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If have computed this Gröbner Basis with Buchberger's algorithm for Degree-Lexicographic-Ordering:

$$\{ x^²y+x+1,xy^2+y+1,x-y \} $$

I want to to transform it into a unique representation form called Reduced Gröbner Basis. Therefore I remove the leading powers which are divided by other leading powers. In this case $x \mid x^2y$ and $x \mid xy^2$ so the resulting base is

$$\{x-y\}.$$ This is not the reduced base yet but a form called Minimal Gröber Basis.

I checked the result with Singular. The reduced base computed by Singular ist:

$$\{x-y, y^3+y+1\}$$

Nevertheless

$$ \langle x-y, y^3+y+1 \rangle \neq \langle x-y\rangle.$$

Any ideas?

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You do not have a Groebner Basis. The S-polynomial of $x-y$, $xy^2+y+1$ is $y^3+y+1$

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You cannot blindly remove elements from the generating set of a Gröbner basis and assume that the resulting set is a generating set of the same ideal.

In particular, the practical approach to find a minimal Gröbner basis is to calculate an interreduced Gröbner basis. Interreduction means that all generators are in normal form w.r.t. the rest of the generators. This is clearly not true here: $x^2y + x + 1$ can be reduced by succesive reduction with $x-y$ into $y^3+y+1$, and $xy^2+y+1$ into $y^3+y+1$ as well. The sets $\{ x^2y+x+1, xy^2+y+1,x-y \}$ and $\{ y^3+y+1, x-y \}$ generate the same ideal (you can write the elemtents of one in terms of the elements of the others), and it is interreduced: $y^3+y+1$ does not reduce $x-y$, and $x-y$ does not reduce $y^3+y+1$. Thus, this is an interreduced Gröbner basis, and it is easy to see that it is also a minimal Gröbner basis.

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