Let $R$ be a commutative ring with a multiplicative identity such that there is a finitely generated $R$-algebra that is Noetherian. Is $R$ Noetherian then?
I tried to prove this using the fact that the homomorphic image of a Noetherian ring is Noetherian but I can not find an ideal the quotient by which is isomorphic to $R$ for general finitely generated algebras. In some special cases, that can be done (e.g. for the free polynomial algebra take the ideal $(X_1, \dots, X_n)$).
Let $R$ be any non-zero ring. If you are willing to accept the existence of maximal ideals, then there's always a finitely generated non-zero $R$-algebra $S$ which is noetherian.
Just take $\mathfrak{m}$ a maximal ideal of $R$ and then $R/\mathfrak{m}$ is, being a field, noetherian.
Therefore every non-noetherian ring will still have finite non-zero morphisms to noetherian rings
Note that noetherianity won't descend down finitely presented morphisms either. Continuing with the reasoning from maximal ideals, there are many non-noetherian rings which have finitely generated maximal ideals (e.g. construct an appropriate rank > 1 valuation domain).