Let $X\subset \mathbb{P}^3$ be an irreducible quartic surface, defined over an algebraically closed field $k$. Suppose that $X$ is rational (i.e. birational to $\mathbb{P}^2$). Is is true that $X$ has one of the following type of singularities?
a) a point $q\in X$ of multiplicity $4$ (hence $X$ is a cone over a rational quartic curve).
b) a point $q\in X$ of multiplicity $3$ (hence the projection away from $q$ gives a birational map $X\dashrightarrow \mathbb{P}^2$).
c) a curve of singular points.
It seems to me that these are the only possible cases, as otherwised) a double point $q\in X$ and a double line $l$ in the canonicalexceptional divisor would be trivial. For instance, if $X$ has only isolated$E_q\simeq \mathbb{P}^2$ of $q$ edit: new case added
e) a double pointspoint $q\in X$, a double point $q_1$ in the canonicalexceptional divisor is trivial. But maybe one could have infinitely near curves$E_q\simeq \mathbb{P}^2$ of singularities that make$q$ and a double line $l$ in the canonicalexceptional divisor non-effective? Is it possible to get a rational surface?$E_{q_1}\simeq \mathbb{P}^2$ of $q_1$ edit: new case added
It seems to me that this should be quite classical. I looked overfrom the internet for "rational quartic surfaces" and found a lot of examples and some articlesclassification of classifications, but I did not find the answer toNoether indicated by abx in
M. Noether, "Ueber die rationalen Flächen vierter Ordnung" (Mathematische Annalen, 1189, page 546-571)
that these are the above question. Thanks for your helponly possible cases. Is there a modern proof of this ?