linear series vs. linear system on algebraic curves Could someone please tell me if there is any  difference between the concepts "linear series" and "linear systems" on algebraic curves? 
Also, for smooth plane curves of degree $n$, what is the main difference between linear system of "curves of degree $d$" in the case $d\leq n-3$, and the case $d>n-3$?
 A: First question:
The terms linear system and linear series are completely interchangeable, it's just a matter of taste. Here's the definition of a linear system of divisors:
Def: A divisor $D$ is linearly equivalent to $D'$ if there exist a globally defined rational function $f:C\to k$ such that $D+(f) = D'$.
Def: A divisor $D$ is effective if the order of every point is non-negative.
Def: Given a divisor $D$ on a curve $C$, the complete linear system (or complete linear series) $|D|$ associated to $D$ is the set of all effective divisors on $C$ which are linearly equivalent to $D$.
Def: A linear system (or linear series) is a linear subspace of a complete linear system.

Second question:
Recall that the Riemann-Roch theorem for smooth curves states that
$$ h^0(D) - h^0(K-D) = \deg(D) - g + 1, $$
where:

*

*$g$ is the genus of the curve $C$ which, by the genus-degree formula for smooth plane curves, is given by
$$ g = \frac{(n-2)(n-1)}{2} $$


*$h^0(D)$ denotes the dimension of the linear series $|D|$ as a vector space over the ground field $k$


*$K$ is any canonical divisor of $C$ and $h^0(K-D)$ denotes the dimension of the linear series $|K-D|$ as a vector space over the ground field $k$
Further, recall that as soon as $\deg(E)<0$ we have $h^0(E) = 0$, i.e. the linear series $|E|$ consists of $E$ only.
Now, since the degree of a canonical divisor $K$ is given by (to see this just plug $D=0$ in the Riemann-Roch formula above)
$$ \deg(K) = 2g-2 = n\cdot(n-3), $$
we deduce that, if $D$ is the divisor of degree $n\cdot d$ consisting of the points of intersection between $C$ and another plane curve of degree $d$, we have
$$ d > n-3 \implies \deg(D)>\deg(K) \implies h^0(K-D) = 0. $$
Therefore in the case $d > n-3$ the dimension of the linear series $|D|$ can be easily computed using the Riemann-Roch formula: in this case indeed we have
$$ h^0(D) = n\cdot d - g + 1 = \frac{n\cdot d - (n-2)(n-1) + 2}{2} = \frac{n\cdot (2d -n+ 3)}{2}. $$
On the other hand, if $d \leq n-3$ the dimension of $|D|$ is harder to compute, because of the tricky term $h^0(K-D)$ appearing in the Riemann-Roch formula.
