Check whether $\mathbb{Q}[X_{1},X_{2},...]/(f_{1},f_{2},...)$ is a Bézout domain, Noetherian, Factorial, or PID with $f_{i} =X_{i}-X_{i+1}^2$.
I don't really have a clue how to start. It would be nice, if you would give me some steps and show me how I can proof those things. I can't even imagine how this ring looks like.
Thank you.
This ring is isomorphic to $$\Bbb{Q}[X, X^{1/2}, X^{1/4}, X^{1/8}, X^{1/16}, ...]$$ which is at least a domain. Note that this is the union of the following chain of subrings $$\Bbb{Q}[X] \subsetneq \Bbb{Q}[X^{1/2}] \subsetneq \Bbb{Q}[X^{1/2^2}] \subsetneq \Bbb{Q}[X^{1/2^3}] \subsetneq ...$$
This is not Noetherian because the chain of ideals $$(X) \subsetneq (X^{1/2}) \subsetneq (X^{1/2^2}) \subsetneq (X^{1/2^3}) \subsetneq ...$$ is not stationary.
Let's prove that it is a Bézout domain: let $I=(f_1 , \dots , f_k)$ be a finitely generated ideal. Let $$n= \min \{ r: f_1 , \dots , f_k \in \Bbb{Q}[X^{1/2^r}] \}$$ Since $\Bbb{Q}[X^{1/2^n}]$ is a PID, there exists $g \in \Bbb{Q}[X^{1/2^n}]$ such that $$(f_1 , \dots , f_k)\Bbb{Q}[X^{1/2^n}] = g \Bbb{Q}[X^{1/2^n}]$$ this $g$ is indeed a generator of $I$, so that $I$ is generated by one element.
With a similar argument you can prove that this ring is a factorial ring (indeed a UFD).
EDIT AFTER 7 YEARS: This ring is not a UFD. The obvious example is $$X= (X^{1/2})^2 = (X^{1/4})^4 = (X^{1/8})^8 = \cdots$$ which is not a product of irreducibles. This was pointed out in the comments, I'm sorry for not having modified my answer in these years.
