Homework 6

Problem 1

Suppose $R$ is a commutative ring and $M$ is an $R$ -module. Prove that for $m, n_{1}, n_{2}, \dots, n_{k} \in M$ one always has

$m \land n_{1} \land n_{2} \land \dots \land n_{k} = (- 1)^{k} (n_{1} \land n_{2} \land \dots \land n_{k}) \land m .$

Solution

This follows by induction on $k$ . When $k = 1$ , the result $m \land n_{1} = - (n_{1} \land m)$ was proven here. Suppose the result holds for some $k \geq 1$ . Then observe that

$\begin{aligned} m_{1} \land n_{1} \land n_{2} \land \dots \land n_{k + 1} & = (m_{1} \land n_{1}) \land n_{2} \land \dots \land n_{k + 1} \\ = - (n_{1} \land m) \land n_{2} \land \dots \land n_{k + 1} \\ = - (n_{1} \land (m \land n_{2} \land \dots \land n_{k + 1})) \\ = - (n_{1} \land ((- 1)^{k} (n_{2} \land n_{3} \land \dots \land n_{k + 1}) \land m)) \\ = (- 1)^{k + 1} (n_{1} \land n_{2} \land n_{3} \land \dots \land n_{k + 1}) \land m . \end{aligned}$

We've thus proven the result holds for $k + 1$ and hence holds for all $k \geq 1$ by mathematical induction.

Problem 2

Suppose $R$ is a commutative ring. Prove that for each cyclic $R$ -module $M$ we have $T (M) ≃ S (M)$ ; i.e., the tensor algebra is already commutative.

Hint

Show the ideal $C (M)$ is trivial.

Solution

Let $m_{0} \in M$ be a generator for $M$ as an $R$ -module. Then every $m \in M$ is of the form $m = r m_{0}$ for some $r \in R$ (not necessarily unique). In particular, every simple 2-tensor $m \otimes m^{'}$ is of the form $(r m_{0}) \otimes (r^{'} m_{0})$ for some $r, r^{'} \in R$ . But then observe that

$m \otimes m^{'} = (r m_{0}) \otimes (r^{'} m_{0}) = r r^{'} \cdot (m_{0} \otimes m_{0}) .$

In particular, we always have

$\begin{aligned} m \otimes m^{'} - m^{'} \otimes m & = r r^{'} \cdot (m_{0} \otimes m_{0}) - r^{'} r \cdot (m_{0} \otimes m_{0}) \\ = (r r^{'} - r^{'} r) \cdot (m_{0} \otimes m_{0}) \\ = 0_{r} \cdot (m_{0} \otimes m_{0}) \\ = 0. \end{aligned}$

Since the ideal $C (M)$ is generated by 2-tensors of the form $m \otimes m^{'} - m^{'} \otimes m$ and these tensors are all zero, the entire ideal $C (M)$ must be trivial. It follows that

$S (M) = T (M) / C (M) = T (M) / 0 ≃ T (M) .$

Problem 3

Suppose $R$ is a commutative ring and $M$ is a free $R$ -module of rank $n$ , i.e., $M ≃ F (X)$ for some set $X$ with $n$ elements. Prove that $\overset{k}{⋀} (M)$ is a free $R$ -module of rank $(\binom{n}{k})$ for $k = 0, 1, 2, \dots, n$ .

Hint

You can use the fact that for $0 \leq k \leq n$ the $R$ -module $\overset{k}{⋀} (M)$ is a free $R$ -module with a basis explicitly given in terms of a basis for $M$ .

Solution

Let $B = {m_{1}, \dots, m_{n}}$ be a basis for $M$ as a free $R$ -module. We know that for $0 \leq k \leq n$ the $R$ -module $\overset{k}{⋀} (M)$ is a free $R$ -module with basis

${v_{i_{1}} \land \dots \land v_{i_{k}} ∣ 1 \leq i_{1} < \dots < i_{k} \leq n}$

It follows that the dimension of $\overset{k}{⋀} (M)$ is equal to the number of $k$ -tuples of integers $(i_{1}, \dots, i_{k})$ satisfying $1 \leq i_{1} < \dots < i_{k} \leq n$ . The number of such $k$ -tuples is exactly $(\binom{n}{k})$ , since any choice of $k$ distinct integers from the set ${1, \dots, n}$ produces a unique such $k$ -tuple (obtained by simply listing the chosen distinct integers in ascending order).

