Math 2.0

statement theorem

Product of primitive polynomials is primitive

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.

Explanations

L1 human

Suppose that no prime divides all the coefficients of ff, and no prime divides all the coefficients of gg. Then no prime divides all the coefficients of fgf g. The proof reduces the coefficients modulo a prime and uses that the result is an integral domain.

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To read this node, read these 15 nodes in this order. The foundations come first, and the list ends with this node.

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  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

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Sources

  1. Web https://en.wikipedia.org/wiki/Gauss%27s_lemma_(polynomials), Statements for unique factorization domains