Proposition

If $I \subset k[x_1, \cdots, x_n]$ is an ideal, prove that $\ev{ \LT(g) \mid g \in I \setminus \{ 0 \} } = \ev{ \LM(g) \mid g \in I \setminus \{ 0 \} }$.

Solution

Let $T = \ev{ \LT(g) \mid g \in I \setminus \{ 0 \} }, M = \ev{ \LM(g) \mid g \in I \setminus \{ 0 \} }$.

Let $\LT(g) \in T$. Then $\LT(g) = (1/\LC(g)) \cdot \LM(g) \in M$.

Let $\LM(g) \in M$. Let $\LM(g) = \LC(g) \cdot \LT(g) \in T$.