# How to prove that a reduced Gröbner basis for $I\subset K(\mathbf t)[\mathbf x]$ is still a Gröbner basis under $\mathbf t\mapsto\mathbf a$?

This is Exercise 7 of Section 6-3 of Cox, Little and O'Shea's Ideals, Varieties, and Algorithms.

Let $I=\langle f_1,...,f_s\rangle\subset K(t_1,...,t_m)[x_1,...,x_n]$ where $K(t_1,...,t_m)$ is the field of rational functions and the leading coefficient of $f_i$ is assumed to be $1\;\forall\,i$. Let $G=\{g_1,...,g_t\}$ be a reduced Gröbner basis for $I$. For each $j$, We can write $g_j$ as $g_j=\sum_{i=1}^sB_{ji}f_i$. Let $(a_1,...,a_m)\in K^m$ such that none of the denominators of the $f_i$, the $g_j$, and the $B_{ji}$ vanish under $(t_1,...,t_m)\mapsto(a_1,...,a_m)$. Then we can show that $\overline G=\{\overline g_1,...,\overline g_t\}$ is a basis for $\overline I=\langle\overline f_1,...,\overline f_s\rangle\subset K[x_1,...,x_n]$ where $\overline g_j$ and $\overline f_i$ are obtained from $g_j$ and $f_i$ under $(t_1,...,t_m)\mapsto(a_1,...,a_m)$, respectively.

But I have problems with proving that $\{\overline g_1,...,\overline g_t\}$ is a Gröbner basis for $\overline I$. I have tried to construct a proof by contradiction (see below). Another approach is welcome. I would appreciate your help with this situation.

Suppose that there were $1\le a,b\le t$ such that $\overline{S(\overline g_a,\overline g_b)}^\overline G=r\neq 0$. I want to get a contradiction with the assumption that $\overline{S(g_a,g_b)}^G=0$ but I have no idea how to proceed. I know that $LT(g_j)=LT(\overline g_j)\;\forall\,j$. Hence if $S(g_j,g_k)=\frac{x^{\gamma}}{LT(g_j)}g_j-\frac{x^{\gamma}}{LT(g_k)}g_k$, then $S(\overline g_j,\overline g_k)=\frac{x^{\gamma}}{LT(g_j)}\overline g_j-\frac{x^{\gamma}}{LT(g_k)}\overline g_k$. Therefore, every monomial of $S(\overline g_j,\overline g_k)$ must be a monomial of $S(g_j,g_k)$. But the converse is false. Some monomials of $g_j,g_k$ may disappear under $(t_1,...,t_m)\mapsto(a_1,...,a_m)$, and so may $S(g_j,g_k)$. Difficulties arise when I try to compare the division processes of $\overline{S(\overline g_a,\overline g_b)}^\overline G$ and $\overline{S(g_a,g_b)}^G$.

## 2 Answers

Let $F=\{f_1, ..., f_s\}$ and $G=\{g_1, ..., g_t\}$. By the Exercise 7.(b), we have $I=\langle F\rangle=\langle G\rangle \stackrel{\mathbf{x}\mapsto \mathbf{a}}{\mapsto} \overline{I}=\langle \overline{F}\rangle=\langle \overline{G}\rangle$. We want to show that $\langle LT(\overline{I})\rangle\subseteq \langle LT(\overline{G})\rangle$.

Let $\overline{h}\in \langle LT(\overline{I})\rangle$, write $$\overline{h}=p_1(\mathbf{x})LT(\overline{h_1}) +p_2(\mathbf{x})LT(\overline{h_2}) +\cdots +p_r(\mathbf{x})LT(\overline{h_r}),$$ where $\overline{h_i}\in \overline{I}$.

Incorrect Assertion: Note that $\overline{h_i}\in \overline{I}\subseteq I$ implies that $LT(\overline{h_i})\in \langle LT(I)\rangle=\langle LT(G)\rangle$.

Second Attempt: $\overline{h_i}\in \overline{I}$ for some $h_i\in I$. Note that $LT(\overline{h_i})=LT\left(\frac{1}{LC(h_i)}h_i\right)$. Since $\frac{1}{LC(h_i)}h_i\in I$, we have $LT\left(\frac{1}{LC(h_i)}h_i\right)\in \langle LT(I)\rangle=\langle LT(G)\rangle$

Thus, $LT(\overline{h_i})=LT\left(\frac{1}{LC(h_i)}h_i\right)$ is divided by some $LT(g_{k_i})=LT(\overline{g_{k_i}})$. Therefore, $$\overline{h} =p_1(\mathbf{x})q_1(\mathbf{x})LT(\overline{g_{k_1}}) +p_2(\mathbf{x})q_2(\mathbf{x})LT(\overline{g_{k_r}}) +\cdots +p_r(\mathbf{x})q_r(\mathbf{x})LT(\overline{g_{k_r}}).$$ Which is in $\langle LT(\overline{G})\rangle$.

