Polynomial Rings

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updates: 14 February 2011

Polynomial rings

Definition.

• Let R be a commutative ring and for each i=1,2,3,… let xi be a formal symbol. A polynomial with coefficients in in R is an expression of the form fx=r0+r1x+r2x2+⋯ such that
  1. ri∈R for i=0,1,2,…,
  2. There exists a positive integer N such that ri=0 for all i>N.
• Let R be a commutative ring. Polynomials fx=r0+r1x+r2x2+⋯ and gx=s0+s1x+s2x2+⋯ with coefficients in R are equal if ri=sifor alli=0,1,2,…
• The zero polynomial is the polynomial 0=0+0x+0x2+⋯.
• The degree, degfx, of a polynomial fx=r0+r1x+r2x2+⋯ with coefficients in R is the smallset nonnegative integer N such that rN≠0 and rk=0 for all k>N. If fx=0+0x+0x2+… then we define degfx=0.
• Let R be a commutative ring. The ring of polynomials withe coefficients in R is the set Rx of polynomials with coefficients in R with the operations of addition and multiplication as defined as follows:

  If fx,gx∈R[x] where fx=r0+r1x+r2x2+⋯ and gx=s0+s1x+s2x2+⋯, then fx+gx=(r0+s0)+(r1+s1)x+(r2+s2)x2+⋯,andf(x)g(x)=c0+c1x+c2x2+⋯,whereck=∑i+j=krisj.

Let R be a commutative ring. Then R[x] is a commutative ring.

Let R be an integral domain. Then R[x] is an integral domain.

The following theorem is a deep theorem which will be proved at the end of this section.

Let R be a unique factorization domain. Then R[x] is a unique factorization domain.

The following is an important theorem which shows that if F is a field then F[x] is a Euclidean domain, and therefore, by Theorem 1.3, that F[x] is a principal ideal domain.

Let F be a field. The ring F[x] is a Euclidean domain with size function given by deg:F[x]→ℕf(x)↦deg(f(x)).

Let R,S be commutative rings and ϕ:R↦S be a ring homomorphism. Then the map ψ:R[x]→S[x]r0+r1x+r2x2+⋯↦φ(r0)+φ(r1)x+φ(r2)x2+⋯ is a ring homomorphism.

Adjoining elements to R, the rings R[α]

Definition.

• Let R be a commutative ring and let α∈R. The evaluation map evα:R[x]→R is the map given by evα:R[x]↦Rf(x)↦f(α), where, if f(x)=r0+r1x+r2x2+… then f(α)=r0+r1α+r2α2+….

Let R be a commutative ring and let α∈R. Then the evaluation homomorphism evα:R[x]→R is a ring homomorphism.

Definition

• Let S be a commutative ring. Let R⊆S be a subring and let α∈S. Let evα:S[x]→S be the evaluation homomorphism given by evaluating at α. The ring R adjoined α is the subring R[α] of S given by R[α]=evα(R[x]).

HW: Prove that R[α]=evα(R[x]). is a subring of S.

HW: Let S be a commutative ring. Let R⊆S be a subring and let α∈S show that the ring R[α] is the subring of S consisting of all elements of the form r0+r1x+r2x2+⋯+rdxd, where ri∈R and d is a nonnegative integer.

Proof of Theorem 1.3

Definition. Let R be a unique factorization domain. A polynomial f(x)=c0+c1x+⋯+ckxk∈R[x] is primitive if there does not exist any p∈R such that p divides all of the coefficients c0,c1,…ck, of f(x).

(Gauss' Lemma) Let R be a unique factorization domain. Let f(x),g(x)∈R[x] be primitive polynomials. Then f(x)g(x) is a primitive polynomial.

Let R be a unique factorization domain. Let F be the field of fractions of R and let f(x)∈F[x]. Then

1. There exists an element c∈F and a primitive polynomial g(x)∈R[x] such that f(x)=cg(x).
2. The factors c and g(x) are unique up to multiplication by a unit.
3. f(x) is irreducible in F[x] if and only if g(x) is irreducible in R[x].

Let R be a unique factorization domain. Then R[x] is a unique factorization domain.

References

[CM] H. S. M. Coxeter and W. O. J. Moser, Generators and relations for discrete groups, Fourth edition. Ergebnisse der Mathematik und ihrer Grenzgebiete [Results in Mathematics and Related Areas], 14. Springer-Verlag, Berlin-New York, 1980. MR0562913 (81a:20001)

[GW1] F. Goodman and H. Wenzl, The Temperly-Lieb algebra at roots of unity, Pacific J. Math. 161 (1993), 307-334. MR1242201 (95c:16020)

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