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I'm current reading Koblitz's "Introduction to Elliptic Curves and Modular Forms," and the author repeatedly mentions that, given a fixed point $P$, points $Q$ of the form $2Q=P$ are found by taking the lines emanating from $-P$ that are tangent to the curve somewhere, and that there are 4 distinct lines with this property.

I understand that if these 4 lines exist, the resulting points at which they are tangent to the curve are indeed exactly the points $Q$ we are after. What I do not understand is why these 4 distinct lines are always guaranteed to exist.

Johnny Apple
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  • Are you familiar with the fact that an elliptic curve over $\mathbb{C}$ is abstractly isomorphic to $S^1 \times S^1$? That makes its properties as an abstract group very clear. – Qiaochu Yuan Oct 22 '20 at 23:47
  • I am, and have considered this, but I do not see how that isomorphism answers my question. If P is the point at infinity, then we know there are 4 points of order 2, so that would answer the question in that case, but the general case eludes me. – Johnny Apple Oct 23 '20 at 00:06
  • The general case is the same. There's at least one square root, and the other square roots must be obtained from it by a square root of the identity. This is a general feature of divisible abelian groups. – Qiaochu Yuan Oct 23 '20 at 00:31
  • I'm not sure what you mean. Square root of what? Would you mind posting an answer? – Johnny Apple Oct 23 '20 at 00:49
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    If you write $2Q = P$ multiplicatively you get $Q^2 = P$, so it makes sense to say that $Q$ is a square root of $P$. I don't know what to call $Q$ if you write it additively (saying "a half of $P$" sounds awkward to me). – Qiaochu Yuan Oct 23 '20 at 01:03

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If you know that an elliptic curve over $\mathbb{C}$ is isomorphic to $S^1 \times S^1$ as a group then you can argue as follows: if $Q$ is any element then there is at least one $P$ such that $2P = Q$. Moreover, if $P_1, P_2$ are two such elements then

$$2(P_1 - P_2) = Q - Q = 0$$

so $P = P_1 - P_2$ satisfies $2P = 0$, and there are exactly four such points, namely $(0, 0), (\frac 12, 0), (0, \frac 12), (\frac 12, \frac 12)$ (thinking of $S^1$ as $\mathbb{R}/\mathbb{Z}$). So $2P = Q$ has $4$ solutions for any $Q$, not just $Q = 0$.

Qiaochu Yuan
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