Difference between revisions of "Prime Ideals"

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:''return to [[817 - Algebra#Prime Ideals| Algebra main page]]
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A ''prime ideal'' of a ring <math>R </math> is proper ideal <math>P </math> such that whenever <math>xy\in P </math> for <math>x,y \in R </math>, <math>x\in P </math> or <math>y \in P </math>. </br>
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An ideal <math>P </math> is prime if and only if <math> R\backslash P</math> is closed under multiplication.
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Maximal implies prime but not conversely.
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An ideal <math>P</math> is prime if and only if <math>R \backslash P </math> is closed under multiplication.
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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.
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The ideal <math>I</math> is prime if and only if <math>R/I</math> is an [[Integral Domains|integral domain]].
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[[Category: Ring Theory]]

Latest revision as of 01:12, 8 March 2023

return to Algebra main page

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.