Is a minimal Gröbner Basis a minimal system of generators? I'm studying Gröbner Bases and I'm trying to show that every finitely generated graded module $M$ over $k[x_1,\ldots,x_n]$ has a minimal graded free resolution of length $l \leq n$. (According to Exercise $5$ of Chapter 6 $\S$3 in Cox, Little & O'sheas  $\textit{Using Algebraic Geometry 2nd ed.}$ pp. 272).
The (graded) Hilbert Syzygy Theorem ensures that every such module has a (graded) free resolution of length $l \leq n$.

Theorem (Graded Hilbert Syzygy Theorem). Let $R = k[x_1,\ldots, x_n]$. Then every finitely generated graded $R$-module has a finite graded resolution of length at most $n$.

The proof is based on Schreyer's Theorem, which basically says that some relations $s_{ij}$ arising from the $S$-vectors of the elements of a Gröbner basis  $G = \{ g_1, \ldots, g_s\}$ form a Gröbner basis of the Syzygy module $Syz(G)$.
Like this, one can construct a (graded) free resolution $$ 0 \hookrightarrow F_l \xrightarrow{\phi_l} F_{l-1} \xrightarrow{\phi_{l-1}} \ldots \xrightarrow{\phi_{2}} F_1 \xrightarrow{\phi_{1}} F_0 \xrightarrow{} M \rightarrow 0, $$
by expanding the resolution with free modules $F_i$ and the surjective homomorphism $\phi_i: F_i \rightarrow F_{i-1}$ that takes the canonical basis elements of $F_i$ to the mentioned Gröbner bases elements of $ker (\phi_{i-1}) = Syz(G_{i-2}) \subset F_{i-1}$. (This procedure will terminate after at most $n$ steps due to a result that says that each step, the leading terms of the Gröbner basis of $ Syz(G_{i-2})$ will contain at least one variable $x_i$ less.)

Lemma. Let $G$ be a Gröbner basis for a submodule $M \subset R^t$ w.r.t. to an arbitrary monomial order, and arrange the elements of $G$ to form an ordered s-tuple $G = (g_1,\ldots,g_s)$ so that whenever $LT(g_i)$ and $LT(g_j)$ contain the same standard basis vector $e_k$ and $i < j$, then $LM(g_i)/e_k >_{lex} LM(g_j)/e_k$, where $>_{lex}$ is the $lex$-order on $R$ with $x_1 >\ldots > x_n$. If the variables $x_1,\ldots,x_m $do not appear in the leading-terms of $G$, then $x_1,\ldots,x_{m+1}$ do not appear in the leading terms of the $s_{ij} \in Syz(G)$ w.r.t. to the order $>_{G}$ induced by $G$.

$\textbf{The Problem:}$
The resolution constructed in this proof is generally not minimal i.e $\phi_i$ does not take the canonical basis elements of $F_i$ to a minimal system of generators of  $ Im (\phi_i) =Syz(G_{i-2})$. (Since it takes it to a Gröbner basis.)
In the book, it says that one can construct a minimal resolution by always taking a minimal system of generators of $ker(\phi_{i-1})$ and then take the usual surjective homomorphism $\phi_i$ from the free module $F_i$ to this minimal set of generators .
This principle should now be used to modify the proof above and show that a minimal free resolution of length $\leq n$ exists.
$\textbf{My Question:} $
The proof (of the non-minimal case) above is based on Schreyer's theorem and constructing the resolution with Gröbner bases at each step.  To construct a minimal resolution I would need minimal systems of generators at each step.
Are minimal (or reduced) Gröbner bases actually minimal system of generators such that this construction still works?
If not, how would I need to use this to construct a minimal resolution of length $\leq n$?
(Edit: A minimal Gröbner basis $G$ of $M$ is defined to be a Gröbner basis such that its leading terms are a minimal set of generators of the leading terms of $M$.)
 A: Minimal Gröbner bases are not minimal set of generators, in general.
Even more, there are ideals such that none of their minimal Gröbner bases is a minimal set of generators (I add an example below).
What the book is suggesting is that you should prune this non-minimal resolution to get a minimal one. Since pruning does not add new steps, you are done.
Consider the ideal $I = (x_1^2-x_2^2, x_3^2-x_4^2, x_1x_2-x_3x_4) \subset \mathbb R[x_1,x_2,x_3,x_4]$. It is generated by 3 polynomials and let us see that it has no Gröbner basis with 3 elements.
One can check that for all $i \neq j$ there is a polynomial $x_i^{a} - x_j^{b} \in I$. Indeed, the following polynomials belong to $I$: $x_1^2-x_2^2, x_1^4-x_3^4, x_1^4-x_4^4, x_2^4-x_3^4, x_2^4-x_4^4, x_3^2-x_4^2$. As a consequence, for every term order, there are at least three different values $i,j,k \in \{1,2,3,4\}$ and three values $a,b,c \geq 2$ such that $x_i^{a},x_j^{b}, x_k^{c}$  are minimal generators of the corresponding initial ideal. Moreover, since $x_1x_2 - x_3x_4 \in I$, then we need a fourth generator in the initial ideal. Thus, every Gröbner basis has at least four elements.
