How is the replacement by the remainder gives a reduced Grobner basis?

Question

Here is the proof I am stuck in understanding:

But how is the replacement by the remainder gives a reduced Grobner basis?

Answer 1

This should be a comment but I don't have enough reputation to comment.

The idea of the proof is that starting from a minimal Groebner basis $G$, each time when we replace $g$ by the remainder $g'$ (that we get from the long division), we still get a minimal Groebner basis. In the proof, this is because $\langle \text{LT}(G') \rangle = \langle \text{LT}(G) \rangle$.

The more important thing is that when we replace $g$ by $g'$, there is no monomials in $g'$ that is divisible by any of $\text{LT}(h)$ where $h\in G'\setminus \{ g' \}$ because of the long division process. Since we replace all elements in the original $G$ by its remainder to get a basis $\tilde{G}$, the above property holds for all elements in $\tilde{G}$. Therefore, $\tilde{G}$ is reduced by definition.