User oshe - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T12:17:35Z http://mathoverflow.net/feeds/user/16151 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/71052/the-number-of-short-vectors-in-a-lattice The Number of Short Vectors in a Lattice OSHE 2011-07-23T06:11:09Z 2011-07-23T14:01:15Z <p>Given a lattice $L = \bigoplus_{i=1}^{m} \mathbb{Z}v_i$ (the $v_i$ are linearly independent vectors in $\mathbb{R}^n$) and a number $c > 0$, can one quickly compute or find a good estimate on the number of lattice vectors $v$ with $|v| \leq c$ without actually enumerating these vectors? The basis $v_1,\ldots, v_m$ of the lattice can be assumed to be LLL reduced.</p> <p>Also asked at: <a href="http://cstheory.stackexchange.com/questions/7488/the-number-of-short-vectors-in-a-lattice" rel="nofollow">http://cstheory.stackexchange.com/questions/7488/the-number-of-short-vectors-in-a-lattice</a></p> http://mathoverflow.net/questions/69416/relating-p-adic-valuations-of-elements-in-mathbbc-and-mathbbc-p Relating $p$-adic Valuations of Elements in $\mathbb{C}$ and $\mathbb{C}_p$ OSHE 2011-07-03T21:23:26Z 2011-07-18T04:22:12Z <p>Let $K = \mathbb{Q}(\theta)$, where $\theta$ is a root of an irreducible polynomial $g \in \mathbb{Z}[t]$. Fix a rational prime $p$. Let $\theta^{(1)}, \ldots, \theta^{(n)}$ be the roots of $g$ in $\overline{\mathbb{Q}}$, and let $\phi^{(1)}, \ldots, \phi^{(n)}$ be the roots of $g$ in $\overline{\mathbb{Q}_p}$ (the algebraic closure of $\mathbb{Q}_p$). Let $f \in \mathbb{Z}[t]$, and put $$\alpha = \frac{f(\theta^{(a)})}{f(\theta^{(b)})}, \qquad \beta = \frac{f(\phi^{(c)})}{f(\phi^{(d)})}.$$ Let $L$ be a finite extension of $K$ containing $\alpha$, and let $\mathfrak{p}$ be a prime ideal of $L$ above $p$.</p> <p>If <code>$\text{ord}_p(\beta) = 0$</code>, is <code>$\text{ord}_{\mathfrak{p}}(\alpha) = 0$</code> also? Why? What is the general relationship between <code>$\text{ord}_p(\beta)$</code> and <code>$\text{ord}_{\mathfrak{p}}(\alpha)$</code>?</p> <p>I know that if $\mathfrak{p}$ stands for a prime ideal of $K$ above $p$, if $g_{\mathfrak{p}}$ is the irreducible factor of $g$ in <code>$\mathbb{Q}_p[t]$</code> which corresponds to $\mathfrak{p}$, and if $\phi^{(\cdot)}$ is any root of $g_{\mathfrak{p}}$, then for any $x \in K$ $$\frac{1}{e}\text{ord}_{\mathfrak{p}}(x) = \text{ord}_p(\sigma(x)),$$ where $e$ is the ramification index of $\mathfrak{p}$ over $p$ and $\sigma$ is the embedding of $K$ into $\mathbb{Q}_p(\phi^{(\cdot)})$ that maps $\theta$ to $\phi^{(\cdot)}$ and fixes $\mathbb{Q}$ pointwise.</p> <p>I have a feeling that I can view $\beta$ as the image of $\alpha$ under some embedding (there is an obvious candidate) and get a similar formula relating <code>$\text{ord}_p(\beta)$</code> and <code>$\text{ord}_{\mathfrak{p}}(\alpha)$</code>, but I am having trouble proving this or making it concrete.</p> <p>Can someone please help?</p>