From Ideals, Varieties and Algorithms: Monomial Ideals

Question

From Cox, Little and O'Shea's book Ideals, Varieties and Algorithms. I really don't understand their proof on the following lemma about monomial ideals.

Let $I=\langle x^{\alpha}|\alpha \in A\rangle$ be a monomial ideal. Then a monomial $x^{\beta}$ lies in $I$ iff $x^{\beta}$ is divisible by $x^{\alpha}$ for some $\alpha \in A$.

Here's the proof.

If $x^{\beta}$ is a multiple of $x^{\alpha}$ for some $\alpha \in A$ then $x^{\beta} \in I$ by definition of an ideal. Conversely, $x^{\beta} \in I$ then $x^{\beta} =\sum_{i=1}^s h_ix^{\alpha (i)}$ where $h_i \in k[x_1,...,x_n]$ and $\alpha (i) \in A$. If we expand each $h_i$ as a linear combination of monomials we see that every term on the right side of the equation is divisible by some $x^{\alpha (i)}$. Hence the left side must have the same property.

Firstly, $x^{\beta}$ is a "monomial" right? If I understood monomials correctly, it should be $x_1^{\alpha _1}x_2^{\alpha _2}...$, so basically, it's a multiple of exponents of variables. However, $x^{\beta} =\sum_{i=1}^s h_ix^{\alpha (i)}$ seems to imply, since it's a "sum" notation, that the monomial $x^{\beta}$ is somehow a "linear combination" of monomials (or polynomials, but in the end, I guess it is essentially the same as linear combinations of monomials).

How? I can only think of expressing a monomial as a linear combination of monomials only with every single coefficient being $0$ except for the term where the exponent $n$-tuple ($\beta$ in this case) coincide. Then, why bother with the "sigma" notation? Isn't it redundant?

Second, what does it mean by "expand $h_i$ as a linear combination of monomials"? Does it literally want me to express each coefficient $h_i$ in terms of a liner combination of monomials? Even if so, how does this immediately tell me that "every term on the right side of the equation is divisible by some $x^{\alpha (i)}$?

I mean, so this proof is saying $x^{\beta}= h_1x^{\alpha (1)}+h_2x^{\alpha (2)}+...+h_sx^{\alpha (s)}$ and further, if I expand each $h_i$, namely, $$x^{\beta}= (h_{1,0}x_1^{\gamma_{1,0}}+h_{0,1}x_2^{\gamma_{1,1}}+...+h_{1,n}x_n^{\gamma {0,n}})x^{\alpha (1)}+(h_{2,0}x_1^{\gamma_{2,0}}+h_{2,1}x_2^{\gamma_{2,1}}+...+h_{2,n}x_n^{\gamma {2,n}})x^{\alpha (2)}+...+(h_{s,0}x_1^{\gamma_{s,0}}+h_{s,1}x_2^{\gamma_{s,1}}+...+h_{s,n}x_n^{\gamma {s,n}})x^{\alpha (s)}$$ where $h_i=h_{i,0}x_1^{\gamma_{i,0}}+h_{i,1}x_2^{\gamma_{i,1}}+...+h_{i,n}x_n^{\gamma {i,n}} \in k[x_1,...,x_n]$. I mean, this expanding business, to me, seems like it's just made it super complicated. How is this supposed to make me "see" that it is divisible by some $x^{\alpha (i)}$? I really can't see anything even close to it.

What's going on with this proof? Can somebody please please help me? Thanks...

Answer 1

I just stumbled across this proof and didn't understand what was going on with that "Hence the left side must have the same property.". I guess it's a bit late but here's my attempt anyway for the record (hopefully somebody will find it useful sometime).

