---- > [!definition] Definition. ([[exterior power]]) > Let $R$ be a [[ring]] and let $M$ be an $R$-[[module]]. Denote by $\mathbb{T}^{\ell}(M)$ the $\ell$th [[tensor product of modules|tensor power]] of $M$. > > The **$\ell$th exterior power of $M$** is a new $R$-[[module]] $\Lambda^{\ell}(M)$ satisfying the [[universal property]] that every $R$-[[alternating multilinear map]] $\varphi:M^{\ell} \to P$ factors through $\Lambda^{\ell}(M)$ via a unique $R$-[[linear map]] $\overline{\varphi}$: > > ```tikz > \usepackage{tikz-cd} > \usepackage{amsmath} > \begin{document} > % https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZABgBpiBdUkANwEMAbAVxiRAFkA9AHW5gYYgAvqXSZc+QigCM5KrUYs2ABWGiQGbHgJEy0+fWatEIXgBk6AWwBGUOjz4CABAAp2ASmHyYUAObwiUAAzACcISyQAJmocCCRZBSM2XhgADyw4HDgnAEInXggaGBCGLDAYYF56ELQACywhEGoGOmt+ZXFtKRAQrF9anDVgsIjEMhBYqOpDJRNeAHcffyaQFraGDq1JNl7+wZFh8KRxycQEmeNTbmq6rC8hIA > \begin{tikzcd} > M^\ell \arrow[d, "\wedge"'] \arrow[r, "\varphi"] & P \\ > \Lambda^\ell (M) \arrow[ru, "\exists ! \overline{\varphi}"'] & > \end{tikzcd} > \end{document} > ``` > where $\wedge$ is alternating. > > This defines $\Lambda^{\ell}(M)$ [[terminal objects are unique up to a unique isomorphism|up to isomorphism]], if it exists. > > Indeed, exterior powers *do* exist in $R$-$\mathsf{Mod}$. Let $W \subset \mathbb{T}^{\ell}(M)$ be the [[submodule]] [[submodule generated by a subset|generated by]] all pure tensors $m_{1} \otimes \dots \otimes m_{\ell}$ such that $m_{i}=m_{j}$ for some $i \neq j$, then $\Lambda^{\ell}(M) \cong \frac{\mathbb{T}^{\ell}(M)}{W}.$ > Here the map $\wedge$ is of course the composition $M^{\ell} \xrightarrow{\otimes} \mathbb{T}^{\ell}(M) \xrightarrow{\pi} \frac{\mathbb{T}^{\ell}(M)}{W}$ > and $W$ is the smallest [[submodule]] making the multilinear map $\otimes$ alternating. > [!definition] Definition. (Pure alternating tensor) > The image of a [[tensor product of modules|pure tensor]] $m_{1} \otimes \dots \otimes m_{\ell}$ under $\pi$ is denoted $m_{1} \wedge \dots \wedge m_{\ell}$ and is called a **pure alternating tensor**. > As was the case in [[tensor product of modules]], not every element in $\Lambda_{R}^{\ell}(M)$ is a pure alternating tensor, but every element is certainly a [[linear combination]] of pure alternating tensors. In fact, we can say more [[alternating multilinear map|by definition]], if any two of the $m_{i}s coincide, then $m_{1} \wedge \dots \wedge m_{\ell}$, and $m_{2} \wedge m_{1} \wedge m_{3}=-m_{1} \wedge m_{2} \wedge m_{3}$ (for example); so $\Lambda_{R}^{\ell}(M)$ is in fact generated [[set of all ascending k-tuples from 1 to n|by the]] $r \choose \ell$ elements $\{ e_{i_{1}} \wedge \dots \wedge e_{i_{\ell}} : 1 \leq i_{1} < \dots < i_{\ell} \leq r \}.$ In particular, if $M$ is [[submodule generated by a subset|finitely generated]] then $\Lambda_{R}^{\ell}(M)=0$ for $\ell \gg0$. If $M$ is a [[free module]], then $\Lambda_{R}^{\ell}(M)$ is a [[free module|free]] as well, with [[rank of a free module|rank]] $r \choose \ell$. > [!intuition] > The idea for defining $\Lambda^{\ell}(M)$ is to look at the [[multilinear map]] $\otimes:M^{\ell} \to \mathbb{T}^{\ell}(M)$ and 'make it [[alternating multilinear map|alternate]]'. To do this, one checks: who would get killed if $\otimes$ *did* alternate? The answer is pure tensors of the form $\otimes(m_{1},\dots,m_{\ell})=m_{1} \otimes \dots \otimes m_{\ell}$ satisfying $m_{i}=m_{j}$ for some $i \neq j$, and any [[linear combination]] thereof. And so we [[kernel iff submodule|kill]] precisely the [[submodule]] $W$ generated by such elements. ^intuition > [!justification] > We have to show that the [[universal property]] is indeed satisfied by $\Lambda^{\ell}(M)$. So let $\varphi:M^{\ell} \to P$ be an arbitrary [[alternating multilinear map]]. > Commutativity of the diagram enforces the definition$\overline{\varphi}( m_{1} \wedge \dots \wedge m_{\ell}) = \varphi(m_{1}, \dots, m_{\ell})$ so if $\overline{\varphi}$ is [[well-defined]] then it is certainly unique. First we show that $\overline{\varphi}$, if it exists, is linear: Let $a, b \in R$ and $[x], [y] \in \Lambda^{\ell}(M)$. We can write: $[x] = [\sum_i \alpha_i (m_{i1} \otimes \dots \otimes m_{i\ell})]$, $[y] = [\sum_j \beta_j (n_{j1} \otimes \dots \otimes n_{j\ell})]$. Then: $\overline{\varphi}(a[x] + b[y]) = \overline{\varphi}([\sum_i a\alpha_i (m_{i1} \otimes \dots \otimes m_{i\ell}) + \sum_j b\beta_j (n_{j1} \otimes \dots \otimes n_{j\ell})])$ $= \sum_i a\alpha_i \varphi(m_{i1}, \dots, m_{i\ell}) + \sum_j b\beta_j \varphi(n_{j1}, \dots, n_{j\ell})$ $= a\sum_i \alpha_i \varphi(m_{i1}, \dots, m_{i\ell}) + b\sum_j \beta_j \varphi(n_{j1}, \dots, n_{j\ell})$ $= a\overline{\varphi}([x]) + b\overline{\varphi}([y])$. > To check well-definition, let $[x] = x + W$ and $[y]=y+W$ be [[coset|cosets]] of $\mathbb{T}^{\ell}(M)$ in $\Lambda^{\ell}(M)$ such that $[x]=[y]$, or equivalently $[x-y]=W$. We need to show that $\overline{\varphi}([x]) = \overline{\varphi}([y])$, or equivalently (using linearity) $\overline{\varphi}([x-y] )=0$. But this is immediate because $[x-y]=W$ and $\overline{\varphi}(W)=0$. --- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```