Math 2.0

Commutative algebra Theorem

R[x] is an integral domain iff R is

Builds on Polynomial ringIntegral domainDegree of a product

Statement

Let RR be a commutative ring. Then the polynomial ring R[x]R[x] is an integral domain if and only if RR is an integral domain.

Remarks

If RR is an integral domain and f,g≠0f, g \neq 0, then the leading coefficient of fgf g is not zero, so fg≠0f g \neq 0. Conversely, RR is a subring of R[x]R[x], and a subring of an integral domain is an integral domain. By induction, the ring R[x1,…,xn]R[x_1, \dots, x_n] of polynomials in several variables over an integral domain is an integral domain.

Reading path

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

Show the path as a graph
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. Polynomial ring definition
  5. Zero divisor definition
  6. Degree of a polynomial definition
  7. Integral domain definition
  8. Leading coefficient definition
  9. Degree of a product statement
  10. R[x] is an integral domain iff R is statement

Where this leads

23 pages build on this one.

Further reading

  1. Web https://en.wikipedia.org/wiki/Polynomial_ring, Properties that pass from R to R[X]
About this page
Tag0B5Z1EC66J
URIm2://0B5Z1EC66J
Kindstatement, theorem
WritingW0 stub
MSC13F20
StatusThis copy of the site has no build data. Run m2 status 0B5Z1EC66J.
UsesPolynomial ring, Integral domain, Degree of a product

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