Definition of circuits in matroid theory I am reading about matroids and I am kind of confused in the following definition. 
A nonempty subset $C$ of $\{0,1,...,n\}$ is a circuit of $I$ if $C= supp(l)$ for some nonzero linear form $l$ in the ideal $I$, and $C$ is inclusion-minimal with this property. 
The part I am confused about is the inclusion minimal part. Could someone give me an intuition or example that shows this idea?
 A: One of the axioms of the circuits of a matroid is that any two are incomparable, that is, if $C_1 \subseteq C_2$ are circuits of a matroid, then $C_1 = C_2$. A useful statement about matroids arising from vector spaces (representable matroids), that may help your intuition, is the following:

For a representable matroid whose circuits are the inclusion-minimal supports of elements of $V \subseteq k^n$, the support of any $f \in V$ is a union of circuits. 

For example, suppose that $V = \langle x_0 + x_1, x_1 + x_2, x_2 + x_3 \rangle \subseteq K[x_0,\ldots,x_3]_1$. Then for $f = x_0 - 2x_2 - x_3$, we can see that $supp(f) = \{0,2\} \cup \{2,3\}$, both of which are circuits (as $x_0 - x_2$ and $x_2 + x_3$ are both in $V$, and no monomials are).
A proof sketch for the general argument: suppose $V \subseteq k^n$, and let $f \in V$. If $supp(f)$ is not minimal, then for $g \in V$ such that $supp(g) \subset supp(f)$ and $supp(g)$ is minimal, there exists some $\alpha \in k$ such that $supp(f + \alpha g)$ is smaller than $supp(f)$. Induction allows us to write $f$ as a linear combination of elements whose support is minimal, and whose supports union to the support of $f$.
