Left identity and maximal left ideal We can use Zorn's lemma to prove that for a non-zero ring $R$ with left(right) identity, every proper right(left) ideal of $R$ is contained in a maximal right(left) ideal. The existence of left (right) identity is important.
But we can't use the same method to prove the existence of a maximal left ideal.
There is a lot of examples of non-zero commutative rings without identity which don't have a maximal ideal. For example, $(\mathbb{Q},+)$ doesn't have a maximal subgroup. Then let $xy=0$ for all $x,y\in\mathbb{Q}$.
But I don't know if there is a ring with left identity but doesn't have any maximal left ideal.
 A: I think there is not, owing to the symmetry of the Jacobson radical in rings without identity.
Jacobson developed a characterization of (what is now called) the Jacobson radical as the intersection of all regular right maximal ideals, equivalent to the intersection of all regular maximal left ideals.
Here “regular” applied to a right ideal $T$ means “there exists an element $e$ such that $ex-x\in T$ for all $x\in R$”. Clearly a left identity would suffice for making all right ideals regular.  The left hand counterpart is apparent.
So if the ring has a left identity, its Jacobson radical is proper, indicating that there must exist some maximal (proper, of course) regular left ideal in $R$ too.
A: @rschwieb's argument shows that if a ring has a right or left identity, then it has maximal right ideals and maximal left ideals.  Sometimes it is useful to know not just that maximal objects exist but that every proper object is contained in a maximal object.  I will give an example to show that even if a ring has a left identity, there may be a proper left ideal that is not contained in any maximal left ideal.
Let's start with a general construction of rings without (two-sided) identities and then specialize it to get the claimed example.  Let $R$ be a commutative ring with identity, $A$ a unital $R$-module, and $\epsilon:A\to R$ an $R$-module homomorphism.  We can make $A$ into an $R$-algebra (without identity) by defining the following multiplication on $A$:  $a*b=\epsilon(a)b$.  If $\epsilon(e)=1$, then $e*a=\epsilon(e)a=a$, and so $e$ is a left identity.  Suppose such an $e$ exists: clearly $\epsilon:A\to R$ is surjective in this case.  It's then easy to see that the left ideals of $A$ are precisely the $R$-submodules of $A$.  Thus maximal left ideals are precisely maximal $R$-submodules.
Now consider the setting $R=\mathbb{Z}$, $A=\mathbb{Q}\oplus\mathbb{Z}$, $\epsilon((q,n))=n$.  Make $A$ into a ring as above; it has left identity $(0,1)$ (or indeed $(q,1)$ for any $q$).  Let $I=0\oplus\mathbb{Z}$.  A maximal left ideal of $A$ containing $I$ is the same thing as a maximal subgroup of $A$ containing $I$, and such subgroups have the form $H\oplus I$ for a maximal subgroup $H$ of $\mathbb{Q}$.  Since $\mathbb{Q}$ has no maximal subgroups, no maximal left ideal contains $I$.
