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A definition of multiplication on the projective $1$-points $(a:b)$ of $P_K^1$ with $a$ and $b$ elements of a field $K$ ( e.g. the real or rational numbers ) can be given by mimicking the multiplication of complex numbers.

Is it known whether there a way to define an addition between projective $1$-points ?

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    $\begingroup$ Your proposed multiplication isn't well-defined if $K$ itself already contains a square root of $-1$: in this case, $K[i]$ has zero divisors, so it's possible for the "product" of two points to have both coordinates zero and hence to not be a well-defined point on the projective line. $\endgroup$
    – Qiaochu Yuan
    Commented Aug 2, 2015 at 17:44
  • $\begingroup$ If $K$ doesn't contain a square root of $-1$, then your group is the quotient of the multiplicative group of $K[i]$ by the multiplicative group of $K$. The closest ring in sight is $K[i]$, which, being a field, has no nontrivial quotients. $\endgroup$
    – Qiaochu Yuan
    Commented Aug 2, 2015 at 18:00

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The topological space $\mathbb{P}^1_{\mathbb{R}}$ is a circle, so it has an abelian group structure just defined by adding angles but this is not algebraic, not compatible with the multiplication you describe, and it can't be extended to $\mathbb{P}^1_{\mathbb{C}}$ (which is a sphere and thus has no topological group structure), nor does it restrict to $\mathbb{P}^1_{\mathbb{Q}}$. EDIT: As Noam Elkies points out, the circle group structure can be made algebraic on $\mathbb{P}^1_{\mathbb{R}}$ by defining it by $\tan(x+y)=\tan(x)+\tan(y)$, instead of by adding angles. This still has all the other defects listed above.

More generally, it's impossible to do this algebraically: it's an extremely well known fact that a projective curve with an algebraic group structure (as addition would be) must be an elliptic curve.

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  • $\begingroup$ It is far beyond my knowledge that "a projective curve with an algebraic group structure (as addition would be) must be an elliptic curve". In what kind of books do I find this remarkable result ? Algebraic Geometry, Elliptic Curves, or Projective Geometry ? $\endgroup$
    – Wolfgang Tintemann
    Commented Aug 2, 2015 at 16:05
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    $\begingroup$ Actually there is an algebraic group structure on ${\bf P}^1_{\bf R}$, namely $x \oplus y = (x+y) \, / \, (1-xy)$. But it can't extend to ${\bf P}^1_{\bf C}$, nor to a ring structure on ${\bf P}^1_{\bf R}$. $\endgroup$
    – Noam D. Elkies
    Commented Aug 2, 2015 at 17:34
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    $\begingroup$ This looks like the addition formula of the $\tan$-function. $\endgroup$
    – Wolfgang Tintemann
    Commented Aug 2, 2015 at 17:46
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    $\begingroup$ @WolfgangTintemann: Yes, it is obtained by conjugating addition on $\mathbb{R}/\pi\mathbb{Z}$ by the homeomorphism $\tan:\mathbb{R}/\pi\mathbb{Z}\to\mathbb{P}^1_{\mathbb{R}}$. $\endgroup$
    – Eric Wofsey
    Commented Aug 2, 2015 at 17:52
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    $\begingroup$ If $\phi: X \to Y$ is a bijection and $X$ carries a group structure (denoted by $+_X$), then $Y$ carries a group structure by "conjugation": $y +_Y y' = \phi(\phi^{-1}(y) +_X \phi^{-1}(y'))$. This I'm sure is what Eric meant. Another term for this (which generalizes to any sort of structure you like) is "transport of structure". It's a general thing all mathematicians pick up on at some point. $\endgroup$
    – Todd Trimble
    Commented Aug 2, 2015 at 19:51
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Here is a less algebraic and more topological answer: it's known that any compact (Hausdorff) topological ring must be totally disconnected. In particular, there's no hope for either $\mathbb{RP}^1$ or $\mathbb{CP}^1$ to have the structure of a topological ring. (As Ben Webster mentions, $\mathbb{CP}^1$ also can't even be a topological group.)

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