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Given an ideal $I$ of $k[x,y]$, the size of the Groebner basis depends on the monomial ordering. For example, if $$ I = \langle x^3y^4 , x^2 + y^2 \rangle, $$ then the Groebner basis with the lexicographical order $x,y$ is $$ x^2 + y^2, xy^6, y^8, $$ and it has 3 elements.
If we choose the lexicographical order $y,x$, the Groebner basis is $$ x^2 + y^2, x^7, $$ that has 2 elements. Equivalently, we can think of the change of coordinates $x, y \to y, x$ and we get that the ideal $$ I' = \langle y^3x^4 , x^2 + y^2 \rangle, $$ with Groebner basis of 2 elements (with the lexicographical order $x,y$). More generally: a change of coordinates $x, y \to x', y'$ changes the cardinality of the Groebner basis.

Now, let $I$ be an ideal in $k[x_1,\ldots,x_n]$, such that the Groebner basis with respect to the monomial lexicographical order $x_1,\ldots,x_n $ has $m$ elements. Let $x_1,\ldots,x_n \to x'_1,\ldots,x'_n$ be a change of coordinates and $I'$ the ideal that we get by changing coordinates on $I$ (replacing $x_i$ by $x'_i$).

The question is: does there exist a permutation $\sigma$ such that the Groebner basis of $I'$ with respect to the monomial lexicographical order $x_{\sigma_1},\ldots,x_{\sigma_n}$ contains exactly $m$ elements?

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Here is a counterexample (assuming that $k$ is a field of characteristic $\neq 2$). Consider the monomial ideal $I = \langle x^2,y^2 \rangle$, whose Gröbner basis $\{x^2,y^2\}$ has $2$ elements. Under the change of coordinates $(x,y) \to (x+y,x-y)$, we get the ideal $I' = \langle x^2+y^2,xy \rangle$. When computing a Gröbner basis of $I'$, we get two different answers depending on whether the chosen monomial ordering satisfies $x > y$ or $y > x$. In the former case we get the Gröbner basis $\{x^2+y^2,xy,y^3\}$, while in the latter we get $\{x^2+y^2,xy,x^3\}$. Therefore, no matter the monomial ordering, the associated Gröbner basis of $I'$ has $3$ elements.

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