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Added the groebner-bases tag as the commutative-algebra tag alone is fairly general. Also did some formatting such as replacing < with \langle.
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edit approved Nov 23, 2014 at 6:57
Michael Albanese
edit approved Nov 23, 2014 at 6:57
Michael Albanese
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I would like to ask the following. Given only the leading terms of an ideal I$I$, namely the set $LT(I),$$LT(I)$, is it possible to find a Groebner Basis of I$I$? If not always, then when is it possible? We We know that $< LT(I) > = < LT(g_1),...,LT(g_n)> $$\langle LT(I)\rangle = \langle LT(g_1), \dots, LT(g_n)\rangle$ for a GrobnerGroebner basis $g_1,...,g_n$$g_1, \dots, g_n$ but can we find exactly one basis $ g_1,...,g_n $$g_1, \dots, g_n$ given only the $LT(I)$  ?

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 $< LT(I) > = < LT(g_1),...,LT(g_n)> $ for a Grobner basis $g_1,...,g_n$ but can we find exactly one basis $ g_1,...,g_n $ given only the $LT(I)$  ?

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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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 $< LT(I) > = < LT(g_1),...,LT(g_n)> $ for a Grobner basis $g_1,...,g_n$ but can we find exactly one basis $ g_1,...,g_n $ given only the $LT(I)$ ?

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 $< LT(I) > = < LT(g_1),...,LT(g_n)> $ for a Grobner basis $g_1,...,g_n$ but can we find exactly one $ g_1,...,g_n $ given only the $LT(I)$ ?

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 $< LT(I) > = < LT(g_1),...,LT(g_n)> $ for a Grobner basis $g_1,...,g_n$ but can we find exactly one basis $ g_1,...,g_n $ given only the $LT(I)$ ?

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Sln
Sln
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Can we find a Groebner Basis?

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 $< LT(I) > = < LT(g_1),...,LT(g_n)> $ for a Grobner basis $g_1,...,g_n$ but can we find exactly one $ g_1,...,g_n $ given only the $LT(I)$ ?