Let $A$ be the disk algebra and $A_0 = \{f \in A \mid f(0) = 0\}$. I am trying to give an example of a maximal non-modular ideal in $A_0$.
I have tried $I=\{f\in A_0 \mid f(1) = 0\}$ and proved that it is maximal. But I cannot seem to show that it is non-modular. I tried to argue by contradiction but having $f(1) - u(1)f(1) =0$ doesn't give the desired contradiction. Is this ideal really modular? I doubt it but then I can't prove that it isn't. Are there any other ideals in $A_0$? I can't come up with any. If $p(x) = (x-1)$ then the ideal $p \cdot A_0$ doesn't help either.
Your $I$ is modular, you can take $u(z) = z$ (or any $u\in A_0$ with $u(1) = 1$), then $f - u\cdot f \in I$.
Basically the same works for any ideal of the form $I = \{ f\in A_0 : f(w_i) = 0, 1 \leqslant i \leqslant n\}$ defined by the vanishing in finitely many points.
What works is an ideal involving the derivatives of $f$ in $0$, for example
$$N = \{ f \in A_0 : f'(0) = 0\} = A_0^2.$$
It is clear that $N$ is an ideal, and since $N = A_0^2$, we have $f - u\cdot f \notin N$ for every $f\in A_0\setminus N$.