Countable Maximal Ideals This may be simple but I can not see a way. I am looking for an uncountable ring (with 1) containing a countable maximal left ideal which is not a direct summand (as a left ideal). 
 A: Eric's result for commutative rings, that an example must have the ring no bigger than the continuum and the maximal ideal infinitely generated, extends to left ideals for non-commutative rings, assuming that by "not a direct summand" you mean "not a direct summand as a left ideal".
Let $A$ be the ring, $I$ the countable maximal left ideal, and $S$ the simple left module $A/I$. Considering these all as left $A$-modules, there can be no non-zero maps $S\to I$ or $I\to S$. It follows that there can be no non-zero map $S\to A$, since if there were, then composing with the natural epimorphism $A\to S$ would give a non-zero map $S\to S$, which must be invertible since $S$ is simple, and so $A\cong I\oplus S$ as left $A$-modules.
Hence the map $A\to A$ given by right multiplication by any $0\neq a\in A$ must restrict to a non-zero map $I\to A$, and therefore to a non-zero map $I\to I$, since there are no non-zero maps $I\to S$.
So, as in Eric's argument, $A$ acts faithfully on $I$.
EDIT: By the way, this shows that $I$ is a two-sided ideal. The quotient ring $S=A/I$ must then be a skew field, since it is simple as a left module for itself. Also, $I/I^2$ is an $S$-module, but is at most countable, so must be zero. So $I$ is an idempotent ideal.
A: Here is an example.  Choose an uncountable algebraically independent set $S\subset \mathbb{R}$, and let $k=\mathbb{Q}(S)$.  For each $x\in S$, choose a sequence $(q_n(x))$ of rationals converging to $x$.  Given any $y\in k$, write $y=f(x_1,\dots,x_m)$ for some rational function $f$ and some $x_i\in S$ and define $q_n(y)=f(q_n(x_1),\dots,q_n(x_m))$.  Note that $q_n(y)$ may not be defined since the denominator of $f$ could vanish, but for fixed $y$ it is defined for all sufficiently large $n$ since $q_n(x_i)$ must converge to $x_i$.
Now let $A$ be the set of all sequences of rationals $(a_n)$ such that for some $y\in k$, $a_n=q_n(y)$ for sufficiently large $n$.  This set has the same cardinality as $S$, and forms a ring under pointwise addition and multiplication.  There is an epimorphism $\varphi:A\to k$ that sends any sequence to its limit.  The kernel of $\varphi$ is a countable maximal ideal of $A$.  Explicitly, $\ker(\varphi)$ is the set of sequences of rationals that are eventually $0$.  This ideal is not finitely generated, and so in particular cannot be generated by an idempotent.
Note that this construction can give an example having any cardinality up to the continuum.  The role of the real numbers in this construction is only to guarantee that $q_n(y)$ is always defined for sufficiently large $n$; you might think you can get $A$ to be arbitrarily large by a slightly more clever construction.  It turns out that this is impossible: there are no examples for which either the ring has cardinality larger than the continuum or the maximal ideal is finitely generated.  Let $A$ be an uncountable ring and let $I\subset A$ be a countable maximal ideal.  Then $A$ acts on $I$ by multiplication, giving a homomorphim $\alpha:A\to End_A(I)$.  If either $A$ is larger than the continuum or $I$ is finitely generated, $A$ will have larger cardinality than $End_A(I)$.  In either case, we can conclude that the kernel of $\alpha$ is uncountable.
In particular, we can find some $k\in \ker(\alpha)\setminus I$.  Now by maximality of $I$, there is some $a\in A$ and $i\in I$ such that $ak=1-i$.  But then for any $j\in I$, $0=akj=j-ij$.  This implies $i$ is an idempotent generator of $I$. (This argument assumes $A$ is commutative; see Jeremy Rickard's answer for a variant that works for noncommutative rings as well.)
