Prove that $\mathbf{x^b}\in \langle\mathbf{x^{a_1},...,x^{a_k}} \rangle\iff \exists j\in \{1,...,k\}:\mathbf{x^{a_j}\mid x^b} $

Question

We want to prove the following lemma:

Lemma. Let $K$ be a field and $I:= \langle\mathbf{x^{a_1},...,x^{a_k}} \rangle$ be an ideal of the polynomial ring $K[x_1,...,x_n]$ (which is generated by the monomials $\mathbf x^{a_i}:=x{_1}^{a_{i1}}\cdots x_n^{a_{in}},\forall i\in 1,...,k) $, with $\mathbf{a_1,...,a_k}\in \Bbb{N}^n$. Then, $$\mathbf{x^b}\in \langle\mathbf{x^{a_1},...,x^{a_k}} \rangle\iff \exists j\in \{1,...,k\}:\mathbf{x^{a_j}\mid x^b} $$

Proof. If $\mathbf{x^b}\in \langle\mathbf{x^{a_1},...,x^{a_k}}\rangle \iff \exists f_1,..,f_k\in K[x_1,...,x_n]: \mathbf{x^b}=f_1\cdot \mathbf{x^{a_1}}+\cdots+f_k\cdot \mathbf{x^{a_k}}$. But how can we conclude the division from the second implication?

For the opposite direction, if $$\exists j\in \{1,...,k\}:\mathbf{x^{a_j}\mid x^b} \iff \exists f\in K[x_1,...,x_n]: \mathbf{x^b}=f\cdot \mathbf{x^{a_j}},\ j\in \{ 1,...,k\} \implies \mathbf{x^b}\in \langle\mathbf{x^{a_1},...,x^{a_k}}\rangle.$$ Does the last implication hold for the opposite direction?

Thank you.

Answer 1

Once you have the equality $$x^b = f_1 x^{a_1} + \cdots + f_n x^{a_n}$$ (I won't bother putting the bold font), you know that $x^b$ must occur as a monomial in the expression $$f_1 x^{a_1} + \cdots + f_n x^{a_n}.$$ But all monomials in this expression occur as multiples of some $x^{a_i}$.

Answer 2

We have: $\mathbf{x^b}=f_1\mathbf{x^{a_1}}+...+f_k\mathbf{x^{a_k}}=(\mathbf{x^{a_{11}}+x^{a_{12}}+...})\cdot \mathbf{x^{a_1}}+...+(\mathbf{x^{a_{k1}}+x^{a_{k2}}+...})\cdot\mathbf{x^{a_k}}$. So, in the RHS we have a monomial in the form $\mathbf{x^{a_{ji}}x^{a_j}}$, for some $j\in \{1,2,...,k\}$ and $i =1,2,...$ and from this, $\mathbf{x^{a_j}|x^b}$. Right?