Lemma 10.13.1. Let R be a ring. Let M be an R-module. If M is a free R-module, so is each symmetric and exterior power.
10.13 Tensor algebra
Let R be a ring. Let M be an R-module. We define the tensor algebra of M over R to be the noncommutative R-algebra
\text{T}(M) = \text{T}_ R(M) = \bigoplus \nolimits _{n \geq 0} \text{T}^ n(M)
with \text{T}^0(M) = R, \text{T}^1(M) = M, \text{T}^2(M) = M \otimes _ R M, \text{T}^3(M) = M \otimes _ R M \otimes _ R M, and so on. Multiplication is defined by the rule that on pure tensors we have
(x_1 \otimes x_2 \otimes \ldots \otimes x_ n) \cdot (y_1 \otimes y_2 \otimes \ldots \otimes y_ m) = x_1 \otimes x_2 \otimes \ldots \otimes x_ n \otimes y_1 \otimes y_2 \otimes \ldots \otimes y_ m
and we extend this by linearity.
We define the exterior algebra \wedge (M) of M over R to be the quotient of \text{T}(M) by the two sided ideal generated by the elements x \otimes x \in \text{T}^2(M). The image of a pure tensor x_1 \otimes \ldots \otimes x_ n in \wedge ^ n(M) is denoted x_1 \wedge \ldots \wedge x_ n. These elements generate \wedge ^ n(M), they are R-linear in each x_ i and they are zero when two of the x_ i are equal (i.e., they are alternating as functions of x_1, x_2, \ldots , x_ n). The multiplication on \wedge (M) is graded commutative, i.e., every x \in M and y \in M satisfy x \wedge y = - y \wedge x.
An example of this is when M = Rx_1 \oplus \ldots \oplus Rx_ n is a finite free module. In this case \wedge (M) is free over R with basis the elements
x_{i_1} \wedge \ldots \wedge x_{i_ r}
with 0 \leq r \leq n and 1 \leq i_1 < i_2 < \ldots < i_ r \leq n.
We define the symmetric algebra \text{Sym}(M) of M over R to be the quotient of \text{T}(M) by the two sided ideal generated by the elements x \otimes y - y \otimes x \in \text{T}^2(M). The image of a pure tensor x_1 \otimes \ldots \otimes x_ n in \text{Sym}^ n(M) is denoted just x_1 \ldots x_ n. These elements generate \text{Sym}^ n(M), these are R-linear in each x_ i and x_1 \ldots x_ n = x_1' \ldots x_ n' if the sequence of elements x_1, \ldots , x_ n is a permutation of the sequence x_1', \ldots , x_ n'. Thus we see that \text{Sym}(M) is commutative.
An example of this is when M = Rx_1 \oplus \ldots \oplus Rx_ n is a finite free module. In this case \text{Sym}(M) = R[x_1, \ldots , x_ n] is a polynomial algebra.
Proof. Omitted, but see above for the finite free case. \square
Lemma 10.13.2. Let R be a ring. Let M_2 \to M_1 \to M \to 0 be an exact sequence of R-modules. There are exact sequences
M_2 \otimes _ R \text{Sym}^{n - 1}(M_1) \to \text{Sym}^ n(M_1) \to \text{Sym}^ n(M) \to 0
and similarly
M_2 \otimes _ R \wedge ^{n - 1}(M_1) \to \wedge ^ n(M_1) \to \wedge ^ n(M) \to 0
Proof. Omitted. \square
Lemma 10.13.3. Let R be a ring. Let M be an R-module. Let x_ i, i \in I be a given system of generators of M as an R-module. Let n \geq 2. There exists a canonical exact sequence
\bigoplus _{1 \leq j_1 < j_2 \leq n} \bigoplus _{i_1, i_2 \in I} \text{T}^{n - 2}(M) \oplus \bigoplus _{1 \leq j_1 < j_2 \leq n} \bigoplus _{i \in I} \text{T}^{n - 2}(M) \to \text{T}^ n(M) \to \wedge ^ n(M) \to 0
where the pure tensor m_1 \otimes \ldots \otimes m_{n - 2} in the first summand maps to
\begin{align*} \underbrace{ m_1 \otimes \ldots \otimes x_{i_1} \otimes \ldots \otimes x_{i_2} \otimes \ldots \otimes m_{n - 2} }_{\text{with } x_{i_1} \text{ and } x_{i_2} \text{ occupying slots } j_1 \text{ and } j_2 \text{ in the tensor}} \\ + \underbrace{ m_1 \otimes \ldots \otimes x_{i_2} \otimes \ldots \otimes x_{i_1} \otimes \ldots \otimes m_{n - 2} }_{\text{with } x_{i_2} \text{ and } x_{i_1} \text{ occupying slots } j_1 \text{ and } j_2 \text{ in the tensor}} \end{align*}
and m_1 \otimes \ldots \otimes m_{n - 2} in the second summand maps to
\underbrace{ m_1 \otimes \ldots \otimes x_ i \otimes \ldots \otimes x_ i \otimes \ldots \otimes m_{n - 2} }_{\text{with } x_{i} \text{ and } x_{i} \text{ occupying slots } j_1 \text{ and } j_2 \text{ in the tensor}}
There is also a canonical exact sequence
\bigoplus _{1 \leq j_1 < j_2 \leq n} \bigoplus _{i_1, i_2 \in I} \text{T}^{n - 2}(M) \to \text{T}^ n(M) \to \text{Sym}^ n(M) \to 0
where the pure tensor m_1 \otimes \ldots \otimes m_{n - 2} maps to
\begin{align*} \underbrace{ m_1 \otimes \ldots \otimes x_{i_1} \otimes \ldots \otimes x_{i_2} \otimes \ldots \otimes m_{n - 2} }_{\text{with } x_{i_1} \text{ and } x_{i_2} \text{ occupying slots } j_1 \text{ and } j_2 \text{ in the tensor}} \\ - \underbrace{ m_1 \otimes \ldots \otimes x_{i_2} \otimes \ldots \otimes x_{i_1} \otimes \ldots \otimes m_{n - 2} }_{\text{with } x_{i_2} \text{ and } x_{i_1} \text{ occupying slots } j_1 \text{ and } j_2 \text{ in the tensor}} \end{align*}
Proof. Omitted. \square
Lemma 10.13.4. Let A \to B be a ring map. Let M be a B-module. Let n > 1. The kernel of the A-linear map M \otimes _ A \ldots \otimes _ A M \to \wedge ^ n_ B(M) is generated as an A-module by the elements m_1 \otimes \ldots \otimes m_ n with m_ i = m_ j for i \not= j, m_1, \ldots , m_ n \in M and the elements m_1 \otimes \ldots \otimes bm_ i \otimes \ldots \otimes m_ n - m_1 \otimes \ldots \otimes bm_ j \otimes \ldots \otimes m_ n for i \not= j, m_1, \ldots , m_ n \in M, and b \in B.
Proof. Omitted. \square
Lemma 10.13.5.slogan Let R be a ring. Let M_ i be a directed system of R-modules. Then \mathop{\mathrm{colim}}\nolimits _ i \text{T}(M_ i) = \text{T}(\mathop{\mathrm{colim}}\nolimits _ i M_ i) and similarly for the symmetric and exterior algebras.
Proof. Omitted. Hint: Apply Lemma 10.12.9. \square
Lemma 10.13.6. Let R be a ring and let S \subset R be a multiplicative subset. Then S^{-1}T_ R(M) = T_{S^{-1}R}(S^{-1}M) for any R-module M. Similar for symmetric and exterior algebras.
Proof. Omitted. Hint: Apply Lemma 10.12.16. \square
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