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

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: $$xy =\big(\frac{1}{2}y+xy^2\big)x+\big(\frac{1}{2} x-x^2y\big) y$$.

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 2

(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 3

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 4

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 $x^\alpha \mid x^\beta \iff \alpha \le \beta \iff \beta - \alpha \in \Bbb Z^n_{\ge 0}$.

---

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.