Let $A \subset B$ be faithfully flat and $B$ Noetherian. Prove that $A$ is also Noetherian.
My idea was to construct an exact sequence: $0 \rightarrow A \rightarrow A\otimes B \rightarrow ? \rightarrow 0$ (The first is oke because of the faithful flatness)
If I have these I can use the proposition that says: when $0 \rightarrow M'\rightarrow M \rightarrow M'' \rightarrow 0$ exact then: M Noetherian iff M' and M'' Noetherian.
I just can't complete it yet. Anyone ideas? Or is there a completely other way to proof it.
If $A \to B$ is a faithfully flat ring map, then for any $A$-ideal $I$, $I = IB \cap A$, i.e. $I = I^{ec}$. Take an ascending chain of ideals $I_0 \subseteq I_1 \subseteq \ldots$ in $A$. Extending to $B$ gives an ascending chain of $B$-ideals $I_0^e \subseteq I_1^e \subseteq \ldots$ which stabilizes by hypothesis. Contracting back to $A$ gives that the original chain $I_0 = I_0^{ec} \subseteq I_1 = I_1^{ec} \subseteq \ldots$ stabilizes, so $A$ is Noetherian.