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[数论] 对于哪个d,𝐐(√d) 上的整数分解不唯一?

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hbghlyj Post time 2024-2-19 21:48 |Read mode
$𝐐(\sqrt{10})$上:
\[6=2 \cdot 3=(4+\sqrt{10})(4-\sqrt{10}) \text {. }\]类似地,
\begin{array}{lll}
\mathbf{Q}(\sqrt{15})  : & 10=2 \cdot 5=(5+\sqrt{15})(5-\sqrt{15}) \\
\mathbf{Q}(\sqrt{26})  : & 10=2 \cdot 5=(6+\sqrt{26})(6-\sqrt{26}) \\
\mathbf{Q}(\sqrt{30})  : & 6=2 \cdot 3=(6+\sqrt{30})(6-\sqrt{30})
\end{array}(以上例子并不是唯一的)

这个$d$的序列与oeis.org/A029702相同吗?

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 Author| hbghlyj Post time 2024-2-19 22:07
1-100所有没有平方因数的 d:
[1, 2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 17, 19, 21, 22, 23, 26, 29, 30, 31, 33, 34, 35, 37, 38, 39, 41, 42, 43, 46, 47, 51, 53, 55, 57, 58, 59, 61, 62, 65, 66, 67, 69, 70, 71, 73, 74, 77, 78, 79, 82, 83, 85, 86, 87, 89, 91, 93, 94, 95, 97]
  1. # Define a function to check if a number is square-free
  2. def is_square_free(n):
  3.     # Factorize the number into prime powers
  4.     factors = factor(n)
  5.     # Check if any of the prime powers has an exponent greater than 1
  6.     for p, e in factors:
  7.         if e > 1:
  8.             return False
  9.     # If none of the prime powers has an exponent greater than 1, the number is square-free
  10.     return True
  11. # Create an empty list to store the square-free integers
  12. square_free = []
  13. # Loop through the numbers from 1 to 100
  14. for n in range(1, 101):
  15.     # Check if the number is square-free and add it to the list if it is
  16.     if is_square_free(n):
  17.         square_free.append(n)
  18. # Print the list of square-free integers
  19. print(square_free)
Copy the Code
如何从这个列表中过滤掉满足唯一分解条件的d?

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睡神 Post time 2024-2-19 22:12 From the mobile phone
太高深了,我连题目都还没看懂…还是先解决我的那个问题吧,至少我的更通熟易懂
除了不懂,就是装懂

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 Author| hbghlyj Post time 2024-2-19 22:32
睡神 发表于 2024-2-19 14:12
我连题目都还没看懂


例如$𝐐(\sqrt{-1})$
$x^2+1=(x+i)(x-i)$
$𝐙(\sqrt{-1})$:
$2=-i(1+i)^2$
Untitled.gif

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 Author| hbghlyj Post time 2024-2-19 22:37
wolframalpha.com/input?i=PrimeQ%5B1%2BI%5D
Input
Untitled.gif
Result
Untitled.gif
$1 + i$ is a prime number

wolframalpha.com/input?i=PrimeQ%5B7%2B24I%5D
Input
Untitled.gif
Result
Untitled.gif
$7+24i$ is not a prime number
$|7+24i|=25=|15+20i|$

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Czhang271828 Post time 2024-2-20 17:26
答案应该是数列 oeis.org/A219361 的补. 因为非唯一分解 Dedekind 整环的类数可以是 $2$ 以及更高, 例如 $\mathbb Q(\sqrt{79})$ 的代数整数环.

关于 $\mathcal O_{\mathbb Q(\sqrt{d})}$ 是否是唯一分解环: $d<0$ 时参考 Heegner numbers, $d>0$ 时没有简单的解.

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 Author| hbghlyj Post time 2024-2-20 17:34

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 Author| hbghlyj Post time 2024-3-26 06:24
Czhang271828 发表于 2024-2-20 09:26
例如 $\mathbb Q(\sqrt{79})$ 的代数整数环.


NumberFieldClassNumber[Sqrt[79]]=3

Dedekind domain $\mathbb{Z}[\sqrt{79}]$ has class group of order 3, representatives $(1), (3,\sqrt{79}+1), (3, \sqrt{79}-1)$.

相关:If $\mathcal{O}_{\Bbb{Q}(\sqrt{d})}$ has class number $2$ or higher, does that mean $\sqrt{d}$ is irreducible but not prime?

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