Homework 4

Problem 1

Show that $C \otimes_{R} C$ and $C \otimes_{C} C$ are both left $R$ -modules but are not isomorphic as $R$ -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 $R^{2} ≃ R \oplus R$ .

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 $Q$ 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 $ϕ ((r, s_{1}) + (r, s_{2})) = ϕ (r, s_{1}) + ϕ (r, s_{2}) .$

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