Let $A$ be a Banach algebra( or a ring). A left ideal $I$ of $A$ is called a left minimal ideal, if $I\neq\{0\}$ and there is no any other non-zero left ideal of $A$ completely lies in $I$. With a similar way right minimal ideal and minimal bi-ideal(two sided ideal) have been defined.
Is there a Banach algebra ( or a ring) which has left minimal ideal or right minimal ideal such that it does not have minimal bi-ideal?
This is example of a Banach algebra ( also a ring) that has not minimal ideal. But I don't have any idea that it has left or right minimal ideal . I found it from this link
Let $\Delta=\{z\in \mathcal{C}| |z|\leq 1\}$. Suppose that $A(\Delta)$ be the set of all elements $C(\Delta)$ which are analytic on the interior of $\Delta$. $A(\Delta)$ is closed subalgebra of $C(\Delta)$. For any $n\geq0$ define $I_n=\{f\in A(\Delta)| f(0)=f^\prime(0)=f^{\prime\prime}(0)=\ldots=f^{(n)}=0\}$. $(I_n)_{n\geq0}$ is decreasing sequense of (primary) ideals. Now if $J$ be a minimal ideal of $A(\Delta)$ then if $0\neq f\in J$ then $0\neq z^{n+1}f\in I_n\cap J$. So $I_n\cap J=J$. Hence $(\cap_{n=0}^\infty I_n)\cap J=J$. But $\cap_{n=0}^\infty I_n=\{0\}$ implies $J=\{0\}$ which is contradiction. So minimal ideal does not exists.