The question is following:

Is there an uncountable reduced ring (i.e., a ring with no non-zero nilpotent element) $R$ (with identity) such that $R[x]$ has only a countable number of maximal left ideals?

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    $\begingroup$ Your question is tagged "commutative algebra", but "left ideals" suggest $R$ is not assumed commutative. And then, what is $R[x]$? $\endgroup$ – Laurent Moret-Bailly Sep 16 '13 at 13:53
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    $\begingroup$ As a "commutativist", I don't know what you mean by polynomials over a noncommutative ring. Specifically, does $x$ commute with $R$? $\endgroup$ – Laurent Moret-Bailly Sep 16 '13 at 20:41
  • $\begingroup$ Please explain, does "reduced" assume no nil radical? or no Jacobson radical? (I think that, in the first case, the answer is trivial.) $\endgroup$ – Sasha Anan'in Oct 8 '13 at 22:28
  • $\begingroup$ @SashaAnan'in Reduced means that there is no non-zero nilpotent element. $\endgroup$ – user39982 Oct 9 '13 at 20:25

Let $r\in R$. By the Zorn lemma, there is a maximal left $R$-submodule $M_r$ in $R+Rx\cong R\oplus R$ such that $M_r\left[\begin{matrix}0&1\\r&0\end{matrix}\right]\subset M_r$ and $1\not\in M_r$. Denote $L_r:=R[x](x^2-r)+R[x]M_r$. As $x^{2n}=(\sum_{i=0}^{n-1}r^{n-i-1}x^{2i})(x^2-r)+r^n$ and $x^{2n+1}=(\sum_{i=0}^{n-1}r^{n-i-1}x^{2i+1})(x^2-r)+r^nx$, every element in $L_r$, being considered modulo $R[x](x^2-r)$, lies in $M_r$ and $L_r=R[x](x^2-r)\oplus M_r$ as a left $R$-module. Moreover, $M_r=L_r\cap(R+Rx)$. Indeed, it suffices to observe that $M_rx=M_r\left[\begin{matrix}0&1\\r&0\end{matrix}\right]\mod R[x](x^2-r)$ and that $p(x)(x^2-r)\in M_r$ implies $p(x)=0$. Now it is easy to see that $L_r$ is a maximal left ideal in $R[x]$. Such ideals are different for many different $r$'s. Indeed, if $L_a=L_b$, then $a-b\in L_a=L_b$, hence, $a-b\in M_a=M_b$. Consequently, $\sum_{k\ge0}R(a-b)a^k+\sum_{k\ge0}R(a-b)a^kx\subset M_a=M_b$. We can take uncountably many choices of $a,b\in R$ and $M_a,M_b$ such that either $M_a\ne M_b$ or $\sum_{k\ge0}R(a-b)a^k+\sum_{k\ge0}R(a-b)a^kx\not\subset M_a=M_b$. [Here is a gap in the proof because of lack of time.] Therefore, there is no ring in question.

  • $\begingroup$ why $L_r$ is maximal and just where did you use the fact that $R$ is reduced? $\endgroup$ – dimo Oct 16 '13 at 5:26
  • $\begingroup$ @ dimo $L_r$ is maximal by construction: if $L'_r\supset L_r$, then $M'_r:=L'_r\cap(R+Rx)\supset M_r$ and $1\in M'_r$ by the choice of $M_r$; I did not use yet that $R$ has no nil elements, it would happen in the omitted part of the proof. $\endgroup$ – Sasha Anan'in Oct 16 '13 at 6:50
  • $\begingroup$ Sorry I did not understand. As you have said, $L_r$ is not contained in another $L'_r$ but why $L_r$ is not in any ideal of $R[x]$ !? $\endgroup$ – dimo Oct 16 '13 at 15:17
  • $\begingroup$ @dimo Sorry for bad notation. Please take, say, $N$ in place of $L'_r$. If $N\supsetneq L_r$, it is possible to conclude that $N\cap(R+Rx)\supsetneq M_r$, implying $1\in N$ by the choice of $M_r$ (because $N\cap(R+Rx)$ is closed relatively the multiplication by the above $2\times2$ matrix). $\endgroup$ – Sasha Anan'in Oct 16 '13 at 15:51

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