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Let $\{A_i\}$ be an inverse system of (commutative, unital) Noetherian rings with a finite index set. Is $\varprojlim A_i$ also a Noetherian ring?

Avi Steiner
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2 Answers2

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The answer is no.

Let $\varphi : k[x,y] \to k[x,y], f \mapsto f(x,0)$ and

$$A = \{ f \in k[x,y] ~|~ f(x,0) \in k \} = \varphi^{-1}(k).$$

$A$ is well known to be non-noetherian - $(y,xy,x^2y,x^3y, \dotsc)$ is not finitely generated - but it fits in the following cartesian square (the horizontal arrows are inclusions):

$$\require{AMScd} \begin{CD} A @>>> k[x,y]\\ @VV\varphi V @VV\varphi V \\ k @>>> k[x,y] \end{CD}$$

MooS
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Yes, because $\varprojlim A_i$ is isomorphic to one of the $A_i$ in the case of a finite directed index set.

In a directed set $I$, any two elements must have a maximum, hence $I$ has a maximum $i_0$ in the case where $I$ is finite. So $i_0 \geq i$ for all $i \in I$, and we have ring homomorphisms $\phi_i = \phi_{i,i_0}: A_{i_0} \rightarrow A_i$ for all $i$. Check that $$\varprojlim A_i = \{ (a_j) \in \prod\limits_{j \in I} A_j : a_i = \phi_i(a_{i_0}) \textrm{ for all } i \in I\}$$ You can then see that the projection homomorphism $\varinjlim A_i \rightarrow A_{i_0}$ is injective and surjective.

D_S
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    The question doesn't assume that $I$ is directed, see the counterexample given by MooS. I would slightly rephrase the first sentence to make it clearer. – porkynator Jul 21 '16 at 12:44
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    I see, I was not aware the notion of inverse limit applied to a nondirected index set. – D_S Jul 21 '16 at 13:23