Exterior power of free module is free Let $M$ be a free module over a commutative ring $R$. Is $\wedge^k(M)$ then also free over $R$?
In one of the example of this, we know this is true when $M$ is finitely generated. Is it also true when it is not finitely generated? It seems that we could not write down the basis of $\wedge^k(M)$ anymore as in the finite case.
 A: Yes, and the proof is essentially identical to the finite case.  Let $B$ be a basis for $M$ and fix a total ordering $\leq$ on $B$.  Then the set of elements of the form $b_1\wedge\dots\wedge b_k$ where $b_1,\dots,b_k\in B$ and $b_1<\dots<b_k$ is a basis for $\bigwedge^kM$.  It is easy to see that these elements generate all of $\bigwedge^kM$, since you can use the alternating property to reorder any $k$-fold wedge product so that the basis vectors are in order (and eliminate any wedge products that repeat a basis vector).
Proving they are linearly independent takes more work and involves using some theory of signs of permutations (but again, this work is identical whether $B$ is finite or infinite).  For instance, you can let $N$ be the set of formal linear combinations of expressions $b_1\wedge\dots\wedge b_k$ where $b_1<\dots<b_k$ (so in $N$ you formally declare these expressions to be a basis). Then there is a multilinear map $f:M^k\to N$ which maps a tuple of basis elements $(b_1,\dots,b_k)$ to $0$ if there are any repetitions and to $\pm b_{\sigma(1)}\wedge \dots\wedge b_{\sigma(k)}$ where $\sigma$ is the permutation that puts $b_1,\dots,b_k$ in increasing order and the $\pm$ is chosen according to the sign of $\sigma$.  You can then check that $f$ is alternating and so induces a linear map $\bigwedge^kM\to N$.  Since our purported basis elements in $\bigwedge^kM$ map to linearly independent elements of $N$, they must indeed by linearly indepenednt in $\bigwedge^kM$.
