Effective classes on blowups Consider $X$, the blowup of $\mathbb{P}^2$ at two points $p_1$ and $p_2$. Then Pic $X$ has generators $L, E_1, E_2$ where $L$ is the class of a line and $E_1$ and $E_2$ are the exceptional curves. 
I usually see the definition that a divisor $D = aL + bE_1 + cE_2$ is effective if $a, b, c$ are nonnegative. But since every two points in the plane have a line going through them, $L - E_1 - E_2$ is effective! This seems to me a contradiction. I will appreciate any clarification.
Also, under what condition is $E_1 - E_2$ effective? Is it only when $p_2$ is infinitely near $p_1$?
Thanks.
 A: The confusion comes from blurring the distinction between single divisors and linear equivalence classes of divisors.
If $D_i$ are fixed codimension-1 subvarieties, its true that the divisor $\sum_i a_i D_i$ is effective if and only if all the $a_i$ are nonnegative.
On the other hand, when you say $L-E_1-E_2$ is effective, you are talking not about a fixed divisor, but a linear equivalence class;  what you mean is that there is an effective divisor $D$ whose linear equivalence class is $L-E_1-E_2$. (As you rightly describe in the question, $D$ is the proper transform of a line through the two points.)
Finally, yes, $E_1-E_2$ is effective (as a linear equivalence class) if and only if $p_2$ is infinitely near to $p_1$. "If" is clear; to see why "only if" is true, suppose that $p_1$ and $p_2$ are distinct points. Let $E$ be an effective divisor in the class $E_1-E_2$, and let $D$ be the proper transform of a line through $p_2$ only: then $D \cdot E= D \cdot(E_1-E_2) <0$. That means $E$ must be supported in $D$. But there is a 1-parameter family of such lines $D$, and two different lines in the family are disjoint; contradiction.
