# Can a minimal generating set for an ideal always be made into a Groebner basis?

Let $$I\subseteq k[x_0,\ldots,x_n]$$ be an ideal, generated by some polynomials $$F_1,\ldots,F_r$$, all homogeneous and of the same degree. Suppose $$r$$ is the smallest number of generators that will suffice to generate the ideal.

Can one choose a monomial order or change coordinates to ensure that this generating set is also a Groebner basis for the ideal?

For example, consider the ideal $$I = (xyz, (x + y + 3z)(x + 7z - y)(y - z + 2x), xy(x - y)).$$

With respect to the degree reverse lexicographic ordering, it has as a reduced Groebner basis

$$\{y^4z - 5y^3z^2 - 17y^2z^3 + 21yz^4, xy^3 - y^4 + 5y^3z + 17y^2z^2 - 21yz^3, x^3 - \frac{1}{2}xy^2 - \frac{1}{2}y^3 + \frac{19}{2}x^2z + \frac{5}{2}y^2z + 16xz^2 + \frac{17}{2}yz^2 - \frac{21}{2}z^3, x^2y - xy^2, xyz\},$$ which includes nontrivially more elements than just the $$F_j$$.

However, after changing coordinates with the matrix (here I just generated several invertible matrices at random until one had the desired property)

$$\begin{pmatrix}0 && -31 && 1\\ 0 && -\frac{5}{6} && 1\\ 1 && -2 && 0\end{pmatrix}$$,

under the same monomial order, the transformed original generators (up to a constant multiple) now make up a reduced Groebner basis for the new ideal:

$$(x^3 + \frac{5281}{126}x^2y - \frac{7}{3}x^2z - \frac{4108577}{3348}xyz + \frac{769192975}{140616}y^2z + \frac{22027}{558}xz^2 - \frac{4095575}{23436}yz^2, xy^2 - \frac{191}{155}xyz + \frac{6}{155}xz^2, y^3 - \frac{191}{155}y^2z + \frac{6}{155}yz^2).$$

Does anyone know whether this is always possible, or of references where this sort of question is addressed? Thanks!

No. Let $$k$$ not have characteristic $$2$$, let $$I = \langle x^2, xy, y^2 \rangle$$ and consider the generating set $$x^2$$, $$(x+y)^2$$, $$y^2$$. After a linear change of coordinates, these are $$(a_1 x + b_1 y)^2$$, $$(a_2 x + b_2 y)^2$$ and $$(a_3 x + b_3 y)^2$$ for some $$a_j$$, $$b_j$$. But the leading term of $$(ax+by)^2$$ will always be either $$x^2$$ or $$y^2$$, so these three quadratics cannot have distinct leading terms.