Let $I \subset R = K[x_1,...,x_n]$ be an ideal and $<$ a monomial term ordering on $R$. Let $w\in \mathbb{R}_{\geq 0}^n$ be a weight vector. Then for each $f = \sum c_\alpha x^\alpha \in I$ we can define the following polynomials:
- $in_w(f) = \sum_{w\cdot \alpha = a}c_\alpha x^\alpha$ where $a = \max{\{w\cdot \alpha\}}$ (this uses the so called max-convention)
- $in_<(f)$ the monomial term of $f$ that is the largest according to the term ordering $<$
Both of these definitions give rise to initial ideals:
- $in_w(I) = \langle in_w(f)\ |\ f\in I\rangle$
- $in_<(I) = \langle in_<(f)\ |\ f\in I\rangle$
A Gröbner basis of $I$ with respect to $w$ (resp. $<$) is then a finite subset $G$ of $I$ such that $in_w(I) = \langle in_w(f)\ |\ f\in G\rangle$ (resp. $in_<(I) = \langle in_<(f)\ |\ f\in G\rangle$).
We can also define the term ordering $<_w$ that first chooses the monomial terms that have exponent vectors $\alpha$ that maximize $w\cdot \alpha$ and in case of ties, uses $<$ to resolve these ties.
Now I want to prove the following:
Let $G$ be a Gröbner basis of $I$ with respect to $<_w$. Then the set $\{in_w(g)\ |\ g\in G\}$ is a Gröbner basis of the ideal $in_w(I)$ with respect to $<$.
In other words, we would want to prove that $$ in_<(in_w(I)) = \langle in_<(in_w(g))\ |\ g\in G\rangle.$$ The $\supseteq$-inclusion is easy to prove: every element of $g\in G$ is an element of $I$ and hence $in_w(g) \in in_w(I)$. Therefore also $in_<(in_w(g)) \in in_<(in_w(I))$.
Now, I have a bit of trouble with the $\subseteq$-inclusion. My attempt was the following: Let $m \in in_<(in_w(I))$. We can assume $m$ is a monomial (because $in_<(in_w(I))$ is a monomial ideal and hence it suffices to check the inclusion for only its monomial elements). It's easy to see that this implies that $m = in_<(f)$ for some $f \in in_w(I)$. Now here lies the problem: I don't think we know for sure that $f = in_w(h)$ for some $h \in I$ because we aren't working with monomials here. The only thing we know is that $f$ can be written as $$ f_1in_w(h_1) + ... + f_r in_w(h_r) $$ for some $f_i \in R$ and $h_i \in I$. I don't see how we can use this expansion to conclude our proof of this inclusion. Can anyone help me?
Thanks in advance!
Given a monomial order $<$ and a finite set of polynomials $G = \{g_1,\dots,g_s\}$, we say that a (nonzero) polynomial $f$ has a standard representation if we can write \begin{equation*} f = \sum_{i=1}^s a_ig_i, \end{equation*} where $\deg(f) \geq \deg(a_ig_i)$ whenever $a_ig_i \neq 0$. In the situation at hand, the key idea is to notice that whenever $f$ has a standard representation (with respect to $<_w$ and $G$), then removing the terms $a_ig_i$ whose weighted degree is less than that of $f$, we get a standard representation for $\text{in}_w(f)$ (with respect to $<$ and $G' = \{\text{in}_w(g)\ |\ g \in G\}$): \begin{equation*} \text{in}_w(f) = \sum \text{in}_w(a_ig_i) = \sum \text{in}_w(a_i)\text{in}_w(g_i). \end{equation*}
As a first application, we get that $G'$ is a basis of $\text{in}_w(I)$. Indeed, since $G$ is Gröbner basis of $I$, every polynomial $f \in I$ has a standard representation, hence every polynomial $\text{in}_w(f) \in \text{in}_w(I)$ can be written as a combination of elements in $G'$.
We now show that $G'$ is a Gröbner basis of $\text{in}_w(I)$ with respect to $<$ using Buchberger's criterion. First, observe that for any $g \in G$, \begin{equation*} \text{in}_<(\text{in}_w(g)) = \text{in}_{<_w}(g). \end{equation*} This means that if we write down the $S$-polynomial \begin{equation*} S(g_i,g_j) = ax^\alpha g_i-bx^\beta g_j, \end{equation*} then we have \begin{equation*} S(\text{in}_w(g_i),\text{in}_w(g_j)) = ax^\alpha\text{in}_w(g_i)-bx^\beta\text{in}_w(g_j). \end{equation*} So whenever the latter is nonzero, we see that \begin{equation*} \text{in}_w(S(g_i,g_j)) = S(\text{in}_w(g_i),\text{in}_w(g_j)). \end{equation*} The result follows at once, since a standard representation for $S(g_i,g_j)$ (which always exists since $G$ is Gröbner) gives a standard representation for $\text{in}_w(S(g_i,g_j))$.