# Curves of Genus one are cubic plane curves (proof doubt)

I have been reading Rick Miranda's Book on Algebraic Curves and Riemann Surfaces and there is a proposition whose proof I don't complete understand.

The proposition states that curves of genus one are cubic plane curves.

Is proof is the following : If $$X$$ is an algebraic curve and we have a divisor with degree $$3$$ this divisor is very ample and so $$dimL(D)=3$$ if $$deg(D)=3$$, using Riemann-Roch, we see that $$\phi_D$$ will map $$X$$ to $$\mathbb{P}^2$$. Since $$deg(D)=3$$ the hyperplane divisor will have degree $$3$$ and so the image is a cubic curve.

Now I get that the degree of the smooth projective curve $$Y=\phi(X)$$ will be $$3$$, but how does he know that this a plane curve? We just know that $$\phi_D$$ gives an embedding, we don't know any more additional structure to the Riemann Surface. What is the part that I am missing? Thanks in advance.

• Well, if you agree that the image $\phi(X)$ in $\mathbb{P}^2$ is degree $3$, then you have a cubic plane curve, since it lies in $\mathbb{P}^2$. May 31, 2020 at 16:33
• But isnt supposed that a smooth projective plane curve has to be defined by non-singular polinomyal $F(x,y,z)=0$? I just that it is going to be embedded in $\mathbb{P}^2$ how can I conclude the rest?@AlekosRobotis, It's just that the author sometimes says smooth projective plane curves and sometimes says smooth projective curves , are these supposed to be the same ? May 31, 2020 at 16:46
• I belive these cant be the same because for a smooth projective curve we just know that it is an holomorphically embedded riemann surface in $\mathbb{P}^n$, I dont know how to get the rest. May 31, 2020 at 16:57

As you say above, $$\phi(X)$$ is embedded in $$\mathbb{P}^2$$, which makes it a smooth plane curve. I should mention that a smooth projective curve in $$\mathbb{P}^2$$ is called a smooth plane curve, as a matter of terminology. I believe that this is all that Miranda is claiming, but citing some theorems we can say a bit more.
We have an (a priori) analytic subvariety of $$\mathbb{P}^2$$, given by $$\phi(X)=Y$$. Using Chow's Theorem we can conclude that $$Y$$ is actually algebraic. Hence, $$Y=Z(f_1,\ldots, f_r)$$ for $$f_1,\ldots, f_r\in \mathbb{C}[x,y,z]$$ homogeneous polynomials. However, a version of the Hauptidealsatz from commutative algebra now implies that $$Y=Z(f)$$, i.e. it is cut out by a single homogeneous polynomial. By a result on the degree (found in Miranda), we know that if $$Y=Z(f)$$, and $$Y$$ is of degree $$3$$, then $$\deg(f)=3$$. So, it follows that $$Y$$ is cut out by a single degree $$3$$ equation, $$f(x,y,z)\in \mathbb{C}[x,y,z]$$.
• Yeah that's those things of algebraic geometry/commutative algebra that I am not aware off, and so it will indeed be an smooth affine plane curve , it just because he defines smoth projective curve as just an embedded riemann surface in $\mathbb{P}^n$ and a smooth projective plane curve we need it to be a cut locus of a non-singular polinomyal, thanks! May 31, 2020 at 17:25