The reduced basis for a binomial ideal is formed by binomials

Question

I was able to prove that a binomial ideal in $K[X_1,\ldots,X_n]$ (generated by $X^\alpha - X^\beta$) has a Gröbner basis formed by binomials by using Buchberger's algorithm.

But how can I prove that the reduced basis needs to be formed by binomials? I should prove that when dividing a binomial by a set of binomials I obtain a binomial? How do I do this?

I checked articles like this one and they do not mention how they deduce it.

Answer 1

Let $>$ be a global monomial order on $R=K[X_1,...,X_n]$, and let $f_1,...,f_r \in R-\{0\}$. Then for every $g\in R$, there exists a uniquely determined expression $$g=g_1 f_1 + \cdots + g_r f_r + h.$$ with $g_1,...,g_r,h \in R$ and such that

• DDR1: For $i>j$, no term of $g_i L(f_i)$ is divisible by $L(f_j)$.
• DDR2: For all $i$, no term of $h$ is divisible by $L(f_i)$.

We define the standard normal form of $g$ w.r.t. $f_1,...,f_r$ by $\mathrm{NF}(g,\{f_1,...,f_r\}):=h$.

Let $I$ be an ideal in $R$ with Gröbner basis $G$. Then we define $\mathrm{NF}(g,I) := \mathrm{NF}(g, G)$.

Now let $m_1,...,m_s$ be a minimal generating system of $L(I)=\langle L(G) \rangle_R$. Then the reduced Gröbner basis of $I$ is given by $$\{m_1-\mathrm{NF}(m_1,I), ..., m_s-\mathrm{NF}(m_s,I)\}.$$

Having that, all you need now is to show that $\mathrm{NF}(m_i,I)$ is monomial for all $i=1,...,s$, i.e., you need to show that the standard normal form of a monomial w.r.t. a set of binomials is monomial. You can show that easily by following the proof of Theorem 2.2.12 in this nice script