I don't think that this statement is true. Take for example, $I = (x^2 + y, x^2 y + 1)$. Clearly $I$ $\subseteq$ $k[x,y]$ and can have a term ordering yet $I$ is not a Grobner basis and since it's not a Grobner it can't have a Grobner basis with respect to a term order $\leq$. Anyone disagree?
Is it true that when $ I$ has a term ordering on $R$ where $I$ is an ideal of $R$ then $I$ has a Grobner basis with respect to $\leq$
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abstract-algebra
ring-theory
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0This question doesn't make sense to me. A term ordering is attached to the polynomial ring $R,$ not to an ideal. Given a term ordering on $R$ and an ideal $I\subseteq R$ one can certainly compute a groebner basis for $I$ with respect to the given term ordering. – 2012-11-06