$0$-th exterior power, empty product of modules and their tensor product

Question

$\def\finiteprod#1#2{#1_{1}\times#1_{2}\times\dots\times#1_{#2}}$In Lang's algebra he defines the tensor product as a universal object in the category of multilinear maps from $\finiteprod En$ where all of these are $R$-modules for some commutative ring $R$. He then defines the k-th exterior power $\bigwedge^k\left(E\right)$ to be a quotient module of the tensor product.

My question is how do I make sense of this when this product of modules in the empty product.

The reason I am asking is that I am reading about exterior algebra and then $0$-th exterior algebra $\bigwedge^{0}\left(E\right)$ should be (if the definition is still consistent for the $0$-th exterior algebra) a quotient module of the tensor product of an empty product of modules. Now as far as I understand the emppty product of modules is the zero module, since the empty product should be an identity element with respect to to the operation (in this case the direct product), and taking the product of a module with the $0$ module (which for finitely many is equivalent to the direct sum) would give the same module back. And so I conclude that the empty product of modules is the $0$ module.

So the tensor product should be a universal object in that category (more correctly it is the codomain of the multilinear map and not the map itself). Now at first I made a mistake here but one of the comments helped me see this implies that the tensor product is itself the zero module.

In the following way: let $T$ be the tensor product, then given a map $E^{0}\overset{0}{\to}F$ to some module $F$ (which is the 0 map), we must have a unique linear map $f$ from $T\overset{f}{\to}F$ such that $\phi\circ f=0$ where $\phi$ is the map into the tensor product (what we called the universal object of the category).

We take $F=T$ and see both the identity map and the zero map fulfill the requirements for $f$ but must be the same by uniqueness, so $T$ is the zero module.

Okay the tensor product is 0, so I do not see how $\bigwedge^{0}\left(E\right)$ which is a quotient of the tesor product can be $R$.

So is the fact that $\bigwedge^{0}\left(E\right)=R$ simply a definition?

Would appreciate any clarification. Thank you

Answer 1

You just have to work with the definitions (i.e. universal properties) in order to answer this question. It is not an extra convention or something like that (unfortunately, many mathematicians believe this).

If $(E_i)_{i \in I}$ is a family of $R$-modules with underlying sets $|E_i|$, and $F$ is some $R$-module, a map $\prod_i |E_i| \to |F|$ is called multilinear if, for every $i \in I$ and every $e \in \prod_{j \neq i} E_j$, the induced map (given by fixing $e$) $E_i \to F$ is linear. We obtain a functor $\mathrm{Mult}((E_i)_{i \in I},-)$ which is represented by the tensor product $\bigotimes_{i \in I} E_i$. By the way, we don't need that $I$ is finite here. Now take $I=\emptyset$. Then $\prod_i |E_i|$ is the empty cartesian product, i.e. the terminal object of $\mathsf{Set}$, i.e. just a point $\{\star\}$. A map $\{\star\} \to |F|$ is just an element of $|F|$. Since $I$ is empty, the multilinear condition above is empty! It follows that every map is already multilinear. Hence, $\mathrm{Mult}((E_i)_{i \in I},F) \cong |F| \cong \mathrm{Hom}(R,F)$, i.e. $\bigotimes_{i \in I} E_i = R$ if $I=\emptyset$.

With the above notations, and $E:=E_i$ for all $i$, a map $\prod_i |E| \to |F|$ is alternating if it is multilinear and it vanishes on every $e$ such that $e_i=e_j$ for some $i \neq j$ in $I$. But if $I=\emptyset$ this condition is again empty! Hence, the representing object $\wedge^{i \in I} E$ is again $R$ when $I=\emptyset$.

Answer 2

$E^{\otimes 0}$ should be the identity of $\otimes$, i.e. the base ring $R$.

