Difference between revisions of "Prime Ideals"

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Maximal implies prime but not conversely.
 
Maximal implies prime but not conversely.
  
An ideal <math>P</math> is prime if and only if <math>R \P </math> is closed under multiplication.
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An ideal <math>P</math> is prime if and only if <math>R \backslash P </math> is closed under multiplication.
  
 
If <math>R</math> is a domain, <math>S</math> is a ring and <math>f : R \to S</math> is a ring homomorphism, then <math>\text{Ker}(f)</math> is a prime ideal.
 
If <math>R</math> is a domain, <math>S</math> is a ring and <math>f : R \to S</math> is a ring homomorphism, then <math>\text{Ker}(f)</math> is a prime ideal.
  
 
The ideal <math>I</math> is prime if and only if <math>R/I</math> is an [[Integral Domains|integral domain]].
 
The ideal <math>I</math> is prime if and only if <math>R/I</math> is an [[Integral Domains|integral domain]].

Revision as of 20:47, 17 January 2023

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A prime ideal of a ring \(R \) is proper ideal \(P \) such that whenever \(xy\in P \) for \(x,y \in R \), \(x\in P \) or \(y \in P \).
An ideal \(P \) is prime if and only if \( R\backslash P\) is closed under multiplication.
Maximal implies prime but not conversely.

An ideal \(P\) is prime if and only if \(R \backslash P \) is closed under multiplication.

If \(R\) is a domain, \(S\) is a ring and \(f : R \to S\) is a ring homomorphism, then \(\text{Ker}(f)\) is a prime ideal.

The ideal \(I\) is prime if and only if \(R/I\) is an integral domain.