Forgot password?
 Register account
View 279|Reply 5

[数论] 求 (6) 的所有约数

[Copy link]

3159

Threads

7941

Posts

610K

Credits

Credits
63770
QQ

Show all posts

hbghlyj Posted 2024-2-19 02:35 |Read mode
Last edited by hbghlyj 2024-2-19 11:35这个帖子中他列出了 6 个不同的
$\mathfrak{p}_{2}=(2,\sqrt{-10})$,
$(2)$,
$\mathfrak{p}_{3}=(3)$,
$\mathfrak{p }_{2}\mathfrak{p}_{3}$,
$2\mathfrak{p}_{3}$,
$(1)$.
它们包含 6。

3159

Threads

7941

Posts

610K

Credits

Credits
63770
QQ

Show all posts

 Author| hbghlyj Posted 2024-2-19 03:19

我尝试画画;以下是正确的吗?箭头代表包含

G (1) (1) p2 𝔭2 (1)->p2 p3 𝔭3 (1)->p3 (2) (2) p2->(2) p2p3 𝔭2𝔭3 p2->p2p3 2p3 2𝔭3 (2)->2p3 p3->p2p3 p2p3->2p3

3159

Threads

7941

Posts

610K

Credits

Credits
63770
QQ

Show all posts

 Author| hbghlyj Posted 2024-2-19 05:03
也许画成平行四边形更好:

48

Threads

771

Posts

110K

Credits

Credits
13880
QQ

Show all posts

Czhang271828 Posted 2024-2-19 10:22
$(6)=(2)(3)=\mathfrak p_2^2\mathfrak p_3$. 和 $\mathbb Z$ 中质因数分解一样(只不过这里变成素理想).

3159

Threads

7941

Posts

610K

Credits

Credits
63770
QQ

Show all posts

 Author| hbghlyj Posted 2024-2-19 16:06
Czhang271828 发表于 2024-2-19 02:22
$(6)=(2)(3)=\mathfrak p_2^2\mathfrak p_3$. 和 $\mathbb Z$ 中质因数分解一样(只不过这里变成素理想). ...
懂了。$\mathfrak p_2^i\mathfrak p_3^j$,$i=0,1,2,j=0,1$,所以6個。

3159

Threads

7941

Posts

610K

Credits

Credits
63770
QQ

Show all posts

 Author| hbghlyj Posted 2024-2-19 17:00
  1. K.<α> = QuadraticField(-10)
  2. I = ideal(6+0*α)
  3. I.factor()
Copy the Code
(Fractional ideal (2, α))^2 * (Fractional ideal (3))

Mobile version|Discuz Math Forum

2025-5-31 10:54 GMT+8

Powered by Discuz!

× Quick Reply To Top Edit