Answer 3

First, let's settle the issue of the empty tensor product. Consider the identity map $id: W\to W$ for an arbitrary $R$-module $W$. This is an $R$-balanced map out of the collection $\{W\}\cup \emptyset$ and so descends to a map $W\otimes E\to W$. But we know that $(W,id)$ satisfies the universal property of the tensor product of $W$, so $W\cong W\otimes E$ canonically for every $R$-module $W$. Therefore as we have $-\otimes E\cong Id_{R-mod}\cong -\otimes R$ as functors which implies that $E\cong R$. Therefore the empty tensor product must be exactly $R$.

---

Here's an alternate definition of $\Lambda^n(V)$: $\Lambda^n(V)$ is the $n$-graded piece of the exterior algebra of $V$.

To understand what this means, first, we construct the tensor algebra for an arbitrary $R-R$ bimodule $V$: Let $T(V)=R\oplus V\oplus (V\otimes_R V)\oplus V^{\otimes 3}\oplus \cdots$ and define a map $T(V)\times T(V) \to T(V)$ by concatenation, ie $(x,y)\mapsto x\otimes y$, which descends to the tensor product $T(V)\otimes T(V)$. Checking the axioms for an $R$-algebra, we see that this map and the canonical inclusion map $R\hookrightarrow T(V)$ make $T(V)$ into an $R$-algebra. $T(V)$ also comes with a canonical grading, where the degree 0 piece is $R$, the degree 1 piece is $V$, the degree 2 piece is $V^{\otimes 2}$, etc.

Next, we define the exterior algebra $\Lambda(V)$ as the quotient of $T(V)$ by the ideal generated by the monomials $\{v\otimes v| v\in V\}$. $\Lambda(V)$ is also naturally graded, as the ideal generated by the monomials $\{v\otimes v| v\in V\}$ is homogeneous. We then say that $\Lambda^n(V)$ is the $n$-th graded piece of $\Lambda(V)$. Since the $0$-th graded piece of $\Lambda(V)$ is just $R$ for any $R$-module $V$, we have that $\Lambda^0(V)=R$ for all $R$-modules $V$.

Answer 4

This can be shown using the universal property that defines an $k$-th exterior power.

Universal Property. Suppose $N$ is a $R$-module. The $k$-th exterior power $(\bigwedge^k N, \wedge_k )$ is up to isomorphism uniquely determined by the following universal property, for any $R$-module $M$ and any alternating $k$-linear mapping $\pmb{\varphi}: N^k \to M$, there exists a unique $R$-linear mapping $\tilde{\pmb{\varphi}}:\bigwedge^k N \to M $ such that $\pmb{\varphi} = \tilde{\pmb{\varphi}} \circ \wedge_k$.

Preliminaries. Suppose $N, M$ are $R$-modules. Then $N^0=R^0 = \{0\}$ and every function from $\{0\}$ to $M$ (resp. $R$) is an alternating $0$-linear mapping (the defining conditions are emptily fulfilled) therefore $$ \mathbf{Alt}^0_R(N,M) = \{ \hat{\pmb{m}}:\{0\}\to M, 0\mapsto \pmb{m} | \pmb{m}\in M\}\cong M,$$

$$ \mathbf{Alt}^0_R(N,R) = \{ \hat{r}:\{0\}\to R, 0\mapsto r | r\in R\} \cong R.$$

On the other hand, a $R$-linear map from $R$ to $W$ is given by $\pmb{m}\in M$ i.e.

$$ \mathbf{Hom}_R(R,M) = \{ \tilde{\pmb{m}}: R \to M, r \mapsto r\pmb{m} | \pmb{m}\in M\} \cong M.$$

Proposition. Suppose $N$ is a $R$-module. Then $(R, \hat{1})$ has the universal property of the $0$-th exterior power of $N$, i.e $\bigwedge^0 N \cong R$.

Proof. For every $\pmb{m} \in M$, we have $\hat{\pmb{m}} = \tilde{\pmb{m}}\circ \hat{1}$ i.e. $$ (\tilde{\pmb{m}}\circ\hat{1})(0) = \tilde{\pmb{m}}(\hat{1}(0))= \tilde{\pmb{m}}(1) = 1\pmb{m} =\pmb{m} =\hat{\pmb{m}}(0).$$