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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. –  Alex Becker Apr 3 '12 at 22:09
    
+1: That was a big step in the right direction, Alex! –  Georges Elencwajg Apr 3 '12 at 22:35

2 Answers 2

up vote 10 down vote accepted

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... –  Asaf Karagila Apr 3 '12 at 22:32

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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You and I have already used the same ring for uncountably many examples on this site :) –  Mariano Suárez-Alvarez Apr 3 '12 at 23:26
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Infinitely true , Mariano: let us hope the others won't notice :) –  Georges Elencwajg Apr 3 '12 at 23:29

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