LEC ALG4 15

Exterior algebra

Definition 427.1(Definition).

Let $R$ be commutative, $M$ be an $R$-module. The exterior algebra $\wedge M$ of $M$ is $T(M)/J$, where $J$ is the ideal of $T(M)$ generated by $\{x\otimes x:x\in M\}$.

Observe that $\wedge M$ is a graded algebra, with ${\wedge }^{0}M\cong R$, ${\wedge }^{1}M\cong M$. For $n\ge 0$, ${\wedge }^{n}M$ is called the $n$th exterior power of $M$. The image of $x_1\otimes \cdots \otimes x_n$ in $\wedge M$ is denoted by $x_1\wedge \cdots \wedge x_n$.

Proposition 427.2(Proposition).

$x\wedge y=-y\wedge x$ for all $x,y\in M$.

Proof.

$0=(x+y)\wedge (x+y)=x\wedge y+y\wedge x$.□

96459f

Prp 2 is only useful when $1\ne -1$ in $R$, i.e, $2\ne 0$, i.e, $R$ is not an ${\mathbb{F}}_2$-algebra.

Remark 427.3(Remark).

If $2\ne 0$ in $R$, then $J$ can also be generated by $\{x\otimes y+y\otimes x:x,y\in M\}$.

Note that $x\otimes y+y\otimes x=(x+y)\otimes (x+y)-x\otimes x-y\otimes y$. …

Proposition 427.4(Proposition).

Suppose that $M$ is generated by $\{x_{\lambda }:\lambda \in \Lambda \}$. Then for all $n⩾0$, ${\Lambda }^{n}M$ is generated by $\{x_{{\lambda }_1}\wedge \cdots \wedge x_{{\lambda }_n}:{\lambda }_1,\dots ,{\lambda }_n\text{ distinct}\}$.

Proof.

Note that $\{x_{{\lambda }_1}\wedge \cdots \wedge x_{{\lambda }_n}:{\lambda }_1,\dots ,{\lambda }_n\in \Lambda \}$ generates $T^{n}M$ and hence ${\wedge }^{n}M$.

The proof proceeds as you’d expect.□

Corollary 427.5(Corollary).

If $M$ can be generated by $m$ elements (as an $R$-module), then ${\Lambda }^{n}M=0$ for all $n>m$.

Proposition 427.6(Proposition).

Let $A\in {\text{AAlg}}_R$, and $\phi \in {\text{Hom}}_{R\text{-Mod}}(M,A)$ such that $(\phi (x){)}^2=0$ for all $x\in M$. Then, there exists unique $\Phi \in {\text{Hom}}_{R-\text{AAlg}}(\wedge M,A)$ such that

commutes.

[!Proof]-
Similar to the case of $\text{Sym}(M)$.

[!Definition]
Say that an $R$-multilinear map $f:M\times \cdots \times M\to N$, where $N\in R\text{-Mod}$, is alternating if …
The set of alternating maps $M\times \cdots \times M\to N$ is denoted by ${\text{Alt}}_R(M\times \cdots \times M,N)$,

[!Proposition]
${\text{Hom}}_R({\wedge }^{n}M,N)={\text{Alt}}_R(M\times \cdots \times M,N)$.

[!Proof]-

sketch

[!Proposition]
$M={⨁}_{i=1}^{m}R{x}_i$. Then ${\wedge }^{n}M$ is a free $R$-module with basis

$$\{x_{i_1}\wedge \cdots \wedge x_{i_n}:1⩽i_1<\cdots <i_n⩽m\}.$$