What is the length of the following local ring Let $f:Y\to X$ be a finite etale cover of smooth projective connected varieties. (Or, just a finite degree connected topological cover of connected Riemann surfaces.)
Let $y\in Y$ and let $x=f(y)$. Let $Y_x$ be the fibre $f^{-1}(x)$. (It contains $y$.)
What is the length of the local ring $\mathcal{O}_{Y_x,y}$?
Is it 1 or $\deg f$?
 A: Let's remain in dimension one: algebraic  curves  over an algebraically closed field $k$ of characteristic zero or  Riemann surfaces.
We have a morphism of discrete valuation rings  rings $f^*:\mathcal O_{X,x}\to \mathcal O_{Y,y}$ sending a uniformizing parameter $z\in \mathcal O_{X,x}$ to $u\cdot w^d \in\mathcal O_{Y,y}$, where $w$ is a uniformizing parameter of $\mathcal O_{Y,y}$ and $u\in \mathcal O_{Y,y}^*$ is a unit.
The local ring $\mathcal O_{Y_x,y}$ at $y$ of the fiber is then $\mathcal O_{Y_x,y}=\mathcal O_{Y,y}/w^d\mathcal O_{Y,y}$.
Ramification occurs when $d\gt1$ and contrarywise, $d=1$ means that $f$ is étale at $y$, the case your question is about.
Well, if $d=1$ the local ring of $y$ in the fiber of $x$ is $ \;\mathcal O_{Y,y}/w^d\mathcal O_{Y,y}=\mathcal O_{Y,y}/w^1\mathcal O_{Y,y}=k$ so that the length you are asking about is $length (\mathcal O_{Y_x,y})= length (k)=1$.
So it is important to distinguish between the length of $\mathcal O_{Y_x,y}$ which is a purely local invariant measuring the ramification of $f$ at $y$ and the global dimension $dim_k\Gamma(Y_x,\mathcal O_{Y_x} )$, which is the sum of these local invariants and is always equal to $deg(f)$ independently of any hypothesis of étaleness of the points in that fiber. 
[I have tried to to write-up the answer so that it applies both to smooth  complete algebraic curves and to Riemann surfaces, for example by using $z$ and $w$ for the uniformizing parameters so that you can think of them also as holomorphic coordinates]
