Math 2.0

Commutative algebra Basic notions Theorem

R/P is an integral domain iff P is prime

Builds on Quotient ringPrime idealIntegral domain

Statement

Let RR be a commutative ring and let PP be an ideal of RR. Then the quotient ring R/PR/P is an integral domain if and only if PP is a prime ideal.

Remarks

The condition 1+P≠0+P1 + P \neq 0 + P in R/PR/P is the condition P≠RP \neq R. The condition that (a+P)(b+P)=0(a + P)(b + P) = 0 implies a+P=0a + P = 0 or b+P=0b + P = 0 is the condition that ab∈Pa b \in P implies a∈Pa \in P or b∈Pb \in P. For example, ℤ/nℤ\mathbb{Z}/n\mathbb{Z} is an integral domain exactly when nn is prime or n=0n = 0.

Reading path

To understand this page from the ground up, read these 10 pages in order. The foundations come first.

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Each page uses the pages above it. The colour of the bar on a box shows how strong the evidence is, and the box also gives it as text. The list has the same pages.
  • no evidence level
  1. Group concept
  2. Ring definition
  3. Commutative ring definition
  4. Ideal definition
  5. Zero divisor definition
  6. Prime ideal definition
  7. Integral domain definition
  8. Two-sided ideal definition
  9. Quotient ring definition
  10. R/P is an integral domain iff P is prime statement

Where this leads

1 page builds on this one.

Further reading

  1. Web https://en.wikipedia.org/wiki/Prime_ideal, Properties
About this page
Tag0AVXBEC05W
URIm2://0AVXBEC05W
Kindstatement, theorem
WritingW0 stub
MSC13A15
StatusThis copy of the site has no build data. Run m2 status 0AVXBEC05W.
UsesQuotient ring, Prime ideal, Integral domain

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