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Here is a counterexample. Let $\mathbb G_{\rm m}$ act on $\mathbb A^2$ by $t\cdot(x,y) = (tx,t^{-1}y)$, and let $f\colon \mathbb A^2 \to \mathbb A^1$ be defined by $f(x,y) = xy$. I am positive that w …

I meant to write this as a comment, but it won't fit. In my opinion, you are confusing the étale analytic with the étale algebraic topology. The étale analytic topology is essentially the same the sa …

There are many counterexamples to this. Suppose that $S$ is a smooth surface over $\mathbb C$. Let $T \to S$ be a finite morphism from another smooth surface $T$, and consider a factorization $T \to V …

As a module, $A[[X]]$ is the product of a countable family of copies of $A$. It is known that the product of flat $A$-modules is flat if and only if the ring $A$ is coherent, that is, every finitely g …

I assume you want $Y \to X$ to be proper. The answer is a definite no, in general. For example, take a polynomial $f: \mathbb A^2 \to \mathbb A^1$; such a $Y$ would have to be finite over $\mathbb A^2 …