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Author: hbghlyj

[数论] $q$ 是有理数, $\cos qπ$ 也是有理数

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 Author| hbghlyj Posted at 2023-3-19 01:35:33
Last edited by hbghlyj at 2024-10-19 14:54:00
abababa 发表于 2023-2-11 11:44
印象里maven证明过,是把多项式弄成一个友矩阵,然后用张量积。同样的方法还可以证明全体代数数构成一个域 ...
搜到了,是不是这个:
求$x=\sqrt[3]{2}+\sqrt[5]{3}$的极小多项式
发网友的解答:$A=\begin{bmatrix}
0 & 0 & 2\\
1 & 0 & 0\\
0 & 1 & 0\\
\end{bmatrix},B=\begin{bmatrix}
0 & 0 & 0 & 0 & 3\\
1 & 0 & 0 & 0 & 0\\
0 & 1 & 0 & 0 & 0\\
0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & 1 & 0\\
\end{bmatrix}$
分别作$A,B$和$5,3$阶单位矩阵的张量积再求和,最后求此矩阵的特征多项式,这就是结果。
我用软件试了一下,确实是,结果是15次多项式。

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 Author| hbghlyj Posted at 2023-6-1 17:06:14
How to prove that the sum and product of two algebraic numbers is algebraic? [duplicate]
Recall that finding a polynomial over which $\alpha+\beta$ or $\alpha \beta$ is a root of $p(x) \in F[x]$ is equivalent to finding the eigenvalue of a square matrix over $F$ (living in some algebraic extension of $F$), since you can link the polynomial $p(x)$ to the companion matrix $C(p(x))$ which has precisely characteristic polynomial $p(x)$, hence the eigenvalues of the companion matrix are the roots of $p(x)$.

If $\alpha$ is an eigenvalue of $A$ with eigenvector $x \in V$ and $\beta$ is an eigenvalue of $B$ with eigenvector $y \in W$, then using the tensor product of $V$ and $W$, namely $V \otimes W$, we can compute
$$
(A \otimes I + I \otimes B)(x \otimes y) = (Ax \otimes y) + (x \otimes By) = (\alpha x \otimes y) + (x \otimes \beta y) = (\alpha + \beta) (x \otimes y)
$$
so that $\alpha + \beta$ is an eigenvalue of $A \otimes I + I \otimes B$. Also,
$$
(A \otimes B)(x \otimes y) = (Ax \otimes By) = (\alpha x \otimes \beta y) = \alpha \beta (x \otimes y)
$$
hence $\alpha \beta$ is an eigenvalue of the matrix $A \otimes B$. If you want explicit expressions for the polynomials you are looking for, you can just compute the characteristic polynomial of the tensor products.

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 Author| hbghlyj Posted at 2024-10-19 22:30:19
abababa 发表于 2023-2-14 13:47
证明:由于$a$是代数整数,所以存在首一整系数多项式$p(x) = x^m+c_{m-1}x^{m-1}+\cdots+c_1x+c_0$使得$p(a) = 0$,假设$a \in \mathbb{Q} \setminus \mathbb{Z}$,则可设$a = \frac{s}{t}, \gcd(s,t) = 1, t \neq 1,0$,代入方程$p(a) = 0$并化简有
\[s^m = -(c_{m-1}s^{m-1}t+\cdots+c_1st^{m-1}+c_0t^m)\]

显然$t$整除右边,所以$t \mid s^m$,由于$t \neq 1$,所以$t$存在一个素因子$q$,从而$q \mid t \mid s^m$,而$q$是素数,由 Euclide 第一定理知$q \mid s$,但$q \mid t$,于是$q \mid \gcd(s,t)$,这与$\gcd(s,t) = 1$矛盾,所以假设$a \in \mathbb{Q} \setminus \mathbb{Z}$错误,因此$a \not\in \mathbb{Q} \setminus \mathbb{Z}$。
这个证明是有理数根定理(RRT)的证明的特殊情况

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 Author| hbghlyj Posted at 2024-11-7 18:18:41
kuing 发表于 2022-4-6 06:08
(公式后面不知为啥多了条竖线……)
scholarpedia.org/article/Calabi-Yau_manifold 也是,公式后面不知为啥多了条竖线
Screenshot 2024-11-07 101855.png

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