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In notes of prof. W.Stein - http://wstein.org/edu/2010/581b/stein-algebraic_number_theory.pdf - the first paragraph of page 112 has the following told:

"When $K$ is a perfect field, the prime ideals correspond to the Galois orbits of affine points of $E(\bar{K})$."

I would be thankful if someone could elaborate what is meant here by the Galois orbits of points of $E$ and what is the implied correspondence.

Thank you!

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  • $\begingroup$ Do you know the Nullstellensatz? $\endgroup$
    – Bruno Joyal
    Jan 28, 2014 at 17:30
  • $\begingroup$ Yes. Here, I fail to see what is the Galois group whose action on $E(\bar{K})$ the term "galois orbit" suggests; it doesn't seem to be the Gal(K(x, y)/K(x)). $\endgroup$
    – Albertas
    Jan 28, 2014 at 18:03
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    $\begingroup$ It is the Galois group of $\overline{K}$ over $K$! Note that since $E$ is defined over $K$, there is an action of this Galois group on $E(\overline{K})$. $\endgroup$
    – Bruno Joyal
    Jan 28, 2014 at 18:13
  • $\begingroup$ Thanks! I'll have to think a bit then; wll ask here again if I won't manage to clarify the rest by myself. $\endgroup$
    – Albertas
    Jan 28, 2014 at 18:33
  • $\begingroup$ Sounds good. Perhaps you'd like to take a look at the first chapter of Silverman's Arithmetic of Elliptic Curves. $\endgroup$
    – Bruno Joyal
    Jan 28, 2014 at 18:44

1 Answer 1

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When $K = \bar{K}$, the correspondence defined by $P \mapsto \{r \in R: r(P) = 0\}$ is the well-known bijection (as follows from the Nullstellensatz, prime ideals correspond to irreducible algebraic sets), while when $K \neq \bar{K}$, any $\sigma(P)$ gets mapped to the same prime ideal when $\sigma$ runs through $Gal(\bar{K}/K)$.

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