# A quartic with an ordinary singular points of multiplicity $3$ hasn't got any other singular point.

Show that an irreducible quartic with a singular point of multiplicity $3$ hasn't got any other singular points.

So to prove that, I thought that I could use the notions of arithmetic and geometric genus. The arithmetic genus of a degree $d$ curve is $$g = \frac{1}{2}(d-1)(d-2)$$ so in our case, when $d=4$, we get that $g=3$. Now, since the geometric genus of a curve $C$ is reduced by $\frac{1}{2}r(r-1)$ when the curve has an ordinary singularity of multiplicity $r$. and by hypothesis, our quartic has a singular point of multiplicity $3$, the geometric genus of the curve si then $0$. So if the curve had another singular point, its geometric genus would be negative, and that would made no sense, right?

Is this argument fine? I've recently began studying the genus notion of an algebraic curve, so I'm doubtful if this proof is okay. Thanks in advance!

• Your argument is fine. For plane curves, such as yours, you can also use Bezout's theorem If $P$ and $Q$ are points with multiplicities $3, a$, then the line joining them will meet the quartic at least $3+a$ times and since the curve is irreducible, this number is four, giving $3+a\leq 4$. So, every other point must be non-singular. – Mohan Feb 4 '17 at 14:53
• @Mohan Thank you very much! I like that alternative proof using Boxout's theorem. Feel free to post it as an answer! – user313212 Feb 4 '17 at 15:48

Your argument is fine. For plane curves, such as yours, you can also use Bezout's theorem. If $P$ and $Q$ are points with multiplicities $3,a$, then the line joining them will meet the quartic at least $3+a$ times and since the curve is irreducible, this number is four, giving $3+a\leq 4$. So, every other point must be non-singular.