Do countable unital rings with uncountably many distinct right ideals have uncountably many maximal right ideals? Suppose we are given a countable unital ring $R$ with uncountably many distinct right ideals. Does it follow from this that $R$ has uncountably many maximal right ideals?
 A: No.   
Take $R= \mathbb Q[X_0,X_1,...,X_n,...]/\langle X_iX_j\mid i,j\in \mathbb N\rangle=\mathbb Q[x_0,x_1,...,x_n,...]$
The only maximal  ideal (actually only prime ideal !) is 
$\langle x_0,x_1,...,x_n,...\rangle$ but $R$ has  a family of distinct ideals indexed by the uncountably many subsets $P\subset \mathbb N$, namely $$ I_P=\langle   x_i\mid i\in P\rangle=\operatorname {vect}_\mathbb Q (x_i\mid i\in P)$$   
Edit
I hadn't seen Mariano's answer when I posted mine a few minutes later, but our rings are actually isomorphic : if his $V$ has basis $(v_i)_{i\in \mathbb N}$ over $\mathbb Q$ we have an isomorphism (of $\mathbb Q$-algebras even)   
$$ \mathbb Q\oplus V \stackrel {\cong}{\to} \mathbb Q[x_0,x_1,...,x_n,...]:(q,\sum q_iv_i)\mapsto q+\sum q_ix_i$$                         
A: Let $V$ be a $\mathbb Q$-vector space of countable dimension and let $R=\mathbb Q\oplus V$ with commutative multiplication such that the injection $\mathbb Q\to R$ is a map of rings, multiplication between $\mathbb Q$ and $V$ is the obvious one, and $v\cdot w=0$ for all $v$, $w\in V$.
Every subspace of $V$ is an ideal of $R$, so there are uncountably many of these, yet $R$ is local.
