Where is the "point of infinity" for elliptic curve over finite fields?

Question

This might be a dumb question, but I can't find the answer. Everywhere I see people say elliptic curve is this set of points plus a point of infinity on a finite field. Where is that point of infinity? Because finite field is finite, is the point of infinity an actual point or just an "imaginary" one? In real space, the point of infinity is clear.

Answer 1

To start with, you need to understand what "projective plane" is. It's easier to work over $\mathbb{R}$ or $\mathbb{C}$ to start with, because that's what we are used to. So let's start there, we'll move to finite fields in a minute.

Let's start with the affine real plane. That's what you know from analytic geometry, the "cartesian plane". We can identify it with the set of all ordered pairs $(a,b)$ with $a,b\in\mathbb{R}$. An algebraic curve is the set of all points $(a,b)$ that satisfy some polynomial equation $f(x,y)=0$.

In affine space things don't always go "generically". For example, two distinct lines "usually" intersect at a single point... but they may be parallel and never intersect. This can be solved by going to "projective plane."

Here is a post in which I gave a way to think about projective plane. But projective plane is identified with triples of points $(a,b,c)$, not all zero, where we identify two points $(a,b,c)$ and $(r,s,t)$ if there exists a (nonzero) number $\lambda$ such that $(a,b,c)=(\lambda r,\lambda s,\lambda t)$. This corresponds to identifying points with "lines through the origin in $3$-space", as I do in the post linked to above.

The equivalence class of a point $(a,b,c)$ in the projective plane is traditionally denoted something like $[a:b:c]$.

Projective plane can be "covered" with three standard copies of the affine plane: all points with $c\neq 0$ correspond to a point of the form $[a:b:1]$, and we can think of these points as the points of the affine plane $\{(a,b)\mid a,b\in\mathbb{R}\}$. All points with $b\neq 0$ can be uniquely represented as a point $[a:1:c]$, and they "look" like an affine plane. And the points with $a\neq 0$ which can be represented as $[1:b:c]$.

So, the points $[a:b:1]$ can be thought of as being in the "affine part" of the projective plane. There are other points, namely all points of the form $[a:b:0]$, with $a$ and $b$ not equal to $0$. All but one of them can be represented as $[1:m:0]$, and following with the analogy of "lines through the origin" in the other post, we can think of this point as being added "at infinity in the direction of lines of slope $m$". The final "point at infinity" is the point $[0:1:0]$, which we can think of as being the point "added at infinity in the vertical direction".

Now, because a point $[a:b:c]$ and the point $[2a:2b:2c]$ are "the same projective point", we cannot define curves on projective space using any old polynomial: we run into problems. For example, the line $x+y-1=0$ would seem to include the point $[0:1:0]$, but not the point $[0:2:0]$, even though they are the same point in the projective plane.

So for projective space we need polynomials with the property that $p(a,b,c)=0$ if and only if $p(\lambda a,\lambda b,\lambda c)=0$ for any nonzero constant $\lambda$. These are called homogeneous polynomials, and they are polynomials in which every monomial has the same total degree.

To "projectivize" an affine curve $p(x,y)=0$, we find the highest total degree monomial, and we add enough powers of $z$ to every other monomial to get a homogeneous one, $P(x,y,z)$. For example, the polynomial $$p(x,y) = y^2 -x^3-2x-xy$$ has highest total degree $3$, so it is projectivized by adding powers of $z$ to every monomial other than $x^3$ to make them of degree $3$: $$P(X,Y,Z) = Y^2Z - X^3 - 2XZ^2 - XYZ.$$ Note that "affine part" of the curve $P(X,Y,Z)=0$, that is, the one with $z=1$, is exactly the same as the affine curve $p(x,y)=0$.

In projective space, things go much better than in affine space; for example, the intersection of two distinct projective lines is always a single point: we do not need to worry about "parallel lines not intersecting". More generally, over the complex numbers we have Bezout's Theorem that says that the intersection of a curve given by a polynomial of degree $n$ and a curve given by a polynomial of degree $m$ always consists of $nm$ points (counting multiplicities to deal with things like "tangential intersections" and the like).

