Let $F$ be a free module (of finite rank) over $S = k[x_1, \dots , x_r]$ with monomial order >. Let $M \subset F$ be a submodule and let $B = \{g_1, \dots , g_t\}$ be a Gröbner basis for $M.$
I want to show the existence of a unique reduced Gröbner basis for $M.$.
I have the following definitions.
$B$ is a minimal Gröbner basis if $\operatorname{in}(g_1), \dots ,\operatorname{in}(g_t)$ is a minimal set of generators for $\operatorname{in}(M).$
$B$ is a reduced Gröbner basis if, for each $i, 1 \leq i \leq t:$
$(i)$ $\operatorname{in}(g_i)$ is a monomial(i.e., the coefficient in $k$ is 1), and
$(ii)$ $\operatorname{in}(g_i)$ does not divide any term of $g_j$ for $i \neq j.$
Also, if I am given that:
1- If $\operatorname{in}(M)$ is generated by $\operatorname{in}(g_1), \dots ,\operatorname{in}(g_s)$ for some $s \leq t,$ then $\{g_1, \dots , g_s\}$ is also a Gröbner basis for $M.$
2- if $B$ is reduced, then $B$ is minimal.
Also, if my previous knowledge is (From Eisenbud "Commutative algebra, with a view toward algebraic geometry" pg.328 and pg. 325)::
If $>$ is a monomial order, then for any $f \in F$ we define the initial term of $f,$ written $in_{>}(f)$ to be the greatest term of $f$ with respect to the order $>,$ and if $M$ is a submodule of $F$ we define $in_{>}(M)$ to be the monomial submodule generated by the elements $in_{>}(f)$ for all $f \in M.$
A Gröbner basis with respect to an order $>$ on a free module with basis $F$ is a set of elements $g_1, \dots , g_t \in F$ such that if $M$ is the submodule of $F$ generated by $g_1, \dots , g_t,$ then $in_{>}(g_1), \dots , in_{>}(g_t)$ generate $in_{>}(M).$
I am not sure how to tackle this problem I am guessing that it is similar to row reduction of matrices and that we may use a monomial that is a linear combination of other monomials should be divisible by those monomials. Still the idea is not fully figured in my mind, and how the prove of the existence and uniqueness should be handled separately.
Can someone help me removing the confusion in my mind regarding the concrete steps of this proof?
Or referring to a reference that contains the proof will also very greatly appreciated!
EDIT:
My trial for the existence part:\
Let $F$ be a free module (of finite rank) over $S = k[x_1, \dots , x_r]$ with monomial order >. Let $M \subset F$ be a submodule and let $B = \{g_1, \dots , g_t\}$ be a Gröbner basis for $M.$
For the existence part:
For each $i,$ such that $1 \leq i \leq t$ do the following:
$(i)$ If $\operatorname{in}(g_i)$ is not a monomial, say it has coefficient $a,$ divide by $a$ to get a monomial term.
$(ii)$ If $\operatorname{in}(g_i)$ does divide any term of $g_j$ for $i \neq j$ then we can discard the other and still have a basis.
Is that a correct trial?
The last paragraph that you wrote in your post is correct way to produce a minimal Groebner basis (and all generators have coefficient $1$) from any Groebner basis. The reason behind that is by doing $(i)$ and $(ii)$, you don't change $in_{>}(M)$.
The second step is to reduce a minimal Groebner basis to a unique reduced Groebner basis. The procedure is the following:
Starting with a minimal Groebner basis $G$. For any $g \in G$, do the long division by all elements in $G\setminus \{ g\}$ to get the remainder $g'$. Consider the set $G'=G\setminus \{ g\} \cup \{g' \}$. The important observation here is:
1/ $G'$ is also a minimal Groebner basis for $M$. To show this, we note that $\operatorname{in}(g)=\operatorname{in}(g')$. This is because the leading term of $g$ goes to $g'$ during the long division, since $\operatorname{in}(g)$ is not divisible by any of $\operatorname{in}(g_j)$, $g_j\in G\setminus \{ g\}$. Thus $(\operatorname{in}(g_j) \ | \ g_j\in G' ) = (\operatorname{in}(g_j) \ | \ g_j\in G )$, hence $G'$ is a minimal Groebner basis.
2/ There is no monomial in $g'$ that is in the module $(\operatorname{in}(g_j) \ | \ g_j\in G'\setminus \{ g' \} =G\setminus \{ g\} )$ (by say prop 15.6 Eisenbud), i.e, for all $g_j\in G'\setminus \{ g' \}$ we have $\operatorname{in}(g_j)$ does not divide any monomial in $g'$.
Therefore, by doing this for all elements $G$ (replacing each element at a time), we will get a reduced Groebner basis.
Now we show that the reduced Groebner basis is unique. Suppose that $G_1$ and $G_2$ are two Groebner bases that satisfy the conditions of reduced Groebner basis. Since reduced Groebner basis is minimal, we have that $$\{ \operatorname{in}(g) \ | \ g\in G_1 \} = \{ \operatorname{in}(g) \ | \ g\in G_2 \}$$ (as two sets), since the monomial module $\operatorname{in}(M)$ has a unique minimal generating set whose all generators have coefficient $1$.
Therefore, for every $g\in G_1$, there is $g' \in G_2$ such that $\operatorname{in}(g)=\operatorname{in}(g')$. We want to show that $g=g'$ and thus $G_1=G_2$.
In fact, note that $g-g' \in M$, so by doing long division by $G_1$ we will get remainder $0$. But we also know that $\operatorname{in}(g)=\operatorname{in}(g')$, meaning these leading terms cancel in $g-g'$. The remaining terms of $g-g'$ are not divisible by any of $ \operatorname{in}(g), g\in G$ since $G$ is reduced. Therefore $g-g'=0$, showing the uniqueness.