Midterm Exam Solutions

Problem 1

Let $R$ be a commutative ring, $X$ a finite set and $F (X)$ the free $R$ -module on $X$ . Prove there is an $R$ -module isomorphism ${Hom}_{R} (F (X), R) ≃ F (X)$ .
Explicitly describe the isomorphism $ϕ : {Hom}_{R} (F (X), R) \tilde{\to} F (X)$ . In other words, given an $R$ -module morphism $f : F (X) \to R$ , what is the corresponding element $ϕ (f)$ in $F (X)$ ?

Solution

We have two options:

A very short proof

The shortest solution is to recall that if $X = {x_{1}, \dots, x_{n}}$ is a set with $n$ elements, then we have $R$ -module isomorphisms

F (X) ≃ \underset{n times}{\underset{⏟}{R \oplus \dots \oplus R}} = ⨁_{i = 1}^{n} R,

and so (using another problem from the study guide with $M = R$ ) we have

{Hom}_{R} (F (X), R) ≃ {Hom}_{R} (⨁_{i = 1}^{n} R, R) ≃ ⨁_{i = 1}^{n} {Hom}_{R} (R, R) ≃ ⨁_{i = 1}^{n} R ≃ F (X) .

If we use this solution to part (1), then for part (2) we need to chain together the actual maps behind the above isomorphisms. When we do so, we'll arrive at the same map described below.

A direct proof

We can also give a direct proof. Define a map $ϕ : {Hom}_{R} (F (X), R) \to F (X)$ by sending each $R$ -module morphism $f : F (X) \to R$ to the element $ϕ (f) = \sum_{i = 1}^{n} f (x_{i}) \cdot x_{i}$ . (Here we are identifying $x_{i}$ with its image in $F (X)$ . Under the isomorphism $F (X) ≃ ⨁_{i} R$ the corresponding element is what we have previously called $e_{i}$ , the formal sum in which all elements are $0_{R}$ except the $i^{th}$ element, which is $1_{R}$ .) This map is additive:

\begin{aligned} ϕ (f_{1} + f_{2}) & = \sum_{i = 1}^{n} (f_{1} + f_{2}) (x_{i}) \cdot x_{i} \\ = \sum_{i = 1}^{n} (f_{1} (x_{i}) + f_{2} (x_{i})) \cdot x_{i} \\ = \sum_{i = 1}^{n} f_{1} (x_{i}) \cdot x_{i} + \sum_{i = 1}^{n} f_{2} (x_{i}) \cdot x_{i} \\ = ϕ (f_{1}) + ϕ (f_{2}) . \end{aligned}

It is also compatible with the left $R$ -action:

ϕ (r \cdot f) = \sum_{i = 1}^{n} (r \cdot f) (x_{i}) \cdot x_{i} = \sum_{i}^{n} r f (x_{i}) \cdot x_{i} = r \cdot \sum_{i = 1}^{n} f (x_{i}) \cdot x_{i} = r \cdot ϕ (f) .

It is therefore an $R$ -module morphism.

As for the inverse map, define $ψ : F (X) \to {Hom}_{R} (F (X), R)$ by sending each formal sum $s = \sum_{i = 1}^{n} r_{i} x_{i}$ to the map $ψ_{s} : F (X) \to R$ defined by $ψ_{s} (x_{i}) = r_{i}$ . We automatically have that $ψ_{s}$ is an $R$ -module morphism and hence an element of ${Hom}_{R} (F (X), R)$ . Moreover, the map $ψ$ we have defined is indeed additive. If $s = \sum_{i = 1}^{n} r_{i} x_{i}$ and $s^{'} = \sum_{i = 1}^{n} r_{i}^{'} x_{i}$ , then $s + t = \sum_{i = 1}^{n} (r_{i} + r_{i}^{'}) x_{i}$ and for all $i$ we have

ψ_{s + s^{'}} (x_{i}) = r_{i} + r_{i}^{'} = ϕ_{s} (x_{i}) + ϕ_{s^{'}} (x_{i}),

hence $ψ_{s + s^{'}} = ψ_{s} + ψ_{s^{'}}$ . It is also an $R$ -module morphism, since if $s = \sum_{i = 1}^{n} r_{i} x_{i}$ then $r \cdot s = r \cdot \sum_{i = 1}^{n} r_{i} x_{i} = \sum_{i = 1}^{n} (r r_{i}) x_{i}$ and so for all $i$ we have

ψ_{r \cdot s} (x_{i}) = r r_{i} = r \cdot ψ_{s} (x_{i}),

hence $ψ_{r \cdot s} = r \cdot ψ_{s}$ . Thus, $ψ$ is indeed an $R$ -module morphism.

Finally, observe that $ϕ$ and $ψ$ are mutual inverses. For each $f \in {Hom}_{R} (F (X), R),$ let $s = \sum_{i = 1}^{n} f (x_{i}) x_{i}$ . Then

ψ (ϕ (f)) = ψ (\sum_{i = 1}^{n} f (x_{i}) x_{i}) = ψ_{s},

and for every $i$ we have $ψ_{s} (x_{i}) = f (x_{i})$ ; i.e., $ψ_{s} = f$ . In other words, for every $f$ we have $ψ (ϕ (f)) = f$ .

