MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

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 $\bigcup_k I_k=\langle \{x_i\}_{i\in{\mathbb N}}\rangle$.

share|cite|improve this answer
$\mathbb{C}[T]$ is the group algebra of any group $T$ with uncountably many elements which you divide into parts $\{x_i\}$ and $\{y_\alpha\}$? – GiroCont Nov 23 '11 at 17:27
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. – Ilya Bogdanov Nov 23 '11 at 18:19
What does it mean $x_{k+1}y_\alpha$ then? – GiroCont Nov 23 '11 at 18:34
Sorry, right. I was still thinking about the group algebra... – GiroCont Nov 23 '11 at 19:29

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.