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

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

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

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

and similarly

**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

where the pure tensor $m_1 \otimes \ldots \otimes m_{n - 2}$ in the first summand maps to

and $m_1 \otimes \ldots \otimes m_{n - 2}$ in the second summand maps to

There is also a canonical exact sequence

where the pure tensor $m_1 \otimes \ldots \otimes m_{n - 2}$ maps to

**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. 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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