Conversely, suppose $s \in F (X)$ , say $s = \sum_{i = 1}^{n} r_{i} x_{i}$ . Then $ψ_{s}$ is the map defined by $ψ_{s} (x_{i}) = r_{i}$ , and so $ϕ (ψ_{s}) = \sum_{i = 1}^{n} r_{i} x_{i} = s$ ; i.e., $ϕ (ψ_{s}) = s$ . Thus $ϕ$ and $ψ$ are mutual inverses and hence isomorphisms.

Problem 2

Suppose $f : N \to P$ is a $(S, T)$ -bimodule morphism.

Show that for every $(R, S)$ -bimodule $M$ there is an $(R, T)$ -bimodule morphism
$1_{M} \otimes f : M \otimes_{S} N \to M \otimes_{S} P,$
defined on simple tensors by $m \otimes n \mapsto m \otimes f (n)$ .
Prove that if $f$ is surjective, then so is $1_{M} \otimes f$ .

Solution

Define a set map $g : M \times N \to M \otimes_{S} P$ by $(m, n) \mapsto m \otimes f (n)$ . This map is linear in $M$ :
$\begin{aligned} g (m_{1} + m_{2}, n) & = (m_{1} + m_{2}) \otimes f (n) \\ = m_{1} \otimes f (n) + m_{2} \otimes f (n) \\ g (m_{1}, n) + g (m_{2}, n) . \end{aligned}$
It is linear in $N$ , since $f$ is additive:
$\begin{aligned} g (m, n_{1} + n_{2}) & = m \otimes f (n_{1} + n_{2}) \\ = m \otimes (f (n_{1}) + f (n_{2})) \\ = m \otimes f (n_{1}) + m \otimes f (n_{2}) \\ = g (m, n_{1}) + g (m, n_{2}) . \end{aligned}$
It is $S$ -balanced, since $f$ is compatible with the left $S$ -actions:
$\begin{aligned} g (m s, n) & = m s \otimes f (n) \\ = m \otimes f (s n) \\ = g (m, s n) . \end{aligned}$
It is compatible with the left $R$ -actions:
$\begin{aligned} g (r m, n) & = r m \otimes f (n) \\ = r \cdot (m \otimes f (n)) \\ = r \cdot g (m, n) . \end{aligned}$
It is compatible with the right $T$ -actions, since $f$ is compatible with the right $T$ -actions:
$\begin{aligned} g (m, n t) & = m \otimes f (n t) \\ = m \otimes f (n) t \\ = (m \otimes f (n)) \cdot t \\ = g (m, n) \cdot t . \end{aligned}$
So, by the universal property of the tensor product the corresponding map $m \otimes n \mapsto m \otimes f (n)$ is a well-defined $(R, T)$ -bimodule morphism.
Since $f$ is surjective, every element in $P$ is of the form $f (n)$ for some $n \in N$ . Since the simple tensors of the form $m \otimes p$ generate $M \otimes_{S} P$ as an $(R, T)$ -bimodule, it follows that the simple tensors of the form $m \otimes f (n)$ also generate $M \otimes_{S} P$ as an $(R, T)$ -bimodule. But this implies that the elements $(1_{M} \otimes f) (m \otimes n)$ generate $M \otimes_{S} P$ as an $(R, T)$ -bimodule. Since those same elements generate the image of $1_{M} \otimes f$ , this proves the image of $1_{M} \otimes f$ is all of $M \otimes_{S} P$ ; i.e., the morphism $1_{M} \otimes f$ is surjective.

Problem 3

Let $V$ be a finite-dimensional vector space over a field $F$ and let $B = {e_{1}, \dots, e_{n}}$ be a basis for $V$ . You may use the fact that ${e_{i} \otimes e_{j} ∣ 1 \leq i, j \leq n}$ is then an $F$ -vector space basis for $V \otimes_{F} V$ .

Prove that every element of $V \otimes_{F} V$ can be written uniquely in the form $\sum_{i = 1}^{n} v_{i} \otimes e_{i}$ where $v_{i} \in V$ .
Let $v, w \in V$ be nonzero vectors. Prove that $v \otimes w = w \otimes v$ in $V \otimes_{F} V$ if and only if $w = a v$ for some $a \in F$ .

Solution

First consider simple tensors $v \otimes w$ . We can write $w = \sum_{i = 1}^{n} d_{i} e_{i}$ and then use the bilinearity properties of the tensor product to write
$v \otimes w = v \otimes \sum_{i = 1}^{n} d_{i} e_{i} = \sum_{i = 1}^{n} (d_{i} v) \otimes e_{i} .$
This shows it is possible to express every simple tensor $v \otimes w$ in the desired form. For a general tensor $\sum_{j = 1}^{m} v_{j} \otimes w_{j}$ we can use the above process to write each summand $v_{j} \otimes w_{j}$ in the desired form, and then properties of the tensor product allow us to write the entire tensor in the desired form:
$\sum_{j = 1}^{m} v_{j} \otimes w_{j} = \sum_{j = 1}^{m} \sum_{i = 1}^{n} v_{j i} \otimes e_{i} = \sum_{i = 1}^{n} (\sum_{j = 1}^{m} v_{j i}) \otimes e_{i} .$
This proves that every tensor can be written in the desired form.

It remains to prove uniqueness of each such expression. To that end, suppose $v \otimes w$ can be written both as $\sum_{i = 1}^{n} v_{i} \otimes e_{i}$ and as $\sum_{i = 1}^{n} v_{i}^{'} \otimes e_{i}$ for some vectors $v_{i}, v_{i}^{'} \in V$ . We then have