I would like to ask the following. Given only the leading terms of an ideal $I$, namely the set $LT(I)$, is it possible to find a Groebner Basis of $I$? If not always, then when is it possible? We know that $\langle LT(I)\rangle = \langle LT(g_1), \dots, LT(g_n)\rangle$ for a Groebner basis $g_1, \dots, g_n$ but can we find exactly one basis $g_1, \dots, g_n$ given only the $LT(I)$?

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    $\begingroup$ The leading term ideals of $\left(x\right)$ and of $\left(x+1\right)$ in $k\left[x\right]$ are identical, yet the ideals are different and don't share a Gröbner basis. Do you mean something like "given both a Hilbert basis of $LT\left(I\right)$ and a set of generators of $I$, can we compute the Gröbner basis of $I$ by an algorithm which does not compare any element of the polynomial ring to $0$" ? This might be an interesting question. $\endgroup$ – darij grinberg Apr 7 '13 at 22:34
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    $\begingroup$ The leading term ideal $LT(I)$ does not generally determine the original ideal $I$ (think about principal ideals in one variable), while a Groebner basis does, so you can't find a Groebner basis for $I$ given just $LT(I)$. But maybe I am misunderstanding your question - what are you trying to do and why might it be possible? $\endgroup$ – Henry Cohn Apr 7 '13 at 22:36
  • $\begingroup$ (This question has a nice particular case when everything in sight is homogeneous of degree $1$: Can we obtain a row echelon form of a matrix over a field without comparing any elements of the field to $0$, if we know in advance which rows and columns the pivots of the row echelon form will be in? Of course, not comparing things to $0$ is not an artificial requirement; it roughly translates into "works over arbitrary commutative rings".) $\endgroup$ – darij grinberg Apr 7 '13 at 22:36
  • $\begingroup$ For example given only the $LT(I)$ of the ideal $ < 12x+8y-20, 3x+7y-10 > $ can we deduce that the Groebner basis is $\{x-1,y-1\}$ ? $\endgroup$ – Sln Apr 7 '13 at 23:09
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    $\begingroup$ Without further assumptions, no. As the other have already pointed out. The Gröbner basis of $I$ could also be $\langle x,y \rangle$ which gives the same $LT(I)$ $\endgroup$ – Thomas Kahle Apr 8 '13 at 6:03

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