What (I think) they're trying to say is:

$$ x^\beta = \sum_{i=1}^s h_i x^{\alpha(i)} =^1 \sum_{i=1}^s \sum_{j=1}^m a_j x^{\alpha'(j)} x^{\alpha(i)} =^2 \sum_{k=1}^{sm} a_k x^{\gamma(k)} $$

Where the first equality is the "expansion" they mention and the second one is a simple rearrangement for the sake of simplicity. Now we have expressed $x^\beta$ as a sum of monomials ($a_k \in K \ \forall k$, with $K$ the field), and it's clear that:

$$ \sum_{k : \gamma(k) \ne \beta} a_k x^{\gamma(k)} = 0 $$

since, otherwise, the result of the sum couldn't possibly be $x^\beta$. So weeding out those $k$ we end up with:

$$x^\beta = \sum_{l=1}^L a_l x^{\delta'(l)}x^{\delta(l)}$$

where $a_l x^{\delta'(l)}$ is a monomial from some $h_i$, $x^{\delta(l)} = x^{\alpha(i)}$ and $\beta = \delta(l) + \delta'(l)$ for every $1 \le l \le L$.

Finally, since $\beta, \delta(l), \delta'(l) \in \Bbb N^n$, $\beta \ge \delta(l)$ with the product order ($ \iff \forall i,l. \beta_i \ge \delta(l)_i$).

Therefore, $\forall l. x^{\delta(l)} | x^\beta$ and taking the corresponding $i$ such that $x^{\delta(l)} = x^{\alpha(i)}$ we have that $x^{\alpha(i)} | x^\beta$. Note that there can be several such $i$ values that work, but when choosing one of them every term in the sum is divisible by $x^{\alpha(i)}$.

Answer 2

For $\alpha, \beta \in \Bbb Z^n_{\ge 0}$, we write $\beta \ge \alpha$ to mean that $\beta_i \le \alpha_i$ for all $1 \le i \le n$, where $\alpha = (\alpha_1, \ldots, \alpha_n)$ and $\beta_i$ defined similarly.
Under this notation, note that $x^\alpha \mid x^\beta \iff \alpha \le \beta \iff \beta - \alpha \in \Bbb Z^n_{\ge 0}$.

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We have $x^\beta = \displaystyle\sum_{i = 1}^s h_i x^{\alpha(i)}$. We can expand the $h_i$ in terms of monomials as well to get

$$x^\beta = \sum_{i = 1}^s\left(\sum_{j = 1}^{n_i}a_{i, j}x^{\beta(i, j)}\right)x^{\alpha(i)}$$

where $a_{i, j} \in \Bbbk$ and $\beta(i, j) \in \Bbb Z^n_{\ge 0}$ for all valid $i, j$.

Now, recalling that the monomials form a $\Bbbk$-basis for $\Bbb k[x_1, \ldots, x_n]$, there must be some $i \in \{1, \ldots, s\}$ and some $j \in \{1, \ldots, n_i\}$ such that $\beta(i, j) + \alpha(i) = \beta$.
(Why? Otherwise, $x^\beta$ would be a linear combination of monomials distinct from $x^\beta$, contradicting linear independence.)

In turn, $\alpha(i) \le \beta$ and thus, $x^{\alpha(i)} \mid x^\beta$, as desired.

Answer 3

(A bit late, anyway ...) To prove the claim, the implication needs to be shown both ways:

1. If $x^\beta$ is a multiple of $x^\alpha$ for some $\alpha \in A$, then $x^\beta \in I$, and,

2. If $x^\beta \in I$, then it is a multiple of $x^\alpha$ for some $\alpha \in A$.

I assume that you understand how to prove the first part. Then, for the second one, since $x^\beta \in I$, $x$ is a linear combination of multiples of the generators of $I$, hence the notation,

$$ x^\beta = \sum_i h_i x^{\alpha(i)} $$

The left hand side is a monomial, but we can not assume that all but one of the $h_i$ must be zero on the right hand side. (If we assumed that, then we are already assuming that $x^\beta$ is a multiple of some $x^\alpha$, the statement we need to prove!) For example, if $n=2$ and $x^2y$, $x^1y^2$ are three of the many generators of $I$, and $\beta = (4,2)$, then the following is a valid way of expanding $x^\beta$.

$$ x^\beta = (x^2y+xy)x^2y + (-x^2)x^1y^2 $$

This is why they use that notation.

Answer 4

The proof in the book is incorrect.

