Is every basis a Gröbner Basis with respect to some monomial order?

Question

Given a polynomial ring $R=k[x_1,\ldots, x_n]$, and an ideal $I=\langle f_1,\ldots, f_m\rangle\subseteq R$, does there exist a monomial order $<$ on $\mathbb N^n$ such that $\{f_1,\ldots, f_m\}$ is a Gröbner basis with respect to $<$?

Answer 1

Assume $\operatorname{char}(k) \neq 2$. Let $I = \langle x + y, x - y \rangle \trianglelefteq k[x,y]$, and let $g_1 = x + y$ and $g_2 = x - y$. I claim there is no monomial order such that $\{g_1, g_2\} = \{x + y, x - y\}$ is a Gröbner basis for $I$.

If $x > y$, then $LT(g_1) = LT(g_2) = x$, so $\langle LT(g_1), LT(g_2) \rangle = \langle x \rangle$. But $y = \frac{1}{2}(g_1 - g_2) \in I$, so $y \in LT(I)$, hence $\langle LT(g_1), LT(g_2) \rangle \neq LT(I)$.

If instead $y > x$, an analogous argument applies: instead show $\langle LT(g_1), LT(g_2) \rangle = \langle y \rangle$ but $x \in LT(I)$.