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? or what are its connected components?

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?

Let $A=k[x_1,\cdots,x_r]/I$ for some prime ideal $I$ and field $k$. Consider the free $A$-module $A^n$. 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 components of $e$ need to generate $A$ but I doubt this is enough. What about instead of one element $e$ we have a set of $A$ linear independent elements $e_1,\cdots , e_k$ for $k<n$. 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.

This was my main question. 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. After taking the quotient under this equivalence is $S$ connected? or what are its connected components?

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? or what are its connected components?

Let $A=k[x_1,\cdots,x_r]/I$ for some prime ideal $I$ and field $k$. Consider the free $A$-module $A^n$. 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 components of $e$ need to generate $A$ but I doubt this is enough. What about instead of one element $e$ we have a set of $A$ linear independent elements $e_1,\cdots , e_k$ for $k<n$. 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.

Let $A=k[x_1,\cdots,x_r]/I$ for some prime ideal $I$ and field $k$. Consider the free $A$-module $A^n$. 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 components of $e$ need to generate $A$ but I doubt this is enough. What about instead of one element $e$ we have a set of $A$ linear independent elements $e_1,\cdots , e_k$ for $k<n$. 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.

This was my main question. 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. After taking the quotient under this equivalence is $S$ connected? or what are its connected components?