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Suppose $\mathfrak{A}$ is an unital algebra over complex numbers and $\mathfrak{J}$ is chain of left-ideals in $\mathfrak{A}$ ordered by inclusion such that none of its elements is countably generated. Clearly, the union $\bigcup \mathfrak{J}$ is a left-ideal. Can it be countably generated? I am interested in the commutative case as well.

Recently, I asked a similar question for Boolean algebras but I prefer these two questions not to be merged.

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Let $T=\{x_i\}_{i\in{\mathbb N}}\cup \{y_\alpha\}_{\alpha\in{\mathbb R}}$. Consider an algebra $A={\mathbb C}[T]$, and denote $I_k=\langle \{x_i\}_{i=1}^k\cup\{x_{k+1}y_\alpha\}_{\alpha\in{\mathbb R}}\rangle$. Then none of $I_k$'s is countably generated, but $\bigcup_k I_k=\langle \{x_i\}_{i\in{\mathbb N}}\rangle$.

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  • $\begingroup$ $\mathbb{C}[T]$ is the group algebra of any group $T$ with uncountably many elements which you divide into parts $\{x_i\}$ and $\{y_\alpha\}$? $\endgroup$
    – GiroCont
    Commented Nov 23, 2011 at 17:27
  • $\begingroup$ It is just the algebra of polynomials with $T$ as a set of variables. Or, if you prefer, the free commutative algebra with $T$ the set of generators. $\endgroup$
    – Ilya Bogdanov
    Commented Nov 23, 2011 at 18:19
  • $\begingroup$ What does it mean $x_{k+1}y_\alpha$ then? $\endgroup$
    – GiroCont
    Commented Nov 23, 2011 at 18:34
  • $\begingroup$ Sorry, right. I was still thinking about the group algebra... $\endgroup$
    – GiroCont
    Commented Nov 23, 2011 at 19:29

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