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I'm reading J.H. Silverman's The Arithmetic of Elliptic Curves, chapter VIII section $1$, where he proves the weak Mordell-Weil theorem.

Right after proving proposition $1.6$, he makes a remark:

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My questions are:

1) The Minkowski theorem I know (from here) says that there are finitely many number fields whose discriminants are $\leq N$ (constant). Does the fact that $[K(Q):K]\leq m^2$ imply that the discriminant $\Delta_{K(Q)/K}$ is bounded? Why? Besides, what is the role of $S$ when applying Minkowski?

(maybe he's refering to a different Minkowski theorem, I don't know)

2) I understand that the Galois conjugates of $Q$ is of the form $Q+T$ for some $T\in E[m]$, but I can't see why this justifies the claim that $[K(Q):K]\leq m^2$.

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Remember that the primes dividing the relative discriminant of $L/K$ are exactly the primes of $K$ that are ramified in $L$. If we furthermore bound the degree of $L/K$, then we can bound the power with respect to which any particular prime divides the discriminant. Thus if we fix a finite set of ramified primes and bound the degree, we bound the discriminant, and so bound the number of fields extensions $L/K$.

As for your second question, there are $m^2$ elements $T$ of $E[m]$, and so $Q$ has at most $m^2$ Galois conjugates. Thus its minimal polynomial has degree at most $m^2$, and hence so does the extension $K(Q)$.

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  • $\begingroup$ "If we furthermore bound the degree of $L/K$, then we can bound the power with respect to which any particular prime divides the discriminant". Could you explain this in more detail? I don't see exactely how the bounded degree implies the powers are bounded. $\endgroup$ – rmdmc89 Oct 26 '17 at 21:59
  • $\begingroup$ By the way, I've just realized that the relative discriminant $\Delta_{L/K}$ is an ideal, right? So what does it mean to say that it is "bounded"? And does the theorem apply to relative discriminants? $\endgroup$ – rmdmc89 Oct 26 '17 at 22:02
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    $\begingroup$ In these situations, bounded just means that Nm$_{K/\mathbb Q}(\Delta_{L/K})$ is bounded. To answer your first question, we know that $\Delta_{L/K}$ is just the relative $L/K$-norm of the different ideal of $L/K$. But the valuation of the discriminant at each prime are bounded in terms of the ramification index and the valuation of the ramification index (Thm 12.26 in the link) so we can get a universal bound for this in terms of the degree of $L/K$. math.mit.edu/classes/18.785/2016fa/LectureNotes12.pdf $\endgroup$ – Ravi Oct 28 '17 at 23:50
  • $\begingroup$ @tracing, what do you mean by minimal polynomial of $Q$? I would understand it if you meant minimal polynomial of some number $a\in K(Q)$, but not a point in $E(K)$ $\endgroup$ – rmdmc89 Nov 14 '17 at 18:49
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    $\begingroup$ @AguirreK: I mean the minimal polys. of its $x$ and $y$ coords. $\endgroup$ – tracing Nov 16 '17 at 12:41

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