I have a very short proof here for the following lemma, and there's one small bit I am not sure why it is true.
Lemma: Let $G$ be a Groebner basis for the polynomial ideal $I$. Let $p\in G$ be a polynoial such that $LT(p) \in <LT(G-\{p\})>$. Then $G-\{p\}$ is also a Groebner basis.
The following is the proof of the lemma I am given,
We know that $<LT(G)>=<LT(I)>$. If $LT(p) \in <LT(G-\{p\})>$ then we have $<LT(G-\{p\})>=<LT(G)>$. Thus by definition, it follows that $G-\{p\}$ is a Groebner basis.
In the above proof, I don't get why " If $LT(p) \in <LT(G-\{p\})>$ then we have $<LT(G-\{p\})>=<LT(G)>$." How does $<LT(G-\{p\})>=<LT(G)>$ follow? I might be missing a basic fact here so please if someone would remind me of it, that would be very much appreciated.
Thank you
Suppose $\DeclareMathOperator\lt{LT} G=\{p_1,\dots, p_{r-1},p_r\}$. If, say, $\lt(p_r)\in\bigl\langle\lt(p_1),\dots\lt(p_{r-1})\bigr\rangle$, it should be clear that any linear combination of $\;\lt(p_1),\dots,\lt(p_{r-1}),\lt(p_r)\;$ is indeed a linear combination of the $r-1$ first leading terms.