I'm interested in Buchberger's criterion for determining if G={g\_1,....,g\_n} is a Grobner basis for the ideal it generates. In the procedure, I consider the S-polynomial S(g\_i,g\_j) and check if it has remainder zero upon division by G *in any order*.


My question is if Buchberger's criterion holds if one use a different order of G for each i,j rather than a fixed but arbitrary choice of order of G for all i,j.


In other words, could G be falsely Grobner even if S(g\_i,g\_j) has remainder 0 by dividing
by G in different orders depending on (i,j)? 



Buchberger algorithm question
Indeed, it is enough to check that each $S(g\_i,g\_j)$ can be reduced to zero in *some* way. Indeed, if you trace carefully the proof of Buchberger's criterion, you will see that one only needs the *existence* of a representation of $S(g\_i,g\_j)=\sum\_{k}a\_{ijk}g\_k$ with each leading term of $a\_{ijk}g\_k$ being less than or equal to the leading term of $S(g\_i,g\_j)$, which is precisely what you can get from the *existence* of long division that produces remainder $0$.


