \to P_{3}$ , for every $(R, S)$ -module morphism $g : M \to {Hom}_{T} (N, P_{1})$ , and for every $m \in M$ we have
$G (f_{2} \circ f_{1})_{g} (m) = (f_{2} \circ f_{1}) \circ g_{m} = f_{2} \circ (f_{1} \circ g_{m})$
and
$(G (f_{2}) \circ G (f_{1}))_{g} (m) = G (f_{2})_{G (f_{1})_{g}} (m) = f_{2} \circ (G (f_{1})_{g} (m)) = f_{2} \circ (f_{1} \circ g_{m})) .$
By associativity of composition we therefore have $G (f_{2} \circ f_{1})_{g} = (G (f_{2}) \circ G (f_{1}))_{g}$ , and as $g$ was arbitrary, this prove $G (f_{2} \circ f_{1}) = G (f_{2}) \circ G (f_{1})$ .
Suppose $f : P_{1} \to P_{2}$ is an $(R, T)$ -bimodule morphism. We need to verify that the diagram below is commutative:

Take any element in the top-left corner; i.e., a $(R, T)$ -bimodule morphism $g : M \otimes_{S} N \to P_{1}$ . The left vertical map sends $g$ to the $(R, T)$ -bimodule morphism $f \circ g : M \otimes_{S} N \to P_{2}$ . The bottom horizontal map then sends $f \circ g$ to the $(R, S)$ -bimodule morphism $τ_{P_{2}} (f \circ g) : M \to {Hom}_{T} (N, P_{2})$ that assigns to each $m \in M$ the right $T$ -module morphism $n \mapsto (f \circ g) (m \otimes n)$ .

On the other hand, the top horizontal map sends $g$ to the $(R, S)$ -bimodule morphism $τ_{P_{1}} (g) : M \to {Hom}_{T} (N, P_{1})$ that assigns to each $m \in M$ the right $T$ -module morphism $n \mapsto g (m \otimes n)$ . The right vertical map then sends $τ_{P_{1}} (g)$ to the $(R, S)$ -bimodule morphism $(G f)_{τ_{P_{1}} (g)} : M \to {Hom}_{T} (N, P_{2})$ that assigns to each $m \in M$ the right $T$ -module morphism $f \circ τ_{P_{1}} (g)_{m}$ , which is the map $n \mapsto f (g (m \otimes n))$ .

Thus, the diagram is indeed commutative.

"Quotient field" is another term for "field of fractions." ↩︎

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

The abelian group structure

The left R′-module structure

The right R-module structure

The bimodule condition

Problem 7

Optional functoriality check

Optional Functoriality Check

The left $R^{'}$ -module structure

The right $R$ -module structure