Let $X$ be an algebraic variety and $A$ is a ample divisor on $X$. Let $m$ be a sufficiently large natural number such that $X \overset{\varphi\_{mA}}{\to} \mathbf{P}H^0(X, \mathcal{O}\_X(mA))$ defined by the linear system $|mA|$ is an embedding. Denote by $Y$ the image. What's a effective upper bound of $Y$ in terms of $\text{dim}X$, $A^{\text{dim}X}$ and $m$, or other invariants of $X$ and $A$. 


This may not be a concrete question. But I am wondering how can we compute the degree of the image of a variety under the (bi)rational morphism given by a linear system.



Degree of an embedded algebraic variety
I'll expand on Felipe's answer. The degree is defined to be $(\dim X)!$ times the lead term of the Hilbert polynomial of the variety. Now, given an ample line bundle $A$, the Hilbert polynomial of the embedding given by $A$ is $n\mapsto\chi(X,A^n)$. First off, this function is in fact a polynomial (check any standard book, Hartshorne for instance). This term will then be the self intersection $A^{\dim X}$, which can be computed in any number of ways (see any book on intersection theory), and so, for the line bundle $mA$, we'll get $(mA)^{\dim X}=m^{\dim X}A^{\dim X}$.


But as Felipe said, it's pretty much by definition. In a real sense, the definition of $A^{\dim X}$ can be taken to be $(\dim X)!$ times the lead term of the Hilbert polynomial.


