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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?

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    I tested $R=\mathbb Q[x_1,x_2,\ldots]$ as a potential counterexample, but it holds there.2012-04-03

2 Answers 2

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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.

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    "such that" is so thauch...2012-04-03
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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$

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    Infinitely true , Mariano: let us hope the others won't notice :)2012-04-03