I am trying to prove this statement and I am very new to monomial ideals.
The definition of monomial ideal is stated like this in my book:
An ideal $I ⊆ k[x_1,..., x_n]$ is a monomial ideal if there is a subset $A ⊆ \mathbb Z^n_{≥0}$ (possibly infinite) such that $I$ consists of all polynomials which are finite sums of the form $\sum_{α∈A}{h_αx^α}$, where $h_α ∈ k[x_1,..., x_n]$.
$I \ne \{0\}$, so we see that if we consider any ideal $I \subset K[x_1,...,x_n]$ then the leading terms of the polynomial will be of the form $h_{\alpha}x^{\alpha}$, where $\alpha = (a_1,...,a_n)$ and $x^{\alpha}=x_1^{a_1}\cdots x_n^{a_n}$.
Here's the answer which has been there in my book,
The leading monomials $LM(g)$ of elements $g ∈ I / \{0\}$ generate the monomial ideal $\langle LM(g) | g ∈ I / \{0\} \rangle$.
Since $LM(g)$ and $LT(g)$ differ by a nonzero constant, this ideal equals $\langle LT(g) | g ∈ I \ {0} \rangle = \langle LT(I) \rangle$ . Thus, $\langle LT(I) \rangle $ is a monomial ideal.
This is something I can't understand. Can someone help me please.