Let $K=\mathbb{Q}(\sqrt d)$, where $d$ is squarefree and greater than $1$. Assume that $5|(d+1)$ or $5|(d-1)$. Let $\mathfrak{P}$ be a prime of $K$ over $5$. Let $\delta _o$ be a fundamental unit of $K$.

How to show that $\delta_o$ is a $5^{th}$ power in $K_{\mathfrak{P}}$ iff $\delta_o^2 \equiv \pm 1$ (mod $\mathfrak{P}^2)$?

Try: I have shown that $5$ splits in $K$ and $\delta_o^2\equiv \pm 1$ (mod $\mathfrak{P})$. Then I tried to use Hensel`s lemma but failed. Thank you for your help.

  • $\begingroup$ What do you mean by the notation, $K$-subscript-whatever-that-weird-symbol-is? $\endgroup$ – Gerry Myerson Sep 14 '15 at 12:31
  • 1
    $\begingroup$ It is the completion of $K$ at prime $\mathfrak{P}$. $\endgroup$ – MathStudent Sep 14 '15 at 12:41

The $5$th power map, $(1+5x) \mapsto (1+5x)^5 = 1 + 5^2x + O(5^3x^2)$ is a bijection between $1 + 5 \Bbb Z_5$ and $1 + 5^2\Bbb Z_5$ (to show this you can use Hensel's lemma on the polynomial with integer coefficients $((1+5X)^5-1)/5^2$)

In particular this means that anything congruent to $1$ modulo $5^2$ is a $5$th power in $\Bbb Z_5$, and so the invertible $5$th powers are an open subgroup of $\Bbb Z_5^*$ : an invertible element of $\Bbb Z_5$ is a $5$th power if and only if it is a $5$th power mod $5^2$.

The $5$th powers mod $5^2$ are $1^5=1,2^5=7,3^5=-7$ and $4^5=-1$.

Next, the squaring map on the invertibles of $\Bbb Z_5$ is a local isometry, and is $2$-to-$1$ because $1^2 = -1$, so $x = \pm 1, \pm 7 \pmod {5^2}$ if and only if $x^2 = 1^2$ or $x^2 = 7^2 = -1\pmod {5^2}$.

You can even go one step further and show that this is also equivalent to $x^4 = 1 \pmod {5^2}$


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