I want to find an integral domain $R$ with ideal $I$ (considered as an $R$-module) such that $\bigwedge^k I\neq 0$ for all nonnegative integers $k$.
Dummit and Foote gave the example of $R=\mathbb Z[x, y], I=(x, y)$, and showed that $\bigwedge^2\neq 0$ by noting that $(ax+by, cx+dy)\mapsto (ad-bc) \ \text{mod} (x, y)$ is a bilinear alternating map that sends $(x, y)\mapsto 1$, and so the 2nd exterior power can't be the zero module.
I see how to extend this to ensure that the $k$th power does not equal $0$: just take a polynomial ring in $k$ variables, and the ideal of polynomials with no constant term. Write elements of $I^k$ as $k\times k$ matrices where a column represents an element of $I$, and the coefficients of the column represent the coefficients of each of the variables. Then the determinant function gives a multilinear alternating homormorphism that sends $(x_1,\ldots,x_k)\mapsto 1$, so the $k$th exterior power is nonzero.
But how do I extend this to make it work for ALL nonnegative integers $k$? I tried a polynomial ring with infinitely many variables, but then elements of the nonconstant ideal can have arbitrarily many variables, so I can't guarantee that each element of $I^k$ can be written as a column vector of size $k$. Is there a way to make this work?
One way to do this is to take the alternating function that first projects every element of $I$ onto the ideal generated by the first $k$ variables (i.e., mod out by all the other variables), and then take the determinant as in the $2\times 2$ case.