Re: abstract algebra......

From: Dae-jung Yoo (no-mail_at_no-mail-address.com)
Date: 12/05/04 

Date: Sun, 5 Dec 2004 23:09:49 +0900

> ring homomorphism
> f: Z -> Z/3Z + Z/5Z + Z/7Z
> which f(x) = (x+3Z , x+5Z , x+7Z)
> (+ is direct sum)
>
> 1) ker f =?
>
> 2) find all integer x such that
> f(x) = (2+3Z , 3+5Z , 4+7Z)
>
> ---------------------------------------------
> i think.......
>
> 1) [3,5,7]=105, so 105*Z
>
> right ??

Right. 105Z is the kernel. You can make it sure by proving it.

>
> 2) um....sorry....i don't know....i need your advice.
>

You want to find a number k such that
(a) k == 2 mod 3 ,
(b) k == 3 mod 5 , and
(c) k == 4 mod 7 ,
and then claim that k+105Z is the answer for 2).
So let's find out what k would be.

Method I: (formula)
By Chinese Remainder Theorem, there is such k with 0< k < 105.
But the usual proof of Chinese Remainder Theorem provides a formula for k.
Use that formula to calculate k.

Method II: (heuristic)
numbers satysfing (a): 2, 5, 8, 11,...
...(b): 3, 8, 13, ...
...(c): 4, 11, ...
we see 11 is a number satysfing both (a) and (c).
So
numbers satisfying (a),(c): 11, 32, 53, ...
But the number 53 satisfy (b).
So k = 53.