[proofplan] We use the basis axiom that a basis is a maximal independent subset of the ground set. Since $e \notin B$, adjoining $e$ to $B$ makes the set dependent, so it contains a circuit. Any such circuit must use $e$, because $B$ itself is independent. Uniqueness follows by applying the circuit elimination axiom to two hypothetical circuits sharing $e$, producing a circuit contained entirely in $B$, which contradicts independence. [/proofplan] [step:Show that $B \cup \{e\}$ contains at least one circuit] Since $B$ is a basis of $M$, it is a maximal independent subset of $E$. Because $e \in E \setminus B$, the set $B \cup \{e\}$ strictly contains $B$. By maximality of $B$, the set $B \cup \{e\}$ is dependent. By the definition of dependence in a matroid, every dependent set contains a circuit. Hence there exists a circuit $C$ of $M$ such that \begin{align*} C \subset B \cup \{e\}. \end{align*} [/step] [step:Prove that every circuit contained in $B \cup \{e\}$ contains $e$] Let $C$ be a circuit of $M$ satisfying $C \subset B \cup \{e\}$. Suppose, for contradiction, that $e \notin C$. Then $C \subset B$. Since $B$ is independent, every subset of $B$ is independent. Therefore $C$ is independent, contradicting the definition of a circuit as a minimal dependent set. Hence $e \in C$. [guided] Let $C$ be a circuit of $M$ with $C \subset B \cup \{e\}$. We want to show that the new element $e$ must actually occur in $C$. The only alternative is that $C$ lies completely inside the old basis $B$. Assume, toward a contradiction, that $e \notin C$. Since $C \subset B \cup \{e\}$ and $e$ is not an element of $C$, we get \begin{align*} C \subset B. \end{align*} Now use the independence of the basis. A basis is independent, and every subset of an independent set is independent by the hereditary axiom for matroids. Therefore $C$ is independent. But $C$ was assumed to be a circuit, and a circuit is dependent by definition. This contradiction shows that the assumption $e \notin C$ is impossible. Hence every circuit contained in $B \cup \{e\}$ contains $e$. [/guided] [/step] [step:Use circuit elimination to prove uniqueness] Suppose $C_1$ and $C_2$ are circuits of $M$ satisfying \begin{align*} C_1 \subset B \cup \{e\} \end{align*} and \begin{align*} C_2 \subset B \cup \{e\}. \end{align*} By the previous step, $e \in C_1 \cap C_2$. If $C_1 \ne C_2$, then the circuit elimination axiom applied to the circuits $C_1$ and $C_2$ and to the common element $e$ gives a circuit $C_3$ of $M$ such that \begin{align*} C_3 \subset (C_1 \cup C_2) \setminus \{e\}. \end{align*} Since $C_1 \subset B \cup \{e\}$ and $C_2 \subset B \cup \{e\}$, removing $e$ gives \begin{align*} (C_1 \cup C_2) \setminus \{e\} \subset B. \end{align*} Thus $C_3 \subset B$. This contradicts the independence of $B$, because an independent set contains no circuit. Therefore $C_1 = C_2$. The circuit whose existence was proved in the first step is consequently unique; denote it by $C_B(e)$. By the second step, this unique circuit contains $e$. [/step]