Math 2.0

Commutative algebra Theorem

Product of primitive polynomials is primitive

Builds on Primitive polynomialUnique factorisation domainPrime element

Statement

Let RR be a unique factorisation domain and let f,g∈R[x]f, g \in R[x] be primitive polynomials. Then fgf g is primitive.

Remarks

The proof: suppose an irreducible pp divides all the coefficients of fgf g. Since pp is prime, R/(p)R/(p) is an integral domain. The images of ff and gg in (R/(p))[x](R/(p))[x] are nonzero, but their product is zero, which is a contradiction. Many texts call this statement Gauss's lemma. The node Gauss's lemma states the form about irreducibility. Over ℤ\mathbb{Z}: 2x+12x + 1 and 3x+23x + 2 are primitive, and so is 6x2+7x+26x^2 + 7x + 2.

Reading path

To understand this page from the ground up, read these 15 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. Unit of a ring definition
  5. Zero divisor definition
  6. Divisibility in a commutative ring definition
  7. Integral domain definition
  8. Associate elements definition
  9. Prime element definition
  10. Irreducible element definition
  11. Greatest common divisor in an integral domain definition
  12. Unique factorisation domain definition
  13. Content of a polynomial definition
  14. Primitive polynomial definition
  15. Product of primitive polynomials is primitive statement

Where this leads

14 pages build on this one.

Further reading

  1. Web https://en.wikipedia.org/wiki/Gauss%27s_lemma_(polynomials), Statements for unique factorization domains
About this page
Tag08C1NWBNHA
URIm2://08C1NWBNHA
Kindstatement, theorem
WritingW0 stub
MSC13F20
StatusThis copy of the site has no build data. Run m2 status 08C1NWBNHA.
UsesPrimitive polynomial, Unique factorisation domain, Prime element

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