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 $\overline{S(g_a,g_b)}^G$.
 A: 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 $\langle LT(\overline{G})\rangle$.
A: 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, $LT(\overline{g_j})\in \langle LT(\overline{g_1}),\dots,LT(\overline{g_t}) \rangle \subseteq  \langle LT<\overline{f_1},\dots,\overline{f_s}>\rangle=\langle LT(\overline{I})\rangle$.
