Suppose I have a coordinate system $t_1,\ldots t_N$ with a lexicographical ordering. Let LT denote choosing the lowest term of a polynomial with respect to this ordering. e.g. LT$(t_1 + t_2)=t_2$.

Given a finite dimensional vector space V, define LT(V):=span$_\mathbb{C}\{LT(w)|w\in V\}$.

I want to find a basis $\{v_1,\ldots , v_n\}$ for V such that LT(V)=span$_\mathbb{C}\{LT(v_1),LT(v_2),\ldots ,LT(v_n)\}$.

I have never worked with Gröbner bases and know very little about them, but this question looks similar to them I think, only here I'm talking about subspaces instead of ideals.

Is there a well-known method for finding such a basis? I've been searching, but haven't found anything yet, so I thought I'd ask on here.

For example, by an algorithm I know $\{1,t_1+t_3,t_2,t_1t_2,t_1t_2(t_1+t_3),t_2(t_1+t_3),t_2(t_2t_3),t_1t_2(t_2t_3)\}$ forms a basis for the space of sections $H^0(X,L)$ of of particular variety $X$ and a line bundle $L$ over it. And for the vector spaces I'm working with I already know $dim(LT(H^0(X,L))=dim(H^0(X,L))$, so I know $\{1,t_3,t_2,t_1t_2,t_1t_2t_3,t_2t_3,t_2(t_2t_3),t_1t_2(t_2t_3)\}$ forms a basis for $LT(H^0(X,L)$.

But in general, if the vector space I begin with is n-dimensional, it may not be the case that the basis I choose has n distinct lowest terms, so I'm looking for a way to find a basis with n distinct lowest terms.

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    $\begingroup$ V is a subspace of the vector space on which your coordinate system is defined? Then a basis $\left(v_1, v_2, \ldots, v_n\right)$ has the property that you desire if and only if it is a permutation of the list of rows of a matrix in row-echelon form. I am fairly sure that this is well-known; many people motivate Groebner basis theory as a noncommutative analogue of classical linear algebra (Groebner basis <~~> row echelon form; reduced Groebner basis <~~> reduced row echelon form). $\endgroup$ Oct 7 '14 at 21:33

Darij is right. Computing Groebner basis for a linear ideal is the same as computing the row echelon form (this can be seen from the Buchberger algorithm). And I think he meant "nonlinear" instead of "noncommutative".

  • $\begingroup$ Thank you for your responses, but in my set up I don't have a linear ideal; I only have a vector space, so I don't have any kind of multiplication defined (unless I'm misunderstanding what you mean by a linear ideal). I tried to clarify my question above with an example. $\endgroup$
    – Lauren
    Oct 8 '14 at 15:40
  • $\begingroup$ For your example, you can replace each monomial that appears with a new variable, then you'll get linear forms. The monomial ordering gives you an ordering of the new variables. Running the Buchberger algorithm on this set of linear forms is the same as computing the row echelon form of your original elements (in the span of the monomials that appear). $\endgroup$
    – Milan
    Oct 9 '14 at 1:12

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