Is a finite inverse limit of noetherian rings noetherian? 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?
 A: 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):
$$\require{AMScd}
\begin{CD}
A @>>> k[x,y]\\
@VV\varphi V @VV\varphi V \\
k @>>> k[x,y]
\end{CD}$$
A: 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.
