What Are the Idempotents of the Ring Z/100Z?

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SUMMARY

The idempotents of the ring Z/100Z are definitively identified as the elements 0, 1, 25, and 76. This conclusion is reached by solving the equation a(a-1) = 0 (mod 100), which leads to the requirement that 100 divides a(a-1). The proof involves demonstrating that for each of these elements, the condition a² ≡ a (mod 100) holds true, confirming their status as idempotents.

PREREQUISITES
  • Understanding of modular arithmetic, specifically Z/nZ
  • Familiarity with the concept of idempotent elements in ring theory
  • Basic algebraic manipulation and proof techniques
  • Knowledge of divisibility rules and properties of integers
NEXT STEPS
  • Study the properties of idempotent elements in different rings, such as Z/nZ for various n
  • Learn about the structure of rings and fields in abstract algebra
  • Explore the application of modular arithmetic in number theory
  • Investigate the relationship between idempotents and the decomposition of rings
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This discussion is beneficial for students of abstract algebra, mathematicians focusing on ring theory, and anyone interested in the properties of modular arithmetic and idempotent elements.

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

I know by trial and error that they 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 do the maths to get a proper reasoning

any help?
thanks
 
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Chose an element [a] of Z/100Z whose representant a is such that [itex]0\leq a \leq 99[/itex] (for simplicity). Then by definition, [a] will be idempotent if [a]²=[a]. That is to say, if a²+100Z = a+100Z. But this will happen if and only if there exists a k such that a² = a + k. Can you go from there?

Between, this question belongs to the Homework help section of Physics Forum (PF): https://www.physicsforums.com/forumdisplay.php?f=152
 

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