Relatively minimal elliptic surfaces which are not minimal A complex surface is called minimal if it contains no $(-1)$-curves, while an elliptic surface $\pi : X \to C$ is called relatively minimal if the fibers of $\pi$ contain no $(-1)$-curves. It follows that there could be elliptic surfaces which are relatively minimal but not minimal. Such surfaces exist.
Example: Choose two conics in $\mathbb{CP}^2$ which intersect at nine points $p_1, \dots, p_9$. Then there is a pencil of cubics through these nine points, and this defines a map $\mathbb{CP}^2\setminus\{p_1, \dots, p_9\} \to \mathbb{CP}^1$ (given $p_1, \dots, p_9$ and a tenth point, there is a unique cubic in the pencil which passes through all ten points). As cubics in $\mathbb{CP}^2$ are elliptic curves by the degree-genus formula, blowing up $\mathbb{CP}^2$ at $p_1, \dots, p_9$ gives rise to an elliptic surface $\pi : \operatorname{Bl}_{p_1,\dots, p_9}(\mathbb{CP}^2) \to \mathbb{CP}^1$. Note that $\operatorname{Bl}_{p_1,\dots, p_9}(\mathbb{CP}^2)$ is not minimal, but $\pi : \operatorname{Bl}_{p_1,\dots, p_9}(\mathbb{CP}^2) \to \mathbb{CP}^1$ is relatively minimal (the fibers of $\pi$ intersect each $(-1)$-curve once).
I have seen it claimed that these are the only examples.

How does one show that a non-minimal, relatively minimal elliptic surface is a blowup of $\mathbb{CP}^2$ at nine points?

Suppose $X$ is the blowup of $Y$ at $k > 0$ points where $Y$ is minimal. If $\pi : X \to C$ is a relatively minimal elliptic surface, then $c_1(X)^2 = 0$, so $c_1(Y)^2 = k > 0$. It follows from the classification of surfaces that $Y$ must be rational, and hence biholomorphic to either $\mathbb{CP}^2$ (in which case $k = 9$) or a Hirzebruch surface $\Sigma_n$ with $n \neq 1$ (in which case $k = 8$). As $\Sigma_1$ is biholomorphic to the blowup of $\mathbb{CP}^2$ at a point, we can combine the previous two possibilities to conclude that $X$ is a blowup of a Hirzebruch surface $\Sigma_n$ at $8$ points.

How can I show that $n$ must be $1$?

 A: Here is a (very sketchy) sketch that $n \leq 2$
Claim: In $\mathbb{\Sigma}_n$, ($n>2$) there is not smooth elliptic curve $C$ with $C^2=8$.
Proof. I follow Beauvilles set up for Hirzebruch surfaces and I prefer to use the anticanonical bundle. By the genus formula the assumptions $g(C)=1, C^2=8$ gives $-K.C=8.$ Let $h$ be the class of the tautological bundle of the Hirzebruch surface and $F$ the fibre of the $\mathbb{P}^1$-bundle. Then, $h,F$ is a basis of $H^2 = $Picard group. Furthermore $h^2=n, h.F=1, F^2=0 $, and $-K=2h - (n-2)F$. Finally there is as  a unique smooth, irreducible and rational curve, with class $D= h-nf$, $D^2=-n$, which is given by one of the sections of the $\mathbb{P}^1$-bundle.
Then, if $c= ah+bF$, then $$c^2 = na^2 + 2ab  =8 (*).$$, $$ -K.C = 2an + 2b- (n-2)a=a(n+2) +2b= 8, $$ so taking $2a$ for each side: $$an +2b = (8-2a) ,$$ multiplying by $a$ and plugging in $(*)$; $$na^2 +2ab = a(8-2a)  = 8.$$
This system has the unique solution $a=2,b=2-n$. So lets compute $$D.c = (2h+(2-n)F).(h-nF) = 2n + (2-n) -2n = 2-n
.$$
Note that since this is an intersection of irreducible smooth curves of different genus it has to be positive.
Now here is the sketchy part. If there was such an elliptic surface, I would guess that by genericity the elliptic fibres will intersect the exceptional fibres transversally, so we may project to the base to give a curve as above and proving $n \leq 2$. It is clear that the elliptic fibre has to intersect all of the exceptional curves (since elliptic fibres cover the surface), so if you can prove the it doesn't intersect any in more than one point, you are done.
Edit. I have realised (unless I made a mistake) there is a way to make the above argument rigourous using the canonical bundle formula for elliptic surfaces (Theorem 6.8 of https://arxiv.org/pdf/0907.0298.pdf).
The canonical bundle formula implies that for any elliptic surface $S$ with holomorphic Euler characteristic $1$ over a smooth rational curve, then $$K_{S} = -Y (*),$$ where $Y$ is a fibre of the elliptic fibration (I already used F for something different above.)
Hence, since for any exceptional curve $E$ of a blow-up, $K_{s}.E=-1$, we have that the fibres of the elliptic fibration intersect all the exceptionbl rational curves in exactly 1 point transversally, hence the argument above can be made rigourous.
Edit: This formula (*) is only true in the absence of multiple fibres (there seems to be a misprint in the paper I quoted).
Note that the standard relatively minimal rational elliptic surface has no multiple fibres so it may be possible to prove this in general.
Final edit: I found a reference which finishes this. By Proposition 1.1 of "On Rational Elliptic Surfaces with Multiple Fibers" by Yoshio Fujimoto,  a rationally elliptic surface with multiple fibres is a blow-up of projective space $\mathbb{CP}^2$. The proof is detailed, and involves a fairly concrete application of the general canonical bundle formula. Hence combining with above the statement is proved.
