Are elliptic curves algebraic varieties?

Question

I got a short question.

Are elliptic cubes also algebraic varietes?

Say we have $E:y^2=x^3+5x=:f(x)$

Then we can $f(x)=x(x^2+5)$

So it can't be an algebraic variety..

I feel like I am totally misunderstanding the definition of algebraic varieties.

Can someone help? Thanks

Answer 1

Elliptic curves are very famous examples of algebraic varieties. You seem to have forgotten that the function can have more than one variable; take the function $f(x,y)=y^2-x(x^2+5)$, then $E$ is just the variety $V(f)$.

Answer 2

You are messing up the notation and confusing yourself. You could write the polynomial $$ f(x) = x^3 + 5x \in \mathbb{Q}[x] \text{ or } \mathbb{C}[x] $$ or you could write the polynomial $$ f(x,y) = y^2 - x^3 - 5x \in \mathbb{Q}[x,y] $$ Once you factor $f(x)$ you find that it equals $x(x^2 + 5) = x(x+i\sqrt{5})(x-i\sqrt{5})$. This should give you the hint that it defines a single point in $\mathbb{Q}$ while giving you three distinct points in $\mathbb{C}$. The other polynomial does define an ellitpic curve as a variety over $\mathbb{Q}$.

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Also, you could ramp up this discussion and frame it into a scheme theoretic language. Then, you should consider the scheme $$ X = \text{Spec}(\mathbb{Z}[x,y]/(y^2 - x^3 - 5x)) $$ If you write this as the functor $X = \text{Hom}_{\textbf{CRing}}(\mathbb{Z}[x,y]/(y^2 - x^3 - 5x),-)$, then \begin{align*} X(\mathbb{F}_q) &= \{ (a,b) \in \mathbb{F}_q^2: b^2 - a^3-5a =0 \} \\ X(\mathbb{Q}_p) &= \{ (a,b) \in \mathbb{Q}_p^2: b^2 - a^3-5a =0 \} \\ X(\mathbb{Q}) &= \{ (a,b) \in \mathbb{Q}^2: b^2 - a^3-5a =0 \} \\ X(\mathbb{C}) &= \{ (a,b) \in \mathbb{C}^2: b^2 - a^3-5a =0 \} \end{align*} and so on...