# If $M$ is a free $R$-module of rank $n$, then the $i^{\operatorname{th}}$ exterior algebra is free of dimension $n$ choose $i$.

I want to show that

If $$M$$ is a free $$R$$-module of rank $$n$$, for a unital, commutative ring $$R$$, then the $$i^{\operatorname{th}}$$ exterior power, $$\bigwedge^{i}(M)$$ is free of dimension $$\binom{n}{i}$$.

My reasoning:

We have $$M \cong \bigoplus_{k = 1}^{n} R.$$

We also have that $$\bigwedge^i(M) := \mathcal{T}^i(M)/\mathcal{A}^i(M)$$ where $$\mathcal{T}^i(M) = M^{\otimes i}$$ and $$\mathcal{A}^i(M) := \langle m_1 \otimes \cdots \otimes m_\ell \otimes\cdots \otimes \cdots \otimes m_j \otimes \cdots \otimes m_i \;|\; m_p \in M, \ell \neq j \ \text{so that}\ m_\ell = m_j \ \text{for some} \ \ell,j\rangle.$$

Now we take not of the fact that we have a surjective $$R$$-module homomorphism, that is the projection onto the quotient, i.e. $$f:\mathcal{T}^i(M) = M^{\otimes i} \twoheadrightarrow \mathcal{T}^i(M)/\mathcal{A}^i(M) = \bigwedge^i(M)$$ defined explicitly by $$m_1 \otimes \cdots \otimes m_i \longmapsto m_1 \wedge \cdots \wedge m_i.$$

Now, assume that we have a basis $$\mathcal{B} := \{m'_1,\ldots,m'_n\} \subset M.$$

Then every simple tensor $$m_1 \otimes \cdots \otimes m_i$$ can be written as $$\sum_{k_1 = 1}^{n}r_{k_1} m'_{k_1} \otimes \cdots \otimes \sum_{k_i = 1}^{n} r_{k_i}m'_{k_i}.$$ Now, we can use the multi-additivity of $$i$$-tensors to find that under $$f$$, we will see that only the tensors where $$k_1 \neq \cdots \neq k_i$$ is true, will be sent to a non-zero element.

We thus get that $$f(m_1 \otimes \cdots \otimes m_i)$$$$= f\Big(\sum_{k_1 = 1}^{n} r_{k_1}m'_{k_1} \otimes \cdots \otimes \sum_{k_i = 1}^{n} r_{k_i} m'_{k_i} \Big) = \sum_{\sigma \in S_i} \operatorname{sgn}(\sigma)(r_{\sigma(k_1)} \cdots r_{\sigma(k_i)})m'_1 \wedge \cdots \wedge m'_i.$$

Now since every element in $$M^{\otimes i}$$ can be written as a finite sum $$\sum_{\gamma = 1}^{N} m^{\gamma}_1 \otimes \cdots \otimes m^{\gamma}_i$$ who:s image will be $$\sum_{\gamma = 1}^{N}\Bigg(\sum_{\sigma \in S_i}\operatorname{sgn}(\sigma)(r_{\sigma(k^{\gamma}_1)} \cdots r_{\sigma(k^{\gamma}_i)})m^{\gamma '}_1 \wedge \cdots \wedge m^{\gamma '}_i\Bigg)$$ and we have a surjection, we can conclude that every element in $$\bigwedge^i(M)$$ is on (this not so nice) form.

Now, we notice that if $$n > i$$, then we can choose $$\binom{n}{i}$$ basis elements for $$\bigwedge^i(M).$$

To conclude, we have that $$\{m'_{k_1} \wedge \cdots \wedge m'_{k_i}\;|\; 1 \leq k_1 < \ldots < k_i \leq n\}$$ for $$k_j \in \{1,\ldots,n\}$$ generates $$\bigwedge^i(M).$$

I have not shown linear independence here, though. Is this idea correct? If not, what is wrong?

If it is correct, is there some nicer (notationally or conceptual) way to show this?

Edit: Thank you Qiachu Yuan. Yes, I realized that $$1.$$ might work, as you said. I think I encountered a similar exercise in differential geometry.

Am my interpretation correct, that what you mean by $$1.$$ above is the following:

If $$\{m'_1,\ldots,m'_n\}$$ is a basis for $$M$$, then it follows that $$\{m^{* '}_{1},\ldots,m^{* '}_{i}\}$$ is a basis for $$M^{*}$$. Then, assume we have an element $$\omega \in \bigwedge^i(M)$$ such that $$\omega = \sum_{J} r_J m^{J '}$$ for $$m^{J'} = m^{'}_{j_1} \wedge \cdots \wedge m^{'}_{j_i}$$ and $$J$$ is the set of all increasing indices $$(j_1,\ldots,j_i)$$ such that $$1 \leq j_1 < \cdots j_i \leq n.$$

Now, if $$\omega = 0$$ we will find that $$\omega(m^{*'}_{j_1},\ldots,m^{*'}_{j_i}) = r_{J'} = 0$$ for that particular choice of $$(j_1,\ldots,j_i)$$ Feeding $$\omega$$ with all possible choices of $$m^{*'}_{j_1},\ldots,m^{*'}_{j_i}$$ shows us that all coefficients $$r_{J} \equiv 0.$$

Comment: here we define $$m^{*’}_j$$ as the element that gives back $$1$$ when given $$m^{’}_j$$ and zero otherwise.

• You have the right idea but notationally it's cleaner to first establish that the tensor products of basis vectors form a basis of the tensorial power, then work from there to reason about the induced basis of the exterior power. Qiaochu's answer addresses linear independence. Commented Oct 10, 2023 at 20:06
• @Ben: I think your argument is in the right direction but could use a bit more detail on the last step, since it's the most important one. In particular I think it's worth giving more detail about what is meant exactly by "feeding $\omega$ with..." Commented Oct 10, 2023 at 20:49

1. You can write down a nondegenerate bilinear pairing $$\Lambda^k(M) \times \Lambda^k(M^{\ast}) \to R$$ where $$M^{\ast} = \text{Hom}_R(M, R)$$ is the $$R$$-linear dual. This will let you show linear independence by acting via wedges of the dual basis in $$M^{\ast}$$.
2. You can argue that the exterior algebra takes direct sums to tensor products in the sense that $$\Lambda^{\bullet}(M \oplus N) \cong \Lambda^{\bullet}(M) \otimes \Lambda^{\bullet}(N)$$. Abstractly this follows from the fact that $$\Lambda^{\bullet}(-)$$ is the free graded-commutative algebra (the tensor product here has to be understood in the graded sense, with Koszul signs). Then we reduce to the fact that tensor products and direct sums of free modules are free.
• Ah, maybe the standard argument is an induction on $i$ via acting by derivations coming from $M^{\ast}$? Commented Oct 10, 2023 at 20:00
• Hi @Qiaochu Yuan. See my edit for a comment on what you wrote (specifically point $1$.). Do you agree? Best, Ben Commented Oct 10, 2023 at 20:31
• You can also reduce to proving that $\Lambda^n M$ is free of rank 1 (generated by the wedge product of the generators of $M$). Then taking the wedge product of $e_{i_1} \wedge \cdots \wedge e_{i_k}$ with the wedge product of the other generators of $M$ gives $\pm$ the generator on that element of the claimed basis of $\Lambda^k M$, and it takes the other elements to 0. And showing that $\Lambda^n M$ is as claimed amounts to defining the determinant function on $M^{\otimes n}$. Commented Oct 10, 2023 at 20:52