Let $k$ be a field. I am unable to find a precise definition of essentially smooth $k$ schemes, but I will stick to this definition below, since this is exactly what I need:

Definition: A $k$-scheme $X$ is called essentially smooth if it is a filtered projective limit $X=\underset{\longleftarrow}{\rm lim}\ X_i $ of smooth $k$-schemes $X_i$ such that all transition maps $\phi_{ij}:X_i\to X_j$ are affine and etale.

Q1: Are essentially smooth schemes is noetherian? (just like Henselisation of the local ring of a point in a variety).

Q2:What if one drops conditions '$X_i$ smooth' and/or '$\phi_{ij}$ etale' from the above definition?

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    $\begingroup$ What about $k[x_0,x_1,x_2,x_3,\dots,x_n,\dots]/\langle x_0x_1-1,x_2^2-x_1,x_3^3-x_2,\dots,x_{n+1}^{n+1}-x_n,\dots \rangle$? This ring is the union of the subrings $R_n$ generated by $x_0,\dots,x_n$, i.e., $R\cong k[x_n,x_n^{-1}]$. Each transition map is etale. Yet the union is not Noetherian. $\endgroup$ May 5 '15 at 9:35

This answer is essentially the same as the comment above. Let $d$ be any positive integer that is prime to the characteristic of $k$. Let $R$ be the $k$-algebra, $$ R=k[x_0,x_1,x_2,\dots,x_n,\dots]/\langle x_0x_1 - 1, x_2^d - x_1, x_3^d - x_2,\dots, x_{n+1}^d-x_n,\dots \rangle. $$ For every integer $n$, define $R_n\subset R$ to be the $k$-subalgebra generated by $x_0,x_1,\dots,x_n$. Then $R_n$ is isomorphic to $k[x_n,x_n^{-1}]$, which is a smooth $k$-algebra. Moreover, the transition map $R_n\to R_{n+1}$ is the same as $$ f_n : k[x_n,x_n^{-1}] \to k[x_{n+1},x_{n+1}^{-1}], \ \ f(x_n) = x_{n+1}^d. $$ This is étale since $d$ is prime to the characteristic. The ring $R$ is not Noetherian


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