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 that if $B$ is reduced, then $B$ is minimal.
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:
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 $M.$
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 show the minimality in our case? I am guessing that I should proceed by contradiction but still the idea is not fully figured in my mind. Can someone help me removing the confusion in my mind regarding this proof?
Since, by the given, $B$ 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 $i \neq j.$
And since the given definition of $B$ being a minimal Grobner basis means that for each $i = 1, . . . , t$, we have the following:
1-$\operatorname{in}(g_i)$ is monomial(i.e., the coefficient in $k$ is 1).
2- $\operatorname{in}(g_i)$ does not divide $\operatorname{in}(g_j)$ for any $j$ different than $i$ because otherwise, the given set will not be a minimal set of generators for $\operatorname{in}(M),$
Then, $1$ is satisfied because $B$ is a reduced Gröbner basis by assumption and hence $i$ is correct, $2$ is satisfied because $(ii)$ is correct and if $\operatorname{in}(g_i)$ does not divide any term of $g_j$ for $i \neq j,$ then sure it will not divide $\operatorname{in}(g_j)$ for any $j$ different than $i$ because if $>$ is a monomial order, then for any $g \in F$ we define the initial term of $g,$ written $in_{>}(g)$ to be the greatest term of $g$ with respect to the order $>.$