# Triple covers of $\mathbb{P}^2$ with Tschirnhausen module $\mathcal{O}(-1)\oplus\mathcal{O}(-1)$

Let $$X$$ be a surface as in the title. Rick Miranda said that $$X$$ is a Steiner cubic in $$\mathbb{P}^4$$, and the cover map is projection. Invariants of $$X$$ can be computed directly, $$p_g(X)=0,K^2_X=8,e(X)=4$$.

My question is,

Question: What is a Steiner cubic? Why $$X$$ is a Steiner cubic?

The followings are what I know: Given $$a,b,c,d\in H^0(\mathbb{P}^2,\mathcal{O}(1))$$, $$X$$ is locally defined by $$\begin{cases} F(z,w)=z^2-aw-bw-2(a^2-bd) \\ G(z,w)=zw+dz+aw+ad-bc\\ H(z,w)=w^2-cz-dw-2(d^2-ac) \end{cases}$$ Does, in fact, $$X$$ globally defined as an intersection of three quadrics in $$\mathbb{P}^4$$?

• A locally complete intersection of three quadrics in $\mathbb{P}^4$ is a curve, not a surface.
– abx
Commented May 9, 2023 at 12:31
• Maybe I should not say it is a locally complete intersection. I mean that $X$ is just like a two-dimensional version of a twisted cubic curve in $\mathbb{P}^3$ which is an intersection of three quadrics. Commented May 9, 2023 at 12:47
• So there are no triple covers in the question? Commented May 9, 2023 at 13:35
• Usually, Steiner surface is defined as the image of a regular liner projection of the Veronese surface into $\mathbb{P}^3$, see mathworld.wolfram.com/SteinerSurface.html Commented May 9, 2023 at 18:03

I find the answer, see "Triple planes with p_g=q=0" Proposition 3.1 $$X$$ is a cubic scroll $$S(1,2)\subset\mathbb{P}^4$$, the image of Hirzebruch surface $$\mathbb{F}_1$$ under the linear system $$|c_0+2f|$$, where $$c_0$$ is the section with $$c_0^2=-1$$ and $$f$$ is the class of fiber.