In any ring $R$ with a multiplicative identity , does every non-unit element belongs to some maximal ideal of $R$ ?

Question

In a ring $R$ with a multiplicative identity , does every non-unit i.e. non-invertible element belongs to some maximal ideal of $R$ ?

Answer 1

No.

Search the site (or better yet, rack your brain) for the examples of rings which have elements $a,b$ such that $ab=1$ and $ba\neq 1$, and you'll have an example.

If you are silently assuming the ring is commutative, then yes. Just look at what it means for $xR$ not being a proper ideal.

Answer 2

No, an element $r$ can be left invertible, but not right invertible (example follows). Any two-sided ideal that contains $r$ is the whole ring.

Example: consider the endomorphism ring of an infinite dimensional vector space with basis $\{e_n:n\in\mathbb{N}\}$. Let $f$ be defined by $f(e_n)=e_{n+1}$. Then $f$ has a left inverse, the map defined by $g(e_{n+1}=e_n$ and $g(e_0)=0$, but no right inverse.

If the ring is commutative, the answer is yes. The ideal generated by a non unit $r$ is proper, so by Zorn's lemma it is contained in a maximal ideal: just consider the family of all proper ideals containing $(r)$ ordered by inclusion and show it satisfies the hypotheses of Zorn's lemma.