Independence complex of a prime ideal is a matroid Let $k$ be a field and $I \subseteq k[x_1, \ldots, x_n]$ be an ideal.

Definition. A subset $ \underline{u} \subseteq \{ x_1, \ldots, x_n\} = \underline{x} $ of variables is independent modulo $I$ if $ I \cap k[\underline{u}] = (0)$. The independence complex of $I$ is the collection:
$$ \Delta(I) = \{ \underline{u} \subseteq \underline{x} : \underline{u} \text{ is independent modulo } I\}.$$

$\Delta(I)$ is a simplicial complex, that is, it is closed under finite intersection and inclusions. In order to be a matroid, it should also satisfy the exchange property, that is, if $\underline{u_1}, \underline{u_2} \subseteq \underline{x}$ are indepedent modulo $I$ and $|\underline{u_1}| < |\underline{u_2}|$, then there is a $v \in \underline{u_2} \smallsetminus \underline{u_1}$ such that $\underline{u_1} \cup \{ v \}$ is independent modulo $I$.
However, this is not always true. In fact, from this, one could get that the maximally independent sets of variables modulo $I$ have all the same cardinality. But considering $I = (xy,xz)$, one has that $\{x\}$ and $\{y,z\}$ are maximally independent modulo $I$ but do not have the same cardinality.
I have been told during a course about Groebner bases and I have read on an article by Bernd Sturmfels that if $I$ is prime, however, $\Delta(I)$ is a matroid. Do you know how can I prove this only using the basic definition of a matroid and some Groebner bases tools?
 A: Okay, I have an attempt of proof which I care to post and I hope someone can double check.
Suppose by contradiction that $\underline{u_1}, \underline{u_2}$ are independent modulo $I$, $|\underline{u_1}| < |\underline{u_2}|$ but $\underline{u_1} \cup \{ v \}$ is dependent modulo $I$ for every $v \in \underline{u_2} \smallsetminus \underline{u_1}$.
Denote $\underline{u_1} = \{ x_1, \ldots, x_m \}$ and $\underline{u_2} = \{ y_1, \ldots, y_n \}$, $n > m$. Since $\underline{u_1} \cup \{y_i\}$ is dependent modulo $I$, there is an $f_i \in I \cap k[\underline{u_1}, y_i]$ such that $f_i \neq 0$ for every $i = 1, \ldots, n$. Moreover, as $k[\underline{u}]$ is a Factorization Domain for every $\underline{u} \subseteq \underline{x}$, we can write each $f_i$ as a product of irreducible elements $f_i = f_{i,1} \cdots f_{i,m_i}$. Now, as $f_i \in I$ and $I$ is prime, there is an irreducible factor $f_{i,j}$ of $f_i$ such that $f_{i,j} \in I \cap k[\underline{u_1}, y_i]$ and $f_{i,j} \neq 0$. In particular, $I \cap k[\underline{u_1}, y_i]$ contains a non-zero irreducible polynomial, so we can just assume that $f_i$ is irreducible for every $i = 1, \ldots, n$.
Now, as $\underline{u_1}$ and $\underline{u_2}$ are independent modulo $I$, we have that $f_i$ must be supported on $y_i$ and $x_{j_i}$ for some $j_i \in \{ 1, \ldots, m \}$. But as the $y$-variables are more than the $x$-variables (because $n > m$), there are $i, \ell \in \{ 1, \ldots, n \}$ such that $i \neq \ell$ and $x_{j_i} = x_{j_\ell} = z$. So, $f_i$ is supported on $y_i$ and $z$ and $f_\ell$ is supported on $y_\ell$ and $z$, and both $f_i$ and $f_\ell$ are irreducible.
By a well known theorem, the resultant of $f_i$ and $f_\ell$ with respect to the variable $z$, denoted by $Res_z(f_i,f_\ell) \in k[y_i, y_\ell]$, is non-zero and belongs to the ideal generated by $f_i$ and $f_\ell$. In particular, $Res_z(f_i,f_\ell) \neq 0$ and $Res_z(f_i,f_\ell) \in I \cap k[y_i, y_\ell]$. This is a contradiction, as $\{ y_i, y_\ell \}$ turns out to be dependent modulo $I$ and  contained into $\underline{u_2}$, which is independent modulo $I$.
[EDIT] I think there is a problem with this proof. My concern is about the fact that $f_i$ and $f_\ell$ are supported in at least one (common) variable $z$ and on $y_i$ and $y_\ell$ respectively, so they can have more than these variables. So the resultant will in general have more variables than just $y_i$ and $y_\ell$.
