Take the tour ×  Les S be a scheme. Does there exist a
faithfully flat morphism T to S with T a locally Noetherian
scheme?  share|improve this question No, there is not
necessarily such a $T$. For instance, let $S$ be
$\text{Spec}(A)$ where $A=k[[x_1,x_2,...]]$. If there is such a
faithfully flat $T$, then there is a point $t$ of $T$ which maps
to the closed point of $S$. Let $B$ be the local ring of $T$ at
$t$. Then there is a faithfully flat local homomorphism $A\to
B$. In particular, the induced map $m_A/m_A^2 \to m_B/m_B^2$ is
injective. Since $m_A/m_A^2$ is infinite dimensional, so is
$m_B/m_B^2$, contradicting that $B$ is Noetherian. –  Jason
Starr Dec 15 '11 at 16:27 @Jason: This is not a comment - it is
an answer. –  Martin Brandenburg Dec 15 '11 at 19:52 @Jason: If
$A \to B$ is a local extension of DVR's (automatically
faithfully flat), then the map $m_A/m_A^2 \to m_B/m_B^2$ may be
zero (e.g. if there is ramification). –  Akhil Mathew Dec 16 '11
at 1:55 @Akhil: You are right. That map may be zero. Bu