Regular sequence and relations I read here (page 19) the following remark: "If $g_1,\ldots,g_k$ form a regular sequence, their only linear relations over $R$ are the trivial ones, so that:
$$ R\left[\frac{g_1}{g_i},\ldots,\frac{g_k}{g_i}\right]\simeq R[t_1,\ldots,t_k]/(g_it_j-g_j,j=1,\ldots,k)" $$
I can't relate that property to my definition of regular sequence that is $g_i$ regular in $R/(g_1,\ldots,g_{i-1})$. Some help?
 A: Here's  a proof that the only linear relations are the trivial relations. Proof by induction on $k$. If $k=1$ there are no trivial relations so there is nothing to show. Let's do the  $k=2$ case just to be safe. Suppose 
$a_1g_1+a_2g_2=0$, then $a_1g_1=-a_2g_2$ the map $R/g_1 \to R/g_1$ given by multiplication by $g_2$ sends $\overline{a_2}$ to $0$ and hence (since $g_2$ is regular on $R/g_1$) we have $a_2=rg_1$ for some $r \in R$. So $$0=a_1g_1+a_2g_2 = a_1g_1 + (rg_1)g_2 = (a_1+rg_2)g_1$$ and since $g_1$ is  regular on $R$ we have $a_1+rg_2=0$. Hence $a_1g_1+a_2g_2=0$ becomes $(-rg_2)g_1 + (rg_1)g_2$ and this is a trivial relation.
The induction hypothesis is that for a regular sequence $g_1, \ldots, g_{k-1}$ if we have a relation $a_1g_1 + \cdots + a_{k-1}g_{k-1}=0$, then each $a_i$ is a linear combination $$a_i=\sum_{j \neq i} r_{i,j}g_j$$
with $r_{i,j}=-r_{j,i}$.
Now for the induction step. Now  assume $g_1, \ldots, g_k$ is a regular sequence, and suppose we had a relation $b_1g_1 + \cdots + b_kg_k=0$. The trick is to move the last term $b_kg_k$ to the other side, just like we did in the $k=2$ case. So
$$b_1g_1 + \cdots b_{k-1}g_{k-1}=-b_kg_k$$
and since the map $R/(g_1, \ldots, g_{k-1}) \to R/(g_1, \ldots, g_{k-1})$ given by multiplication by $g_k$ sends $\overline{b_k}$ to $0$ (since $b_kg_k$ is a linear combination of the $g_1, \ldots, g_{k-1}$, and so
$$b_k = c_1g_1 + \ldots c_{k-1}g_{k-1}$$
Substituting this into $b_1g_1 + \cdots + b_kg_k=0$ gives
$$(b_1+c_1g_k)g_1 + \cdots + (b_{k-1}+c_{k-1}g_k)g_{k-1}=0$$
By the induction hypothesis we have for all $1 \leq i \leq k-1$,
$$b_i+c_ig_k = \sum_{j=1, j \neq i}^{k-1} r_{i,j}g_j$$ 
with $r_{i,j}=-r_{j,i}$.
So to summarize for all $1 \leq i \leq k-1$,
$$b_i = -c_ig_k + \sum_{j=1, j \neq i}^{k-1} r_{i,j}g_j$$ 
and 
for $i=k$ we have
$$b_k = c_1g_1 + \ldots c_{k-1}g_{k-1}$$
which is what we wanted to show. 
I don't see immediately the explanation for the second part of the statement you quoted.
