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?

In fact you can show that every proper ideal can be enlarged to a maximal ideal.
The proof is a classical application of Zorn's lemma, combined with the lattice isomorphism theorem.
Let $I \subset R$ be a proper ideal, then apply the lattice isomorphism theorem :
$$\{J\ \text{ ideals in }R\mid  I \subset J \subset R \} \cong \{\text{ideals of }\  R/I\}$$
We know by the Zorn's lemma, every rings has maximal ideal,  therefore there is a maximal ideal in the RHS set, which one to one correspond to the maximal ideal of $R$ contains $I$ in the LHS.
Therefore apply to this question ,since $a$ is no unit, $(a)$ is proper ideal ,there always exist such maximal ideal contains it.
