Is Every Ideal Being Prime Indicative of a Commutative Ring Being a Field?

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    Prime Ring Theory
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SUMMARY

In the discussion, it is established that in a commutative ring with unity, if every ideal is prime, then the ring must be a field. The proof hinges on demonstrating that the only ideals present are trivial, specifically that the ideal containing only the zero element leads to the conclusion that the ring is an integral domain. Furthermore, it is shown that every non-zero element in the ring is a unit, thus completing the proof that the ring is indeed a field.

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  • Understanding of commutative rings with unity
  • Knowledge of prime ideals in ring theory
  • Familiarity with integral domains
  • Concept of units in ring structures
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  • Study the properties of prime ideals in commutative rings
  • Learn about integral domains and their characteristics
  • Investigate the concept of units in ring theory
  • Explore the relationship between ideals and fields in algebra
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Mathematics students, algebra enthusiasts, and anyone studying abstract algebra, particularly those focusing on ring theory and field properties.

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Homework Statement



Given a commutative ring with unity, show that if every ideal is prime than the ring is a field.

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The Attempt at a Solution



I think that I can show that a ring is a field iff it has no nontrivial ideals. So I guess I need to show that if a ring has only prime ideals than these ideals must be trivial. I'm not sure how to do this though.
 
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Can you show that the ring must be an integral domain? (Hint: Consider the ideal that consists just of the zero element.) Next show that every non-zero element is a unit to complete the proof. (Hint: Let r be a non-zero element. Consider the ideal generated by [itex]r^2[/itex].)

Petek
 

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