## Exterior powers of subdomains of a field of fractions

Let $R$ be an integral domain, and let $F$ be its field of fractions.

1. Considering $F$ as an $R$-module in the usual way, prove that $\bigwedge^2(F) = 0$.
2. Let $I \subseteq F$ be a submodule. Prove that $\bigwedge^k(I)$ is torsion for all $k \geq 2$.
3. Exhibit an integral domain $R$ and an $R$-submodule $I \subseteq F$ such that $\bigwedge^k(I) \neq 0$ for all $k \geq 0$.

Let $\frac{a}{b} \otimes \frac{c}{d} \in \mathcal{T}^2(F)$ be a simple tensor. Note that $\frac{a}{b} \otimes \frac{c}{d} = \frac{ad}{bd} \otimes \frac{cb}{bd}$ $= abcd(\frac{1}{bd} \otimes \frac{1}{bd}) \in \mathcal{A}^2(F)$. In particular, we have $\frac{a}{b} \wedge \frac{c}{d} = 0$ in $\bigwedge^2(I)$.

Suppose $\frac{a_1}{b_1} \wedge \frac{a_2}{b_2} \wedge \cdots \wedge \frac{a_k}{b_k} \in \bigwedge^k(I)$ be nonzero; then the $a_i$ are nonzero, and certainly the $b_i$ are nonzero. Now $a_1a_2b_1b_2 \neq 0$ in $R$, and evidently $a_1a_2b_1b_2(\frac{a_1}{b_1} \wedge \frac{a_2}{b_2} \wedge \cdots \wedge \frac{a_k}{b_k}) = \frac{a_1a_2}{1} \wedge \frac{a_1a_2}{1} \wedge \cdots \wedge \frac{a_k}{b_k} = 0$. So every element of $\bigwedge^k(I)$ is torsion, and thus $\bigwedge^k(I)$ is torsion as an $R$-module.

Now consider $R = \mathbb{Z}[x_1,\ldots,x_n]$, and let $I = (x_1,\ldots,x_n)$. We claim that if $\sum \alpha_ix_i = \sum \beta_ix_i \in I$, then there exist $h_i \in I$ such that $\alpha_i - \beta_i = h_i$. To see this, choose some $j$ and consider $\alpha_jx_j - \beta_jx_j = \sum_{i \neq j} (\beta_i - \alpha_i)x_i$. Since $R$ is a domain, $x_j$ divides the right hand side of this equation; say $\sum_{i \neq j} (\beta_i - \alpha_i)x_i = x_jh_j$. Note that every term on the left hand side is divisible by some $x_i$ other than $x_j$. Hence every term in $h_j$ is divisible by some $x_i$ other than $x_j$. In particular, $h_j \in I$, and $\alpha_i - \beta_i = h_i$.

Now consider the elements of $\prod^k I$ as “column vectors”- that is, write $\sum \alpha_i x_i$ as $[\alpha_1\ \alpha_2\ \ldots\ \alpha_n]^\mathsf{T}$. Note that this does not give a unique representation of the elements of $I$. However, if two column vectors $A$ and $B$ represent the same element in $I$, then there exists a third column vector $H$ such that $A = B+H$, and moreover the entries of $H$ are in $I$.

Now write the elements of $R^k$ as (square) matrices. Let us consider the determinant of such a matrix $A$, as an element of $R$, reduced mod $I$. This is an alternating bilinear map on the set of matrices over $R$; we claim that this is also well-defined up to our nonunique identification of matrices over $R$ with elements of $I^k$. To that end, suppose matrices $A$ and $B$ represent the same element of $I^k$; then we have a matrix $H$ such that $A = B+H$. Consider the determinant of both sides mod $I$. Using the combinatorial formula for computing determinants, we have $\mathsf{det}(B+H) = \sum_{\sigma \in S_n} \epsilon(\sigma) \prod (\beta_{\sigma(i),i} + h_{\sigma(i),i})$. Note that each $h_{i,j}$ is divisible by some $x_i$, and so goes to zero in the quotient $R/I$. So in fact $\mathsf{det}(A) = \mathsf{det}(B+H) \equiv \mathsf{det}(B) \mod\ I$; thus the map $\mathsf{det} : I^k \rightarrow R/I$ is a well-defined alternating bilinear map. Since $\mathsf{det}(x_1 \otimes \cdots \otimes x_n) = 1$ is nonzero, this map is nontrivial. Thus $\bigwedge^k(I) \neq 0$ for all $k$.