Now, a (projective) elliptic curve is a projective curve defined by a polynomial of the form $$P(X,Y,Z) = Y^2Z + a_1XYZ + a_3YZ^2 -X^3 - a_2X^2Z-a_4XZ^2+a_6Z^3$$ and we look for projective points with $P(X,Y,Z)=0$. You can also think of it as the projective points $[a:b:c]$ that satisfy the equation $$Y^2Z + a_1XYZ + a_3YZ^2 = X^3 + a_2X^2Z+a_4XZ^2+a_6Z^3.\tag{1}$$ (In characteristics different from $2$ and $3$, a change of variables can bring into a form with $a_1=a_3=0$, yielding the more familiar form.) For such a curve, the affine part consists of solutions to the equation $$y^2 + a_1xy + a_3y = x^3 + a_2x^2+a_4x+a_6.$$ In addition, we also need to check the points "at infinity" (that is, with $Z=0$) that satisfy this equation. Going back to equation $(1)$, we see that if $Z=0$, then we just get the equation $0=X^3$, so it consists of the points $[0:b:0]$. Since $b\neq 0$, all of these points are represented by $[0:1:0]$. This is the only "point at infinity" that lies on the projective curve defined by equation $(1)$. If we keep our idea that points at infinity represent "directions" on the affine plane then this point is the point in "the vertical direction".

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The same thing happens over any finite field $k$. We have the affine $k$-plane $$\mathbb{A}^2_k\colon \{(a,b)\mid a,b\in k\},$$ and we have the projective $k$-plane $$\mathbb{P}^2_k = \{ [a:b:c]\mid a,b,c\in k\text{ not all zero}\},$$ where $[a:b:c]=[r:s:t]$ if and only if there exists $\lambda\in k$, $\lambda\neq 0$, such that $[a,b,c]=[\lambda r:\lambda s:\lambda t]$. An elliptic curve is a projective curve given by equation $(1)$, but where now $X$, $Y$, and $Z$ take values in $k$, not $\mathbb{R}$. And just as before, the only non-affine point that lies in the curve is the projective point $[0:1:0]$. So the projective elliptic curve has exactly one "point at infinity", namely $[0:1:0]$.

Note that in a field with $q$ elements, affine plane has $k^2$ points; "straight lines" (solutions to equations of the form $ax+by=0$ with $a,b$ not both zero) have $q$ points.

Projective $k$-plane has $q^3-1$ triples, but each triple has $q-1$ "names" corresponding to the $q-1$ possible values of $\lambda\neq 0$; so you have $(q^3-1)/(q-1) = q^2+q+1$ projective points. Straight lines (solutions to equations of the form $aX+bY=0$) have $q+1$ points: the $q$ points with $Z=1$, plus a single point with $Z=0$.

Answer 2

Look at the example of a projective line $X = \mathbb P^1(k)$ over a field $k$. Formally, $X$ consists of all equivalence classes of points $(x_0,x_1) \in k \times k$, where $(x_0, x_1)$ is equivalent to $(y_0, y_1)$ if there is a constant $0 \neq c \in k$ such that $cx_0 = y_0$ and $cx_1 = y_1$.

There is a canonical injective mapping of $k$ into $X$ where we send $x \in k$ to the equivalence class of $(x,1)$. So we may think of $k$ as a subset of $X$. Then $X$ is the disjoint union of this copy of $k$ and one more point $\infty$, namely the class of $(1,0)$. This is the point at infinity.

This intuition makes sense in the case where $k = \mathbb R$. Indeed, for $x \in \mathbb R$ large, the point $(x,1)$ is equivalent to $(1, \frac{1}{x})$, which approaches the point $\infty$ as $x$ tends to infinity. Even though this intuition breaks down when $k$ is a finite field, the algebraic construction of $\mathbb P^1(k)$ and its point at infinity may be carried out in exactly the same way.