Commutative rings with identity

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SUMMARY

A finite nonzero commutative ring with no zero divisors must possess a multiplicative identity. The proof involves selecting a nonzero element 'a' from the ring and examining its powers. By defining a mapping from the ring to itself via multiplication by 'a', one can establish that this mapping is injective and thus surjective due to the finiteness of the ring. Consequently, there exists an element 'x'' such that 'a = ax'', which demonstrates that 'x'' serves as the multiplicative identity.

PREREQUISITES
  • Understanding of commutative rings
  • Knowledge of zero divisors in ring theory
  • Familiarity with finite sets and their properties
  • Basic concepts of injective and surjective functions
NEXT STEPS
  • Study the properties of finite commutative rings
  • Explore the concept of injective and surjective mappings in algebra
  • Learn about the role of identity elements in ring theory
  • Investigate examples of rings with and without zero divisors
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Mathematicians, algebra students, and anyone studying abstract algebra, particularly those focusing on ring theory and its foundational concepts.

Marinela
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I have a trouble proving that a finate (nonzero) commutative ring with no zero divisors must have an identity with respect to multiplication. Could anybody please give me some hints?
I do know all the definitions (of ring, commutative ring, zero divisors, identity) but have no idea how to go from there.
Thanks!
 
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Take some nonzero element a in your ring, and look at its various powers: a, a^2, a^3, ... Now use the fact that the ring has finitely many elements.

Edit:
Maybe this is not the easiest approach. It might be cleaner if you define a map x->ax on your ring. This is an injection, and hence a surjection (by finiteness). In particular, a=ax', for some x in the ring. Claim: x' is the multiplicative identity.
 
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