Constructing a basis for a matroid with a circuit in it.

Question

Here is the question I am trying to solve (Matroid Theory, second edition by Martin Oxley, Chapter 1, section 2):

Prove that if $C$ is a circuit of a matroid $M$ and $e \in C,$ then $M$ has a basis $B$ such that $C = C(e, B).$

My thoughts:

My idea for solving this question is that, given a circuit $\mathcal C,$ subtract from it an element $x,$ then $\mathcal{C} - x \in \mathcal{I}(M)$ which is the set of independent sets of the matroid M, then extend this set to a basis, say $\mathcal{B}$ (can any independent set in a matroid be extended to a basis always?) then add the $x$ back to this basis. But then how can I be sure that this constructed basis when $x$ is added to it will give me the circuit $\mathcal C$? Can anyone explain this to me or at least give me the correct way of proving this?

EDIT:

I am now reading this question Circuits in a matroid to try to come up with a proof to my question above.

Answer 1

You basically have the right idea (I will use the notation as in the original problem statement):

• As you said, since $C$ is a circuit ("minimally dependent"), $C - e$ must be independent.
• There exists a basis $B \supseteq C - e$: A basis is by definition maximal independent sets. So, if $C - e$ isn't already a basis, it means that there is a strictly bigger, independent set $I$. So, by "iterating" this type of argument, you get a desired basis $B$ since $E$, the set on which our matroid $M$ lives, is finite.
• $B$ satisfies $C = C(e, B)$: Recall that $C(e,B)$ is defined to be the unique circuit in $B + e$, the so-called fundamental-circuit of $e$ w.r.t. $B$. See Corollary 1.2.6 for that. As $C \subseteq B + e$ and $C$ is a circuit, it must therefore already be $C(e,B)$ by uniqueness.