Suppose $B$ is a boolean algebra and $M$ is an ideal. Suppose $b\in B$ and $\neg b\notin M$. Then I want to show that the ideal $I$ generated by $M\cup \{b\}$ is proper.
Note that my definition of an ideal generated by some set $S$ is the intersection of all ideals containing $S$.
I know that an ideal is improper if and only if it contains $1$ but am still unsure of how to proceed.
I have tried to suppose otherwise and derive the contradiction that $\neg b\in M$. I realize that if $I$ is improper, then the only ideal containing $M\cup \{b\}$ is $B$.
But I have not had success and would appreciate if someone could explain a proof.
You may want to characterize the ideal $I$ generated by $M \cup \{b\}$ as follows:
$$ I = \{\, b_0 \vee a : b_0 \leq b \text{ and } a \in M \,\} \enspace. $$
This result is proved in, say, Givant and Halmos, Introduction to Boolean Algebras, Chapter 18.
You can then prove that $\neg b \not\in I$ by noting that $\neg b \in I$ implies the existence of $b_0 \leq b$ and $a \in M$ such that
$$ \neg b = b_0 \vee a \enspace. $$
In this answer you find a proof that this can only be if $\neg b = a$, contradicting the assumption that $\neg b \not\in M$.
Since ideals are down sets, $I = B$ implies $\neg b \in I$; since $\neg b \not\in I$, it must be $I \neq B$.