Math 2.0

Commutative algebra Counterexample

Z[x] is not a principal ideal domain

Builds on Principal ideal domainPolynomial ringPrincipal ideal

Counterexample

The ring ℤ[x]\mathbb{Z}[x] is an integral domain, but it is not a principal ideal domain. The ideal (2,x)(2, x), the set of the polynomials with an even constant term, is not principal. Suppose (2,x)=(f)(2, x) = (f). Then f∣2f \mid 2, so ff is a constant cc with c∈{±1,±2}c \in \{\pm 1, \pm 2\}. If c=±1c = \pm 1, then 1∈(2,x)1 \in (2, x), but 11 has an odd constant term. If c=±2c = \pm 2, then xx is a multiple of 22 in ℤ[x]\mathbb{Z}[x], which is false.

Remarks

The quotient ℤ[x]/(2,x)\mathbb{Z}[x]/(2, x) is 𝔽2\mathbb{F}_2, so (2,x)(2, x) is a maximal ideal. The ring ℤ[x]\mathbb{Z}[x] is still a unique factorization domain, by the theorem on polynomial rings over a UFD. Over a field KK, the ring K[x]K[x] is a principal ideal domain.

Reading path

To understand this page from the ground up, read these 11 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. Polynomial ring definition
  6. Zero divisor definition
  7. Ideal generated by a subset definition
  8. Integral domain definition
  9. Principal ideal definition
  10. Principal ideal domain definition
  11. Z[x] is not a principal ideal domain counterexample

Further reading

  1. Web https://en.wikipedia.org/wiki/Principal_ideal_domain, Non-examples
About this page
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Kindcounterexample
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MSC13F20
UsesPrincipal ideal domain, Polynomial ring, Principal ideal

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