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There is a basic theorem in the geometry of schemes saying that the Spec of a Noetherian ring is a Noetherian topological space. It can be formulated as the ACC condition implies the ACCR condition (the ascending chain condition for radical ideals). I'm wondering if the converse is true.

In case it is tangentially relevant, there is also the ACCP condition (the ascending chain condition for prime ideals). Clearly ACCR implies ACCP, but ACCP does not imply ACCR, e.g., an infinite product of $\mathbb F_2$.

If it isn't clear, we assume AC all the time.

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The converse is not true, i.e., there exists a non-noetherian affine scheme whose underlying topological space is noetherian. (Examples: Valuation rings of rank 2.)

More results and examples about such questions can be found for example in the following articles:

J. Ohm, R. L. Pendleton, Rings with noetherian spectrum, Duke Math. J. 35 (1968), 631-639.

W. Heinzer, J. Ohm, Locally noetherian commutative rings, Trans. Amer. Math. Soc. 158 (1971), 273-284.

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Here's one example: Let k be a field and V an infinite dimensional vector space over k. Let R=k⊕V, with (a,v)(b,w)=(ab,aw+bv). Then R has one prime = (0)+V, (0)+W is an ideal for any subspace W⊂V, and there are infinite chains of subspaces so R isn't Noetherian.

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  • $\begingroup$ Geometrically, this can be considered as a single point with thickening in infinitely many directions, so the single radical ideal doesn't really capture the geometry of the thickening. $\endgroup$ – Fan Zheng Mar 4 '15 at 18:54
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Here's another example. $k$ any field. Let $x:=x_1$.

$$A:=k[x_1,x_2,\ldots]/(x_i^2-x_{i-1})$$

Then your infinite ascending chain is of course

$$(x_1)\subset (x_2)\subset (x_3)\subset\cdots$$

but for a radical ideal $\mathfrak a$, one has $\mathfrak a=\operatorname{rad}\mathfrak a=\operatorname{rad}(\mathfrak a\cap k[x])$, and for two radical ideals $\mathfrak a,\mathfrak b$, $\mathfrak a\subset\mathfrak b\Leftrightarrow \mathfrak a\cap k[x]\subset \mathfrak b\cap k[x]$ and $\mathfrak a\subsetneq\mathfrak b\Leftrightarrow \mathfrak a\cap k[x]\subsetneq \mathfrak b\cap k[x]$, so by intersecting an ascending chain of radical ideals $$\mathfrak a_1\subset \mathfrak a_2\subset\mathfrak a_3\subset\cdots$$ in $A$ with $k[x]$, one obtains an ascending chain of radical ideals in $k[x]$, with strict inclusions maintained. Now $k[x]$ is noetherian.

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