3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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 $x^4 = 1 \pmod {5^2}$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.