Let $R=k[x_1,\ldots,x_n]$ for some field $k$. Let $I\subset R$ be an ideal, and let $G,G'$ be two minimal Gröbner bases for $I$. I want to show that $\text{LT}(G)=\text{LT}(G')$. That is, the set of leading terms of elements of $G$ is the same as the set of leading terms of elements of $G'$. Recall the definition of minimal:
A Gröbner basis $G$ is minimal provided:
- The leading coefficient $\text{LC}(p)=1$ for all $p\in G$.
- For all $p\in G$, we have $\text{LT}(p)\not\in\langle\text{LT}(G-\{p\})\rangle$.
My attempt: It suffices without loss of generality to show inclusion in one direction. So pick $\text{LT}(g)\in\text{LT}(G)$. We know, by definition of Gröbner bases, that $\langle\text{LT}(G)\rangle=\langle\text{LT}(I)\rangle=\langle\text{LT}(G')\rangle$, so $\text{LT}(g)\in\langle\text{LT}(G')\rangle$. But $\langle\text{LT}(G')\rangle$ is a monomial ideal, so $\text{LT}(g)$ is divisible by some $\text{LT}(g')\in\text{LT}(G')$. But how can I go from here? I need to somehow get $\text{LT}(g)\in\text{LT}(G')$. Any hints?
Best Answer
Let $G=(g_1,\ldots,g_r)$ and $G'=(g'_1,\ldots,g'_s)$ be two minimal Gröbner bases for $I$ (w.r.t. the same monomial ordering).
Applying the fact that $G'$ is a Gröbner basis to $LT(g_i)$ yields $g'_k$ such that $LT(g'_k) \mid LT(g_i)$. Conversely, applying the fact that $G$ is a Gröbner basis to $LT(g'_k)$ yields $g_\ell$ such that $LT(g_\ell) \mid LT(g'_k)$. Putting the two together gives $LT(g_\ell) \mid LT(g_i)$. Since $LT(g_i)\not\in\langle LT(G-\{g_i\})\rangle$ by minimality of $G$, it follows that $\ell = i$. Now we have $LT(g_i) = cLT(g'_k)$ for some constant $c \neq 0$. From the minimality of both $G$ and $G'$ it follows that $1 = LC(g_i) = cLC(g'_k) = c$, so $LT(g_i) = LT(g'_k)\in LT(G')$.