Every non-unit is in some maximal ideal I am trying to prove that every non-unit of a ring is contained in some maximal ideal. I have reasoned as follows: let $a$ be a non-unit and $M$ a maximal ideal. If $a$ is not contained in any maximal ideal, then the ideal $\langle M, a \rangle$ (that is, the ideal generated by $M$ and $a$) strictly contains $M$, a contradiction.
However, there is a detail I'm unsure about. It's easy to see that $\langle a \rangle$ is a proper ideal for any non-unit $a$, but how can I be sure that $\langle M,a \rangle$ is also a proper ideal? Couldn't there, for example, exist some $m \in M$ such that $1=m+a$, so that $\langle M,a \rangle$ is the whole ring?

Yes, it can happen that $\langle M,a\rangle$ is the whole ring, so your proof doesn't work. Try to reason as follows: let $\Sigma$ be the collection of ideals which contain $a$. Now use Zorn's lemma.
