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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.

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    $\begingroup$ 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 $\mathbb{P}^2$. $\endgroup$ May 31, 2020 at 16:33
  • $\begingroup$ 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 ? $\endgroup$
    – Someone
    May 31, 2020 at 16:46
  • $\begingroup$ 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. $\endgroup$
    – Someone
    May 31, 2020 at 16:57

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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, $f(x,y,z)\in \mathbb{C}[x,y,z]$.

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  • $\begingroup$ 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! $\endgroup$
    – Someone
    May 31, 2020 at 17:25

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