Let $A=k[x\_1,\cdots,x\_r]/I$ for some prime ideal $I$ and some field $k$. Consider the free $A$-module $A^n$. 


**Question 1.** Given an element $e\in A^n$, is there a method to tell whether $e$ can be admitted as an element of some $A$-basis for $A^n$? 


Obviously, the components of $e$ need to generate $A$, but I doubt this is enough. What about if instead of one element $e$ we have a set of $A$-linearly independent elements $e\_1, \dots, e\_k$ for $k < n$. 


**Question 2.** 
Is there a method to tell whether these can be admitted as basis elements? 


I'm looking for more practical methods than theoretical ones. For example, you can consider the determinant polynomial with one column equal to $e$ and equate it to $1$ and say whenever this has a solution in $A$, then $e$ is an element of some basis. But this method is not practical.


These were my main questions. 


Let $S$ be the set of elements in $A^n$ that can be part of some basis. Now define an equivalence relation on $S$ by setting two elements to be equivalent iff there is some basis containing both of them as basis elements. 


**Question 3.** After taking the quotient under this equivalence, is $S$ connected? If not, what are its connected components?



Basis for free modules over an affine domain
1) In the positive direction: If any entry in $e$ is an $(n-1)!$ power, then $e$ is an element of a basis. (More generally, it's enough to have $e=(z\_1^{m\_1},\ldots z\_n^{m\_n})$ with $(n-1)!$ dividing the product of the $m\_i$.) The case $n=3$ appears in a paper of Swan and Towber and is proved by explicitly writing down a basis for the complement. Explicitly, if $p\alpha+q\beta+r\gamma=1$ then


$$\left|\matrix{\alpha^2&\beta&\gamma\cr
\beta+r\alpha&-r^2+p r \beta& -p+q r-p q \beta\cr
\gamma-q\alpha&p+q r+p r \gamma&-q^2-p q \gamma\cr}\right| = 1$$


(I remember Dick Swan saying "I have no idea how Towber came up with this".) I think the general case is probably due to Mohan Kumar and/or Nori.)


1A) The divisibility condition in 1) is best possible in the following sense: Let $k$ be the complex numbers, $A=k[X\_1,Y\_1,\ldots X\_n,Y\_n]/(\Sigma X\_iY\_1-1)$, $Z\_j=X\_j+iY\_j$. Then if the row $e=(Z\_1^{m\_1},\ldots Z\_n^{m\_n})$ can be completed, it follows that $(n-1)!$ divides the product of the $m\_i$. This is also due to Swan and Towber.


2) Also in the positive direction, Bass's stable range theorem tells you that if $n-1\ge$ the projective stable range of $A$, then $e$ is part of a basis; and the projective stable range is bounded above by $1+dim(A)$.


3) Still in the positive direction one has (from Plumstead's proof of the Eisenbud-Evans conjectures) that if $A$ is $d$-dimensional and of the form $R[t]$, and if $n-1\ge d$, then $e$ is part of a basis. 


4) In the negative direction, at least if the characteristic of $k$ is not 2 and if 
$$A=k[X\_1,Y\_1,\ldots X\_n,Y\_n]/(\Sigma X\_iY\_i-1)$$
then $e=(X\_1,X\_2,\ldots X\_n)$ is **not** part of a basis; this was first proved by Michele Raynaud. The proof comes down to showing that the projection from $GL\_n$ to $GL\_n/GL\_{n-1}$ has no section; Raynaud shows that no section can be compatible with the Steenrod operations in etale cohomology. A later and very different proof is due to Mohan Kumar and Nori, using characteristic classes.


5) Of course one way to attack the problem is to try to show that more generally, **all** stably free $A$-modules (or even more generally, all projective $A$-modules) are free; Quillen, Suslin and Vaserstein all made this work for the case $I=0$.