• Dear @bfhaha: In line 6, you claim that $\overline I\subseteq I$. Let $I_a\subseteq K[x_1,...,x_n]$ be the set of all $\overline f$ obtained from $f\in I$ under $(t_1,...,t_m)\mapsto(a_1,...,a_m)$, then I can show that $\overline I\subseteq I_a$. But how to prove that $\overline I\subseteq I$? By ring homomorphism between $K(t_1,...,t_m)[x_1,...,x_n]$ and $K[x_1,...,x_n]$? Sorry, I am not familiar with it. – lightweight Dec 7 '16 at 0:19
• @Karl You are right. It is a incorrect assertion. Let $I=\langle tx+1\rangle$. Then $x+1\in \overline{I}$ but $x+1\notin I$. – bfhaha Dec 7 '16 at 1:59
• (2) If your $h_i$ in line 1 of "Second Attempt" is defined as $h_i=\sum q_jg_j$ whenever $\overline h_i=\sum q_j\overline g_j$, then I have a question about your claim that $LT(\overline h_i)=LT(\frac{1}{LC(h_i)}h_i)$. Every monomial of $\overline h_i$ must be a monomial of $h_i$, but I think the converse is false in general. Since some coefficients in $h_i$ may become $0$ under $(t_1,...,t_m)\mapsto(a_1,...,a_m)$, I think it is possible that $LM(\overline h_i)\neq LM(h_i)$. – lightweight Dec 7 '16 at 10:54

Here is my sketch of proof:

Note that the elements of $K(t_1,\dots,t_m)$ whose denominators do not vanish under the specialization $(t_1,\dots, t_m)\mapsto (a_1,\dots, a_m)$ form a subring and the specialization is a ring homomorphism between a subring of $K(t_1,\dots,t_m)[x_1,\dots,x_n]$.

Before we start observe the following two facts: Given $f,g \in F[x_1,\dots,x_n]$ where F is a field, under a certain monomial order we have the following:

1). $LT(fg)=LT(f)LT(g)$

2). $LT(f+g)\in \langle LT(f), LT(g) \rangle$

Now, since $\overline{f_i}=\Sigma_{j=1}^t \overline{A_{ij}} \overline{g_j}$, we have

$LT(\overline{f_i})=LT(\Sigma_{j=1}^t \overline{A_{ij}} \overline{g_j})$

$\hspace{10mm}$ $\in \langle LT(\overline{A_{i1}}\overline{g_1}),\dots,LT(\overline{A_{it}}\overline{g_t})\rangle$ by (2)

$\hspace{10mm}$ $=\langle LT(\overline{A_i})LT(\overline{g_1}),\dots,LT(\overline{A_{i1}})LT(\overline{g_t})\rangle$ by (1)

$\hspace{10mm}$ $\subseteq \langle LT(\overline{g_1}),\dots,LT(\overline{g_t})\rangle$ for all $0\leq i \leq s$

For the reverse inclusion, note that $\overline{g_j}=\Sigma_{i=1}^{s}\overline{B_{ji}} \overline{f_1}$ for each $0\leq j \leq t$. That means the leading term of $\overline{g_j}$ falls in $\Sigma_{i=1}^{s}\overline{B_{ji}} \overline{f_1}$, thus the leading term for each $\overline{g_j}$ falls in the ideal generated by $(\overline{f_1},\dots,\overline{f_s})$. Therefore, $LT(\overline{g_j})\in \langle LT(\overline{g_1}),\dots,LT(\overline{g_t}) \rangle \subseteq \langle LT<\overline{f_1},\dots,\overline{f_s}>\rangle=\langle LT(\overline{I})\rangle$.

• Dear @Alan Lao: It is possible that $LM(\overline I)\not\subset LM(I)$ as far as $x_1,...,x_n$ are concerned. I have no idea how you get your 1st inclusion. – lightweight Dec 14 '16 at 10:35
• $LT(\overline{I})$ – Alan Lao Dec 14 '16 at 10:47
• Under the specialization $(t_1,\dots,t_1) \mapsto (a_1,\dots,a_m)$. Some of the monomials of $f_i$ may vanish, and those monomials remained "unvanished" would still be in the ideal generated by $(f_1,\dots, f_s)$ – Alan Lao Dec 14 '16 at 10:59
• Actually, the hint on the book is pretty clear. What matter the most are $x_1,\dots,x_n$. Under the specialization some of the $x_i$ term in $I$ may "disappear" but those remained there would fall inside the $I$. – Alan Lao Dec 14 '16 at 11:04
• Dear @Alan Lao: In your 2nd comment, you said that [those monomials remained "unvanished" would still be in the ideal generated by $(f_1,...,f_s)$]. Consider $I=\langle x+1\rangle$. $1$ is a monomial of $x+1$, but $1\notin I$. – lightweight Dec 14 '16 at 12:04