题号:14622    题型:解答题    来源:2024年丘成桐大学生数学竞赛(代数与数论类)-无答案
Let $n \geq 1$ be an integer and write $\Phi_n(X)$ the $n$-th cyclotomi c polynomial, that is, the minimal polynomial of a primitive $n$ th root of unity in $\mathbb{C}$ over $\mathbb{Q}$. Write also $\varphi(n)=\operatorname{deg}\left(\Phi_n(X)\right)$
(1) Let $q$ be a power of a prime number such that $(q, n)=1$. Show that $\Phi_n$, viewed as an element in $\mathbb{F}_q[X]$, can be decom posed as a product of $\varphi(n) / d$ irreducible polynomials of deg ree $d$, with $d$ the order of $q$ in the multiplicative group $(\mathbb{Z} / n \mathbb{Z})^{\times}$.
(2) From now on, assume $n=2^{r+1}$ for some integer $r \geq 1$. L et $\zeta=\zeta_n$ be a primitive $n$-th root of unity and $K=\mathbb{Q}[\zeta]$. Let $p$ be a prime with $p \equiv-3(\bmod 8)$.
(a) For $x, y \in K=\mathbb{Q}[\zeta]$, define
$$
(x, y):=\sum_\tau \tau(x) \cdot \overline{\tau(y)}
$$
where $\tau$ runs through all the embeddings $K \hookrightarrow \mathbb{C}$ of $K$ into the field $\mathbb{C}$ of complex numbers. Write $K_{\mathbb{R}}=K \otimes_{\mathbb{Q}} \mathbb{R}$, and we use the same notation to denote the (à priori $\mathbb{C}$-valued) bi linear form on $K_{\mathbb{R}}$ obtained by extension of scalars. Show tha $\mathrm{t}(\cdot, \cdot)$ gives an inner product on $K_{\mathbb{R}}$ and for $0 \leq i, j < 2^r$,
$$
\left(\zeta^i, \zeta^j\right)= \begin{cases}2^r, & \text { if } i=j \\ 0, & \text { otherwise }\end{cases}
$$

In particular, we obtain an Euclidean space $K_{\mathbb{R}}$ and $\left(\zeta^i / \sqrt{2^r}\right)_{0 \leq i < 2^r}$ is an orthonormal basis.
(b) Decompose $p \mathcal{O}_K$ into a product of prime ideals.
(c) Let $\mathfrak{p} \subset \mathcal{O}_K$ be a prime ideal of $\mathcal{O}_K$ containing $p$. Show th at for every $\alpha \in \mathfrak{p},|\alpha|^2 \in 2^r p \mathbb{Z}$, and compute the length of $\mathrm{t}$ he shortest non-zero vector in the prime ideal $\mathfrak{p} \subset K_{\mathbb{R}}$.
0 人点赞 纠错 ​ 15 次查看 ​ 我来讲解
答案:

解析:

答案与解析:
答案仅限会员可见 微信内自动登录手机登录微信扫码注册登录 点击我要 开通VIP