Recall that an elliptic curve over a field $k$ i.e a proper smooth connected curve of genus $1$ equipped with a distinguished $k$-rational point, i'll be really grateful for any help in understanding the following part of our course

(1) Why does $E$, the closure of the vanishing locus of the equation (4), define an irreducible algebraic curve?

(2) Why is $E$ smooth over $k$ if and only if $E-\{0\}$ is smooth over $k$?

(3) Given an affine and smooth Weirstrass equation, why does its shematic closure in $\mathbb{P}_k^2$ define an elliptic curve? It is stated that this is consequence of Bezout theorem but i failed to understand how

(4) Let $(E,0)$ be an elliptic curve, using Riemann-Roch we construct an isomorphism into $\mathrm{Proj}\,k[X,Y,Z]/Y^2Z+a_1XYZ+a_3YZ^2-X^3-a_2X^2Z-a_4XZ^2-a_6Z^3$, why does $0$ map to the infinity point $O=[0:1:0]$?

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1 Answer 1


Several things before I answer.

a) You should really put more effort into these questions. Put them as four separate questions and show your thinking for all of them.

b) What course notes are these from? I'm just curious.

(1) As the note writer notes, $E$ is irreducible since


is irreducible. For simplicity let's assume the characteristic $k$ is different than $2$. To see this note that if we write


that $g$ and $h$ must be monic (up to scalars in $k$) as polynomials in $y$ since $f$ is. This then implies that each $g(x,y)$ and $h(x,y)$ if not constant, have degree at least $1$ in $y$. This then implies that $g(x,y)$ and $h(x,y)$ are degree $1$ in $y$. But this is absurd since that implies that

$$\frac{-(a_1x+a_3)\pm \sqrt{(a_1x+a_3)^2+4(-a_3+x^3+a_2x^2+a_4x+a_6)}}{2}$$

is a polynomial in $x$, which is clearly impossible considering that the argument of the square root is a monic odd-degree polynomial.

Now, since $f$ is irreducible we know that $V(f)\subseteq \mathbb{A}^2_k$ is irreducible. Since $E$ is the closure of $V(f)$ in $\mathbb{P}^2_k$, and closure preserves irreducibility, we deduce that $E$ is irreducible.

(2) Let $F$ denote the homogenization of $f$. So,

$$F(x,y,z)=y^2z+a_1 xyz+a_3yz^2-(x^3+a_2x^2z+a_4xz^2+a_6z^3)$$

so then $E=V(F)\subseteq\mathbb{P}^2_k$. We then know from the Jacobian criterion that $E$ is smooth iff


has no common solution in $\overline{k}$. Note though that $0$ of $E$ is the point $[0:1:0]$ and plugging this into $F_z$ results in $1$. So, $0=[0:1:0]$ can never be a singular point. Thus, it suffices to check the smoothness of $E-\{0\}$ which is the affine curve $V(f)\subseteq\mathbb{A}^2_k$.

(3) I think the note writer means 'Bezout's formula' which says that if $C$ is a geometrically integral smooth curve in $\mathbb{P}^2_k$ of degree $d$ then


This formula, as the quoted sentence suggests, comes the from the classifiation of line bundles on $\mathbb{P}^2_k$ and a cohomology calculation. In particular, if $d=3$ we get that $g(C)=1$. So, in our case $E$ has degree $3$ so that $E$ has genus $1$, thus $(E,0)$ is an elliptic curve.

EDIT: Oh, the note taker is claiming to that Bezout's formula as I said above follows from Bezout's theorem. I understand. My suggested method above calculates the arithmetic genus of $C$ (which is the same as the geometric genus by Serre duality). Namely, the adjunction formula says that

$$\omega_C=i^\ast(i_\ast\mathcal{O}_C\otimes \omega_{\mathbb{P}^2_k})$$

where $i$ is the inclusion of $C$ into $\mathbb{P}^2_k$. So then, one sees that using the degree of the canonical bundle is $2g-2$ and that $\omega_{\mathbb{P}^2_k}=\mathcal{O}_{\mathbb{P}^2_k}(-3)$ that

$$2g-2=\deg(\omega_C)=\deg(i^\ast(i^\ast(i_\ast\mathcal{O}_C\otimes \omega_{\mathbb{P}^2_k})))=\deg(C .(C-3))$$

But, if $C$ is cut out be a degree $d$ curve then $\deg(C)=d$ and so applying Bezout's theorem to the above we get


solving for $g$ gives


(4) Your sections are $(x,y,1)$. The map $E\to\mathbb{P}^2_k$ can then imprecisely be written as

$$E \ni e\mapsto [x(e):y(e):1(e)]$$

where even though $x,y,1$ are only sections of a line bundle, they make sense since scalar multiplication doesn't affect points in $\mathbb{P}^2_k$ and thus it doesn't matter what chart you compute this on.

Anyways, $x(e)$ and $y(e)$ have poles of order $2$ and $3$ respetively at $0$ and $1$ has no pole at $0$. So, to evaluate $[x(0):y(0):1(0)]$ you need to multiply through by a uniformizer cubed. Let's call this uniformizer $\pi$. So then, really what $[x(0):y(0):1(0]$ means is something like $[\pi^3 x(0),\pi^3 y(0):\pi^3 1(0)]$ where now since $\pi^3x, \pi^3y$ and $\pi^3 1$ no longer have poles at $0$ it makes sense to evaluate them there. But, note that $\pi^3x$ and $\pi^3 1$ now have poles of order $-1$ and $-3$ at $0$ or, in other words, zeros at $0$. So, $\pi^3x(0)=\pi^31(0)=0$. Since $y$ had a pole of order $3$ we see that $\pi^3y$ is non-vanishing at $0$. So $[x(0):y(0):1(0)]$ becomes something like $[0:c:0]$ where $c$ is non-zero. This is just $[0:1:0]$.

  • $\begingroup$ Thank you so much for your response, i appreciate the time and effort you put in answering my question, and i'll try to work more on formulating my questions. What i still don't understand is how do you view $x,y,1$ as functions with zeros and poles? Here is the link of the notes imo.universite-paris-saclay.fr/~schraen/CoursM2/Cours.pdf $\endgroup$ Aug 20, 2020 at 16:57
  • $\begingroup$ @YassineElkaouni This then is just a lack of background on your part for how to think of maps ot projective space. Read Section 16.4 of the rising sea <math.stanford.edu/~vakil/216blog/FOAGnov1817public.pdf> or this <citeseerx.ist.psu.edu/viewdoc/…> $\endgroup$ Aug 20, 2020 at 17:12
  • $\begingroup$ If you're concerned about the quality level of the question, it may be a better move to let the OP know via a comment before answering a question which you believe needs improvement. Just a suggestion. $\endgroup$
    – KReiser
    Aug 20, 2020 at 17:39

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