# Example of a ring with an infinite inclusion chain of ideals [closed]

I'm trying to track down an example of a ring in which there exists an infinite chain of ideals under inclusion. (i.e. $I_1 \subsetneq I_2 \subsetneq I_3 \subsetneq...$)

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## closed as off-topic by quid, Najib Idrissi, Silvia Ghinassi, heropup, wythagorasFeb 29 at 18:39

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I've removed very many comments from this question, as they were deviating increasingly away from math. For meta discussion, I encourage using Meta, or when appropriate, Chat. – mixedmath Feb 29 at 1:38
This question was recently merged with another question, although the two had significant differences. In particular, the answers below do not answer the merged question. – RghtHndSd Mar 2 at 16:30
@RghtHndSd Discussed here: meta.math.stackexchange.com/questions/22698/… – user26857 Mar 6 at 20:24

By definition, such a ring is non-Noetherian. A good example of a non-Noetherian ring is $F[X_1, X_2, X_3,...]$, the ring of polynomials over a field F in countably infinite indeterminates. In this ring, we have the infinite chain of generated ideals $(X_1) \subsetneq (X_1, X_2) \subsetneq (X_1, X_2, X_3) \subsetneq...$.

Claim: $X_{n+1} \notin (X_1, ..., X_n)$

Proof: Suppose for contradiction that $X_{n+1} \in (X_1, ..., X_n)$. Then, we would be able to write $X_{n+1} = a_1 \cdot X_1 \,+ ... + \,a_n \cdot X_n$ for some $a_1, ..., a_n \in F[X_1, X_2, X_3,...]$. If we take all the indeterminate arguments of each $a_i$ to be $0$ (there are finitely many arguments for each), it follows that $X_{n+1} = a_k \cdot X_k\, + ... + \,a_l\cdot X_l$, where $a_k,...,a_l$ are the constant functions among these $a_1, ..., a_n$. This is a contradiction because $X_{n+1}$ can now be written as a linear combination of certain other indeterminates, when its choice should be unconstrained.

To see this last step more clearly, it can help to actually consider $f(X_{n+1}) = X_{n+1}$ and $g(X_k,...,X_l) = a_k \cdot X_k\, + ... + \,a_l\cdot X_l$. Note that both of these polynomials reside in $F[X_1, X_2, X_3,...]$. Moreover, by the results above, $f(X_{n+1}) \equiv g(X_k,...,X_l)$. Now take $X_{n+1} = 1$ and $X_k= ...=X_l=0$. Then, it follows that

$$1 = f(1) = g(0,...,0) = a_k \cdot 0\, + ... + \,a_l\cdot 0 = 0$$ Hence, $1=0$, which is a contradiction because F is a field and taken to have nontrivial multiplication. The claim follows.

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For the same example and a correct proof of $X_{n+1}\notin(X_1,\dots,X_n)$ the reader is referred to this answer. – user26857 Feb 29 at 12:02
My proof is indeed correct. Appreciate the link though. The more cross references the better. – Will Byrne Feb 29 at 12:31
No, it's not correct: you are confusing $X_{n+1}$ by $X_n$ repeatedly. – user26857 Feb 29 at 12:33
As I said yesterday, it was pretty late when I wrote this up initially. Fixed. – Will Byrne Feb 29 at 12:51

The classical example is the ring of polynomials in a countable number of indeterminates: $k[x_1,x_2,\dots,x_n,\dotsc]$ and the chain is $$0\subsetneq (x_1)\subsetneq(x_1,x_2)\subsetneq\dotsb$$

Note that $x_{n+1}\notin(x_1,\dots,x_n)$ by a standard argument: suppose $$x_{n+1}=f_1 x_1 + f_2x_2 + \dots + f_nx_n$$ where $f_i\in k[x_1,x_2,\dots,x_n,\dotsc]$ (for $i=1,2,\dots,n$). Substitute $x_i=0$ for $i=1,2,\dots,n$; this gives $x_{n+1}=0$, a contradiction.

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For a similar answer the reader is referred to math.stackexchange.com/a/500898/121097. – user26857 Feb 29 at 12:39
@user26857 Yes, it's somewhat folklore. I added the proof because the one in the accepted answer is not well worded, in my opinion. – egreg Feb 29 at 12:42

Edit: By "ring" I mean commutative ring with identity.

Every ring is a quotient of an infinite polynomial ring. Let $R$ be any ring, and let $\{r_i\}_{i \in I}$ be a set of generators (under the ring operations). For example, we could simply take every single element of the ring $R$. Then define a ring homomorphism $\varphi : \mathbb{Z}[x_i : i \in I] \rightarrow R$ by sending $x_i \rightarrow r_i$. This is a surjective homomorphism, and thus $$R \cong \mathbb{Z}[x_i : i \in I]/\ker \varphi.$$

In particular, any non-Notherian integral domain is a quotient of an infinite polynomial ring. So in some sense, your example is a fundamental one.

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For a bi-infinite chain of ideals (and bounded transcendence degree), start from $k[X]$ and adjoin $2^n$-th roots of $X$ for all $n$.

The inclusion ordering of the principal ideals $(X^{2^i})$ for $i \in \mathbb{Z}$ is equivalent to the reversed ordering of integers, via the correspondence $2^i \leftrightarrow i$.

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Consider the ring $R = C([0,1], \mathbf{R})$ of continuous functions $f \colon [0,1] \to \mathbf{R}$.

For $0 \lt t \lt 1$, let $I_t = \{ f \in R \mid f(x) = 0 \text{ for all } 0 \leq x \leq t \}$ be the ideal of functions vanishing on all of $[0,t]$. Then $I_s \supsetneq I_t$ whenever $s \lt t$.

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