Relating degree of projective varieties and a finite morphism between them Let $f:X\to Y$ be a finite, dominant morphism of projective varieties. I suspect that $\deg(f)\cdot\deg(Y)=\deg(X)$ always holds, where $\deg(f)=[K(X):K(Y)]$. If required, we may assume that $X$ and $Y$ are projective over an algebraically closed field (of characteristic zero even). My question is whether this is true and if so, I would love to have a reference.
 A: This  is not  true in general. 
In fact, let $H_X$ be a hyperplane section of $X$ and $H_Y$ be a hyperplane section of $Y$, with respect to the fixed embeddings $X \subset \mathbb{P}^N$ and $Y \subset \mathbb{P}^M$. Then
$\deg X= (H_X)^n, \quad \deg Y =(H_Y)^n$,
where $n= \dim X = \dim Y$.
Now requiring 
$\deg X = (\deg f) \cdot \deg Y \quad (*)$ 
is equivalent to require $(H_X)^n=(\deg f) \cdot (H_Y)^n$. This is true for instance if the map $f \colon X \to Y$ is induced by a subsystem of the complete linear system $|H_X|$, otherwise it is false in general.
For instance, let $X \subset \mathbb{P}^5$ be the Veronese surface and $Y=\mathbb{P}^2$. Since $X$ and $Y$ are isomorphic, there is a map $f \colon X \to Y$ of degree $1$, but $\deg X=4$ and $\deg Y=1$.  
Analogously, let us consider $X=\mathbb{P}^1 \times \mathbb{P}^1 \subset \mathbb{P}^3$. Then $X$ is a quadric and any $2$-dimensional, base-point free subsystem of the complete linear system $|H_X|$ induces a finite morphism  $f \colon X \to \mathbb{P}^2$ of degree $2$, so in this case $(*)$ is satisfied. However, you can compose $f$ with any isomorphism $g \colon X' \to X$, where $X' \subset \mathbb{P}^N$ is a Segre embedding of $\mathbb{P}^1 \times \mathbb{P}^1$ with $N >3$, obtaining a map $f' \colon X' \to \mathbb{P}^2$ of degree $2$ which does not satisfy $(*)$, since $\deg X' >2$. 
