Let $A$ be the disk algebra (i.e. the algebra of all functions that are continuous on the closed unit disk and analytic on the open unit disk) and let $A_{0}=\{f\in A:f(0)=0\}$. Then $A_{0}$ is a closed subalgebra of $A$. I am trying to show that $A_{0}$ has a non-modular maximal ideal. (An ideal $I$ in $A$ is modular if there exists $u\in A$ such that $a-au\in I$ and $a-ua\in I$ for all $a\in A$.)
I considered $J=\{f\in A:f(0)=f'(0)=0\}$ and if I checked things correctly, it seems that $J$ is a non-modular ideal in $A_{0}$. However I'm not sure whether it is maximal.
Indeed, $J$ is an ideal of $A_0$. Any product of two elements of $A_0$ belongs to $J$, since $(f\cdot g)'(0) = f'(0)g(0) + f(0)g'(0) = f'(0)\cdot 0 + 0\cdot g'(0) = 0$, and hence $J$ is non-modular, as we have $au\in J$ for all $a,u\in A_0$, and hence $a-au = a-ua \in J \iff a\in J$. Also, $J$ is a $1$-codimensional (closed) linear subspace of $A_0$, hence it is a maximal ideal.