Problem 4

Suppose $F$ is a field of characteristic not 2. Show that for every $F$ -vector space $V$ we have an $F$ -vector space isomorphism
$V \otimes_{F} V ≃ S^{2} (V) \oplus \overset{2}{⋀} (V) .$
In other words, show that every 2-tensor may be written uniquely as a sum of a symmetric and an alternating tensor.

Hint

For each 2-tensor $z$ , show that $z = \frac{1}{2} Sym (z) + \frac{1}{2} Alt (z)$ , where $Sym (z)$ and $Alt (z)$ are the symmetrization and skew-symmetrizationof $z$ , respectively.

This will show $V \otimes_{F} F$ is the sum of the submodules of symmetric and alternating tensors. To show it is the direct sum, verify that the only 2-tensor that is both symmetric and alternating is $0$ .

Finally, you can use the fact that there is an isomorphism between the symmetric power (respectively, exterior power) and the submodule of symmetric tensors (respectively, alternating tensors).

Solution

Let $m_{1} \otimes m_{2}$ be any simple tensor in $V \otimes_{F} V$ . The symmetrization and skew-symmetrization of $m_{1} \otimes m_{2}$ are then

$\begin{aligned} Sym (m_{1} \otimes m_{2}) & = m_{1} \otimes m_{2} + m_{2} \otimes m_{1} \\ Alt (m_{1} \otimes m_{2}) & = m_{1} \otimes m_{2} - m_{2} \otimes m_{1} . \end{aligned}$

Observe that $Sym (m_{1} \otimes m_{2})$ is symmetric, $Alt (m_{1} \otimes m_{2})$ is alternating, and

$\begin{aligned} m_{1} \otimes m_{2} & = \frac{1}{2} (m_{1} \otimes m_{2} + m_{2} \otimes m_{1}) + \frac{1}{2} (m_{1} \otimes m_{2} - m_{2} \otimes m_{1}) \\ = \frac{1}{2} Sym (m_{1} \otimes m_{2}) + \frac{1}{2} Alt (m_{1} \otimes m_{2}) . \end{aligned}$

The symmetrization and skew-symmetrization maps are actually $R$ -module endomorphisms on $V \otimes_{F} V$ , which just means they extend $R$ -linearly to the general 2-tensors; i.e., if $z = \sum_{i = 1}^{n} (m_{i_{1}} \otimes m_{i_{2}})$ then

$\begin{aligned} Sym (z) & = \sum_{i = 1}^{n} ((m_{i_{1}} \otimes m_{i_{2}}) + (m_{i_{2}} \otimes m_{i_{1}})) = \sum_{i = 1}^{n} Sym (m_{i_{1}} \otimes m_{i_{2}}), \end{aligned}$

and similarly for $Alt (z)$ . It follows that for every 2-tensor $z$ we have

$z = \frac{1}{2} Sym (z) + \frac{1}{2} Alt (z) .$

We've thus shown that every 2-tensor can be written as the sum of a symmetric 2-tensor and an alternating 2-tensor. This proves that $V \otimes_{F} V$ is at least the sum of the submodules of symmetric and alternating 2-tensors.

To show we have a direct sum decomposition we need to show that the submodules of symmetric and alternating 2-tensors have trivial intersection. To that end, suppose that a 2-tensor $z$ is both symmetric and alternating. Let $z = \sum_{i = 1}^{n} (m_{i_{1}} \otimes m_{i_{2}})$ . Then because $z$ is symmetric we have

$z = \sum_{i = 1}^{n} (m_{i_{1}} \otimes m_{i_{2}}) = \sum_{i = 1}^{n} (m_{i_{2}} \otimes m_{i_{1}}) .$

On the other hand, because $z$ is alternating we also have

$z = \sum_{i = 1}^{n} (m_{i_{1}} \otimes m_{i_{2}}) = - \sum_{i = 1}^{n} (m_{i_{2}} \otimes m_{i_{1}}) .$

Summing these two equations then gives $2 z = 0$ and hence $z = 0$ (since the characteristic of our field $F$ is not 2).

Thus, the $F$ -vector space $V \otimes_{F} V$ is (isomorphic to) the direct sum of the submodules of symmetric and alternating 2-tensors. It now follows that $V \otimes_{F} V ≃ S^{2} (V) \oplus \overset{2}{⋀} (V)$ .