1

Consider the field $K=\mathbb Q(\sqrt{2})$. Let $mp=\frac{(2+\sqrt{2})^{p}-1}{1+\sqrt{2}}$.

For the field extension, $K(-1+2\sqrt{2})/K$, and $p\equiv 5\bmod{6}$ how can one show

$ \frac{(2+\sqrt{2})^{p}-1}{1+\sqrt{2}}\equiv 1\pmod{-1+2\sqrt{2} }$ and hence, Artin symbol corresponding to $ K(-1+2\sqrt{2})/K$, is trivial,

where as over $ K(-1-2\sqrt{2})/K$

$ \frac{(2+\sqrt{2})^{p}-1}{1+\sqrt{2}}\not\equiv 1\pmod{-1-2\sqrt{2} }$. I workedout as to take, $-7=(-1+2\sqrt{2})(-1-2\sqrt{2})$ and evaluate $mp$ mod at 7. Any Hint is also welcomed.

Math123
  • 1,243
  • We have $\sqrt2\in K$, so $K(-1+2\sqrt2)=K$, and the extension is trivial, no? – Jyrki Lahtonen Oct 28 '13 at 07:50
  • 1
    @JyrkiLahtonen I got this doubt, while evaluating norms of Mersenne primes, to represent it in the form $x^2+7y^2$ – Math123 Oct 28 '13 at 08:19
  • @JyrkiLahtonen is this statement true: $Frob_{7} (2+\sqrt{2})\equiv 2+\sqrt{2} \pmod{-1+2\sqrt{2}}$ is identity, hence the Artin symbol is trivial – Math123 Oct 29 '13 at 08:34

1 Answers1

3

The Galois group of the extension $K(-1+2\sqrt2)/K$ is trivial, so all the Frobenius automorphisms are equal to the identity for lack of alternatives.

In the specific example from comments we can also check it as follows. You already showed that $(-1+2\sqrt2)\mid 7$. Therefore $$2+\sqrt2\equiv2-6\sqrt2=-1-3(-1+2\sqrt2)\equiv-1\pmod{(-1+2\sqrt2)}.$$ Hence $$ (2+\sqrt2)^7\equiv(-1)^7\equiv(-1)\equiv2+\sqrt2\pmod{(-1+2\sqrt2)}, $$ that is compatible with the Frobenius being the identity.

Jyrki Lahtonen
  • 133,153