Reading first time about monomial ideals and I am stuck with one of the very first results:
Let $\mathbb{K} $ be a field and $\Lambda \subseteq \mathbb{Z}_{\ge 0}^n$. Given a monomial ideal $$ I= \left\{ f_1 x^{\alpha_1} + \dots + f_s x^{\alpha_s} : \alpha_1, \dots, \alpha_s \in \Lambda; f_1, \dots, f_s \in \mathbb{K}[x_1, \dots, x_n]\right\},$$ a monomial $ x^{\beta} $, $\beta \in \mathbb{Z}_{\ge 0}^n$, is in $I$ iff $x^{\beta}$ is divisible by some $x^{\alpha} $, where $\alpha \in \Lambda$.
Here I used the standard notation that when $\alpha=(\alpha_1, \dots, \alpha_n)\in \mathbb{Z}_{\ge 0}^n$, we write $x^{\alpha} =x_1^{\alpha_1} \dots x_n^{\alpha_n} $.
The proof is presented in Wikipedia. It is clear to me that if $x^{\beta} $is divisible by $x^{\alpha}$ for some $\alpha \in \Lambda$, then $x^{\beta}\in I$. However, I don't understand the other direction.
Let $x^{\beta} \in I$. Then there are polynomials $f_1, \dots, f_s \in \mathbb{K}[x_1, \dots, x_n] $ so that $$x^{\beta}= f_1 x^{\alpha_1} + \dots + f_s x^{\alpha_s}.$$
Since $f_1, \dots, f_s $ are polynomials, they are linear combinations of monomials. We write $$ f_i = \sum_{j=1}^{t_i} g_{ij} x^{\gamma_{ij}}, $$ where $g_{ij} \in \mathbb{K}$ and $\gamma_{ij} \in \mathbb{Z}_{\ge 0}^n$. Therefore
$$ x^{\beta}= \sum_{j=1}^{t_1} g_{1j} x^{\gamma_{1j}} x^{\alpha_1} + \dots + \sum_{j=1}^{t_s} g_{sj} x^{\gamma_{sj}} x^{\alpha_s} = \sum_{j=1}^{t_1} g_{1j} x^{\alpha_1+\gamma_{1j}}+ \dots + \sum_{j=1}^{t_s} g_{sj} x^{\alpha_s +\gamma_{sj}} . $$
Now it's unclear to me, how is it 'obvious' that the RHS is divisible by some $x^{\alpha}$ with $\alpha \in \Lambda $?
The reasoning is that each monomial or term on the right-hand side is divisible by some $x^{\alpha_i}$ and so this holds also for the left-hand side. Done.