In the very commonly used J. Silverman's AEC an elliptic curve is defined as a genus 1 projective curve with a fixed point 0. In all the other books I looked at it is defined to be (also) smooth. By the way in AEC it is given a proof of the fact that a genus 1 curve is smooth, but I have the feeling that there is an important gap.
To show such a fact (chapter 3) the book anticipates in section 2 that a genus 1 curve will be shown to be isomorphic to a plane cubic (Weierstrass or equivalently Legendre form), then using elementary computations one can find different proofs of smoothness.
In section 3 it is shown then that a genus 1 curve is isomorphic to a plane cubic in Weierstrass form, but unfortunately this is done by using Riemann-Roch Theorem, which is stated only for smooth curves. The proof concludes by using smoothness in another passage (birational equivalence of curves is actually isomorphic if the curves are smooth).
Therefore the book shows that a genus 1 plane cubic is smooth and that a genus 1 smooth curve is plane cubic, but there is no proof of the fact that a genus 1 curve is BOTH plane cubic and smooth. May someone indicate me a good reference for a proof of this fact?
I moreover point out that at the end of the second proof (Proposition 3.1, page 64, line -9) it is stated that by a non-existing (1.4d) somewhere, if a curve $C$ has a singularity, there exists a degree 1 rational map $\Psi:C\to \mathbb{P}_{1}$. I can not deduce such a statement, which would solve my problem, from any of the results listed before and I am not able to find it nowhere on internet either. In the errata available in J. Silverman's home page this bad reference problem does not appear. If moreover such a statement is true, that would mean that any curve with a singularity is birational with $\mathbb{P}_{1}$. So any non-zero curve would be smooth. Is this true? I can not find this stated nowhere again.
Best Answer
The genus $g$ of a curve $C$ can be defined as follows:
If $C$ is non-singular, then $g=g(C)$ is the number that appears in the Riemann-Roch theorem.
If $C$ is singular, then we find a curve $C'$ that is non-singular, and birational to $C$ (you can find $C'$ by resolution of singularities, for example). In this case, $g(C)$ is defined to be $g(C')$.
In particular, a curve of genus $1$ does not need to be smooth. The fact that $g(C)=1$ only implies that $C$ is birational to a smooth curve $C'$ of genus $1$, and if $C'$ has a rational point, then $C'$ is an elliptic curve and it has a Weierstrass form. Since $C$ and $C'$ are birational, then $C$ is birational to a Weierstrass form (which still does not imply that $C$ is smooth).
I found the answers to this question quite useful here.