Let us take the simplest example possible $I=\langle x,y \rangle $. We let $x^\beta = xy$ but rather than writing it as $x^\beta = xy = (x)y$, we express it in non-standard form: $$xy =\left(\frac{1}{2}y+xy^2\right)x+\left(\frac{1}{2} x-x^2y\right)y.$$

Written in this form, we see the problem with the above proof. It is not apparent that our $x^\beta$ can be written as the product of a polynomial and a single $x^{\alpha(i)}$. We can correct this as follows.

Proof 2: If $x^\beta \in I$ , then $x^\beta = \sum^s_{i=1} h_i\,x^{\alpha(i)}$, where $h_i \in k[x_1 ,\ldots, x_n ]$ and $\alpha(i)\in A$. We may write $h_i = h_{i\gamma}\,x^\gamma$. We expand and may split the sum into two parts: $$ \begin{align*} x^{\beta} & =\sum h_{i\gamma}x^{\gamma}x^{\alpha(i)}\\ & =\sum_{\gamma+\alpha(i)=\beta}h_{i\gamma}x^{\gamma}x^{\alpha(i)}+\sum_{\gamma+\alpha(i)\neq\beta}h_{i\gamma}x^{\gamma}x^{\alpha(i)}\\ & =\left(\sum_{\gamma+\alpha(i)=\beta}h_{i\gamma}\right)x^{\beta}+\sum_{\gamma+\alpha(i)\neq\beta}h_{i\gamma}x^{\gamma}x^{\alpha(i)}. \end{align*} $$ The terms of the second sum must all cancel. In contrast, the first sum must add to $1$. Because of this, at least one term of the sum must be non-zero. Let $h_{i^\prime\gamma^\prime}\neq 0$ with $x^{\gamma^\prime}x^{\alpha(i^\prime)}=x^\beta $. Thus, $x^\beta$ is divisible by $x^{\alpha(i^\prime)}$.

Answer 5

I had difficulty understanding the proof of this lemma (the converse part). Here is how I managed to clarify it. I use the same notation as in the book.

If $x^{\beta}\in I$ then $x^{\beta} = \sum^{s}_{i=1}h_{i}x^{\alpha(i)}$ for some $h_{i}\in k[x_{1}, \ldots, x_{n}]$ and $\alpha(i)\in A$. Because every $h_{i}$ is a linear combination of its terms (which are monomial), so we write $h_{i} = \sum_{j}c_{i, j}x^{\beta(i, j)}$, which is a finite sum. \begin{align*} x^{\beta} = \sum^{s}_{i=1}h_{i}x^{\alpha(i)} = \sum^{s}_{i=1}\left(\sum_{j}c_{i,j}x^{\beta(i, j)}\right)x^{\alpha} = \sum_{i,j}c_{i,j}x^{\beta(i,j)}x^{\alpha(i)}. \end{align*}

Now we add up terms of the same multidegree and identify the coefficient in $x^{\beta}$ and the coefficients in $\sum_{i,j}c_{i,j}x^{\beta(i,j)}x^{\alpha(i)}$ (after simplifying the expression) and obtain \begin{align*} x^{\beta} = \sum_{\stackrel{i,j}{\beta(i,j) + \alpha(i) = \beta}}c_{i,j}x^{\beta(i,j)}x^{\alpha(i)} \end{align*}

The proof was confusing because "$i$", on the one hand, is the index for the finite sum, and on the other hand, is some index that we look for.

Let's choose a fixed pair of indices $(i,j)$ in the above finite sum, where $\beta(i,j) + \alpha(i) = \beta$. So for every other $(i', j')$ in the finite sum, where $\beta(i',j') + \alpha(i') = \beta$, we have $x^{\beta(i,j)}x^{\alpha(i)} = x^{\beta} = x^{\beta(i',j')}x^{\alpha(i')}$, which means $x^{\beta(i',j')}x^{\alpha(i')}$ is divisible by $x^{\alpha(i)}$. Hence $x^{\beta}$ is divisible by $x^{\alpha(i)}$.