While reading an introduction on elliptic curves, I stumbled upon something called reduction modulo $p$.

The definition states that we want to create a group homomorphism that maps an elliptic curve on the rational field to the integers modulo a prime $p$, i.e.: $$E(\mathbb{Q}) \rightarrow \tilde{E}(\mathbb{F}_p), \;\;\;\; P \mapsto \tilde{P}$$

Where $P = (x,y) \in E(\mathbb{Q}),$ and $\tilde{P} = (\tilde{x},\tilde{y}) \in \tilde{E}(\mathbb{F}_p)$.

Such a homomorphism is worked out trivially when $x$ and $y$ are in $\mathbb{Z}$, while in the rational case the reader has to write $x = \frac{a}{b}$. Then, if $p \nmid b$, then $\tilde{b}$ has an inverse in $\mathbb{F}_p$, hence $\tilde{x} = \tilde{a}\tilde{b}^{-1}$ in $\mathbb{F}_p$.

The last step is quite obscure to me. The paper provides some examples:

Let $E: y^2 = x^3 + 2x + 4.$

Then, $P = (2,4)$ and $Q = \left(\frac{1}{4},\frac{17}{8}\right)$ are points in $E(\mathbb{Q})$.

Then the reduction modulo 11 map $E(\mathbb{Q}) \rightarrow \tilde{E}(\mathbb{F}_{11})$ gives:

$\tilde{P} = (2,4)$ and $\tilde{Q} = (3,9)$.

I feel like I'm missing something trivial, but I really have no idea of how the author found the value of $\tilde{Q}$. Can someone help?

  • 3
    $\begingroup$ $1/4\equiv3\bmod{11}$ because $1\equiv4\times3\bmod{11}$. Similarly, $17/8\equiv9$ because $17\equiv8\times9$, working modulo $11$. $\endgroup$ – Gerry Myerson Mar 24 '15 at 10:18
  • $\begingroup$ Well, if I didn't miss something trivial... Thank you very much! $\endgroup$ – Luigi D. Mar 24 '15 at 10:25

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