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I was reading the book $\textit{Ideals, Varieties, and Algorithms}$ by Cox, Little, and O'Shea, and on Chapter 2 page 71 the have the following lemma

Lemma 3: Let $I$ be a monomial ideal, and let $f \in k[x_1, \ldots, x_n]$. Then the following are equivalent:

i) $f \in I$

ii) Every term of $f$ lies in $I$.

iii) $f$ is a $k$-linear combination of the monomials in $I$.

I do not know why ii) $\implies$ iii). Clearly, we can express $f$ as $p_1 x^{\alpha_1} + \cdots p_n x^{\alpha_n}$ where $p_i \in k[x_1, \ldots, x_n]$ and $x^{\alpha_j} \in I$, but I do not "see" why we can express $f$ as a $k$-linear combination (i.e. $f = k_1 x^{\alpha_1} + \cdots + k_m x^{\alpha_m}$ where $k_i \in k$ and $x^{\alpha_t \in I}$). Thank you for your help!

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    $\begingroup$ This should be tagged (commutative-algebra) instead of (algebraic-geometry) or (affine-varieties). $\endgroup$ – Trevor Gunn May 13 '17 at 17:31
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Let

$$f = \sum_{\mathbf{u} \in \mathbf{N}^n} a_\mathbf{u} \mathbf{x}^\mathbf{u}$$

and suppose that for each $\mathbf{u} \in \mathbf{N}^n$,

$$a_\mathbf{u} \mathbf{x}^\mathbf{u} = \sum_{\mathbf{v} \in \mathbf{N}^n} p_\mathbf{u,v}(\mathbf{x}) \mathbf{x}^\mathbf{v} $$

where $p_\mathbf{u,v}(\mathbf{x}) \in k[\mathbf{x}]$ and is nonzero only if $\mathbf{x}^\mathbf{v} \in I$. Since, $p_\mathbf{u,v}(\mathbf{x})$ is a polynomial, we can expand it in terms of its monomials. Thus,

$$ a_\mathbf{u} \mathbf{x}^\mathbf{u} = \sum_{\mathbf{v} \in \mathbf{N}^n} b_\mathbf{u,v} \mathbf{x}^\mathbf{v} $$

where $b_\mathbf{u,v} \in k$ and $b_\mathbf{u,v} \ne 0$ only if $\mathbf{x}^\mathbf{v} \in I$. Notice, that if $\mathbf{x}^\mathbf{v} \in I$ then $\mathbf{x}^\mathbf{v}\mathbf{x}^\mathbf{w} \in I$ for any monomial $\mathbf{x}^\mathbf{w}$.

Putting this together,

$$ f = \sum_{\mathbf{u}, \mathbf{v}} b_\mathbf{u,v} \mathbf{x}^\mathbf{v} $$

where $b_\mathbf{u,v} \ne 0$ only if $\mathbf{x}^\mathbf{v} \in I$.

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