I find myself becoming confused whenever I try to think about this. In the following, $K$ is a field.

An elliptic curve $\mathcal{C}$ is defined to be a nonsingular projective cubic curve over $K$, with a $K$-rational point. Now, I am told that the specified $K$-rational point is always the identity $\mathbf{o}$, which is the point at infinity, but I don't understand this. Isn't a $K$-rational point by definition supposed to be an affine point? I thought that $K$-rational means an element of $\mathbb{A}^2(K)$. The point $\mathbf{o}$ is simply not in affine space!

Another problem I have, which might stem from the above misunderstanding, is what the Mordell-Weil subgroup $\mathcal{C}(K)$ actually is. I understand that the points on $\mathcal{C}$ form a group under the usual law. But if the point at infinity is $K$-rational, and clearly all other points are affine and on a curve over $K$, then isn't the Mordell-Weil group just going to be all the points on $\mathcal{C}$? How is it ever a proper subgroup?


For ease of exposition, consider an elliptic curve with Weierstrass equation $y^2=x^3 + ax +b$. Note that the elliptic curve is actually the projective curve defined by

$y^2 z = x^3 + ax^2 z + b z^3$

If we look at the affine patch where $z \neq 0$, we get back the Weierstrass equation I wrote down originally.

If we look instead at the intersection of our curve with the line $z=0$, we see that the only (projective) point coming from this patch is our point at infinity $\cal{O}$=$[0,1,0]$, and certainly $0,1 \in K$ so this point is K-rational.

Edit: The definition of a K-rational point in $\mathbf{P}^n$ is a point $[x_0,\ldots, x_1] \in \mathbf{P}^n$ such that all $x_i \in K$. That's why $\cal{O}$ above is K-rational.

Not all points on an elliptic curve are K-rational. For example, consider the elliptic curve over $\mathbf{Q}$ defined by the Weierstrass equation

$E:y^2 = x^3 + x$

Then the point $[1,\sqrt{2},1] \in E$ is clearly not $\mathbf{Q}$-rational.

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  • $\begingroup$ Right, okay. So being $K$-rational refers in this case to projective space... I think I was looking at definitions in affine space and thinking that the two were identical. But then, as for my second question, doesn't this mean that the Mordell-Weil group is just all points on the curve? $\endgroup$ – Sputnik Jun 8 '11 at 19:56
  • $\begingroup$ No, see my further addition. (Sorry for all the edits!) The Mordell-Weil group is the group $E(K)$. In the example at the bottom of my post, I exhibited a curve E and a point which was in $E(\mathbf{\bar{Q}})$ but not in $E(\mathbf{Q})$. $\endgroup$ – Jeff Jun 8 '11 at 20:04
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    $\begingroup$ I think I get it now. So the idea is that the group law, as originally defined, applies to all points on the curve regardless of whether they are in $K$; so in your example, you can still 'add' points with irrational (or complex) coordinates. The M-W subgroup just refers to the ones that actually are in $K$. $\endgroup$ – Sputnik Jun 8 '11 at 20:17
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    $\begingroup$ Exactly! =D A nice elliptic curve has lots of points in general...if you look at the points over $\mathbf{C}$, they form a torus, in fact, and the points on this torus have a group structure. But if we look only at points over $\mathbf{Q}$, we get a subgroup with far fewer points, and in fact the M-W theorem tells us that this subgroup is finitely generated. $\endgroup$ – Jeff Jun 8 '11 at 20:20

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