(Toric) Intro: (Superflous, just to consider the initial data)
From a toric point of view, the blow up of $\mathbb{P}^2$ at two points (which we can assume are $[0:1:0]$ and $[0:0:1]$ is given by a fan whose maximal cones $$ \langle e_1,e_2, \rangle, \langle e_2,-e_1\rangle, \langle -e_1,-e_1-e_2 \rangle,\langle -e_1-e_2,-e_2 \rangle, \langle -e_2,e_1 \rangle, $$ And the two exceptional divisors are preicsely the rays $-e_1$,$-e_2$ (don't worry, I won't use toric geometry from this point on).
Question:
The blow up at two points in $\mathbb{P}^2$, which from now on we call $X$, is a smooth variety, hence $\text{Cl}(X)=\text{Pic}(X)$, and I've managed to prove (via toric theory) they're isomorphic to $\mathbb{Z}^3$. Thus 3 of the divisor $D_1,D_2,\ldots,D_5$ are a basis for $\text{Pic}(X)$. But now I'm kinda stuck: which are the basis generators?
Since $\text{Pic}(\mathbb{P}^2)=\mathbb{Z}H$, where $H$ is any hyperplane (in particular any of $D_1,D_2,D_3$), and via blow-up I'm replacing two points with 2 exceptional divisors (which in my ugly notation are $D_4$ and $D_5$), I think $\text{Pic}(X)=\mathbb{Z}D_1\oplus\mathbb{Z}D_4\oplus\mathbb{Z}D_5$ (while actually since $D_1$ is linearly equivalent to $D_2$ and $D_3$ there's no difference). But for istance, how can I prove $D_4$ is not linealry equivalent to any linear combination of the other $D_i$'s?
I'd like to understand my the above argument make any sense, but i'm open to any hint/deeper look at the problem without using necessairly toric geometry, I'd like to have a broad picture (I also sense this is basic blow-up theory, but the only time I saw it was indeed in the toric geometry course). Thanks in advance!
Here are the intersection numbers. $D_4^2=D_5^2=-1, D_i\cdot D_j=0, i\neq j$ where subscripts are one of $1,4,5$. $D_1^2=1$. If $D_4=aD_1+bD_5$, one gets say intersecting with $D_1$, $0=a$, so $D_4=bD_5$. Intersect with $D_4$ to get a contradiction.