Nsryjdtyk

Is a scheme is Noetherian equivalent to that the underlying topological space is Noetherian and all its stalks are Noetherian?

This is false. The easiest counterexample I could come up with is the following "affine line with embedded points at every closed point":

Example. Let $k$ be an infinite field, let $R = k[x]$, and for each $alpha in k$ let $R_alpha = R[y_alpha]/((x-alpha)y_alpha,y_alpha^2)$. Then $R_alpha$ is an affine line with an embedded prime $mathfrak p_alpha = (x-alpha,y_alpha)$ at $x = alpha$, sticking out in the $y_alpha$-direction. Finally, let
$$R_infty = bigotimes_{alpha in k} R_alpha = operatorname*{colim}_{substack{longrightarrow\I subseteq k\text{finite}}} bigotimes_{alpha in I} R_alpha$$
be their tensor product over $R$ (not over $k$); that is
$$R_infty = frac{k[x]left[{y_alpha}_{alpha in k}right]}{sum_{alpha in k}((x-alpha)y_alpha, y_alpha^2)}.$$
This is not a Noetherian ring, because the radical $mathfrak r = ({y_alpha}_{alpha in k})$ is not finitely generated. But $operatorname{Spec} R_infty$ agrees as a topological space with $operatorname{Spec} R_infty^{operatorname{red}} = mathbb A^1_k$, hence $|!operatorname{Spec} R_infty|$ is a Noetherian topological space.

On the other hand, the map $R to R_alpha$ is an isomorphism away from $alpha$, and similarly $R_alpha to R_infty$ induces isomorphisms on the stalks at $alpha$. Thus, the stalk $(R_infty)_{mathfrak q_alpha} = (R_alpha)_{mathfrak p_alpha}$ at $mathfrak q_alpha = mathfrak p_alpha R_infty + mathfrak r$ is Noetherian. Similarly, the stalk at the generic point $mathfrak r$ is just $R_{(0)} = k(x)$. Thus, we conclude that all the stalks of $R_infty$ are Noetherian. $square$