Proving Idempotents of Z/100Z: Solutions and Explanation

[chivhone]

Homework Statement

How do you proove what the idempotents of the commutative ring Z/100Z are?

Homework Equations

elements 'a' are idempotent if a^2 = a i.e a(a-1)=0
Find the integers 0 < a < 99 (inclusive), such that 100 divides a(a-1)

The Attempt at a Solution

I know by trial and error that the answers are the elements 0,1,25,76 but have no proof as to why this is.
i know i have to find the integers 'a' such that 100 divides a(a-1)=a^2 -a but can't do the maths to get a proper reasoning

100 = 2x2x5x5, so does that mean if 100 divides a(a-1) that 2 divides a(a-1) as would 5.
the problem is this wouldn't lead me to any of the known solutions either?


Thanks for any help

 

[mighty2000]

!

To prove the idempotents of the commutative ring Z/100Z, we need to find all the elements 'a' such that a^2 = a. As you correctly stated, this means a(a-1) = 0. We also know that in the ring Z/100Z, the number 100 is equivalent to 0. Therefore, we can rewrite the equation as a(a-1) = 0 (mod 100).

To find all the solutions, we can consider the factors of 100: 2, 2, 5, and 5. This means that a(a-1) must be divisible by 2, 2, 5, and 5. We can break this down into four cases:

1) a is divisible by 2 and 5, but not 2 and 5: In this case, a must be equivalent to either 0 or 50 (mod 100). This means that a = 0 or 50.

2) a is divisible by 2, but not 5: In this case, a must be equivalent to either 1 or 51 (mod 100). This means that a = 1 or 51.

3) a is divisible by 5, but not 2: In this case, a must be equivalent to either 25 or 75 (mod 100). This means that a = 25 or 75.

4) a is not divisible by 2 or 5: In this case, a must be equivalent to either 2, 26, 76, or 99 (mod 100). However, these values do not satisfy the condition a(a-1) = 0 (mod 100).

Therefore, the idempotents of the commutative ring Z/100Z are 0, 1, 25, and 76. I hope this helps clarify the reasoning behind the solutions. If you have any further questions, please don't hesitate to ask.