Homework 5

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.

Note: Our notation is intentionally a bit sloppy here, since this is how you'll often see statements like this written. In the first tensor product, the copy of $C$ on the right is the $R$ -module of complex numbers (i.e., the additive group of complex numbers along with scaling by real numbers), while in the second tensor product it is the $C$ -module of complex numbers (i.e., the ring of complex numbers).

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!)

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 $\sum q_{i} \otimes m_{i}$ .

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.

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 $D$ .

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.

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?

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 $F, G : (R, T) - M o d \to S e t$ such that on objects
$\begin{aligned} F (P) & = {Hom}_{(R, T)} (M \otimes_{S} N, P) \\ G (P) & = {Hom}_{(R, S)} (M, {Hom}_{T} (N, P)) . \end{aligned}$
In other words, what are the maps on arrows?
For every $(R, T)$ -bimodule $P$ there is a set bijection
$τ_{P} : {Hom}_{(R, T)} (M \otimes_{S} N, P) \tilde{\to} {Hom}_{(R, S)} (M, {Hom}_{T} (N, P)) .$
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 $τ_{P} (f) : M \to {Hom}_{T} (N, P)$ 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 $G$ .

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