Gröbner basis

Remarks

$\operatorname{LT}(I)$ is a monomial ideal. Equivalent forms: for every nonzero $f \in I$, some $\operatorname{LT}(g_i)$ divides $\operatorname{LT}(f)$. Or: for every $f \in S$, the remainder of $f$ under the multivariate division algorithm by $G$ is zero if and only if $f \in I$. A Gröbner basis of $I$ generates $I$, and every ideal has a Gröbner basis; this gives another proof of Hilbert's basis theorem. A Gröbner basis is reduced if each $\operatorname{LT}(g_i)$ has coefficient $1$ and no monomial of $g_i$ is divisible by $\operatorname{LT}(g_j)$ for $j \ne i$. For a given monomial order, each ideal has a unique reduced Gröbner basis. Buchberger's algorithm computes a Gröbner basis from any finite set of generators.

Example: $G = \set{x, y}$ is a Gröbner basis of $I = (x, y)$ in $k[x, y]$, for any monomial order. Every nonzero $f \in I$ has no constant term, so each of its monomials, in particular $\operatorname{LT}(f)$, is divisible by $x$ or by $y$.

Non-example: for the lexicographic order with $x > y$, the set $G = \set{x + y, \; x}$ generates the same ideal $I$, and both leading terms equal $x$. But $y = x - (x + y) \in I$ and $\operatorname{LT}(y) = y \notin (x)$, so $G$ is not a Gröbner basis of $I$.