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Polynomial Rings

The polynomial ring R[x]R[x] consists of all formal sums ∑i=0naixi\sum_{i=0}^{n} a_i x^i with ai∈Ra_i \in R. It is a ring under the usual addition and multiplication of polynomials.

Theorem 10.1 (Division Algorithm). If FF is a field and f,g∈F[x]f, g \in F[x] with g≠0g \neq 0 Then There exist unique q,r∈F[x]q, r \in F[x] such that f=qg+rf = qg + r with deg⁡(r)<deg⁡(g)\deg(r) \lt \deg(g) or r=0r = 0.

Theorem 10.2 (Factor Theorem). a∈Fa \in F is a root of f∈F[x]f \in F[x] if and only if (x−a)(x - a) divides ff.

Proposition 10.3. A polynomial of degree nn over a field has at most nn roots (counting multiplicity).

A non-constant polynomial f∈F[x]f \in F[x] is irreducible if it cannot be factored as f=ghf = gh With both gg and hh of degree less than deg⁡(f)\deg(f).

Proposition 10.4. Every polynomial in F[x]F[x] factors uniquely into irreducible polynomials (up to Reordering and multiplication by units).

Theorem 10.5 (Eisenstein’s Criterion). Let f(x)=anxn+⋯+a1x+a0∈Z[x]f(x) = a_n x^n + \cdots + a_1 x + a_0 \in \mathbb{Z}[x]. If there exists a prime pp such that:

  1. pp divides a0,a1,…,an−1a_0, a_1, \ldots, a_{n-1}.
  2. pp does not divide ana_n.
  3. p2p^2 does not divide a0a_0.

Then ff is irreducible in Q[x]\mathbb{Q}[x].

Proof. Suppose f=ghf = gh with g,h∈Z[x]g, h \in \mathbb{Z}[x] (by Gauss’s lemma), deg⁡(g)=r≥1\deg(g) = r \geq 1 deg⁡(h)=s≥1\deg(h) = s \geq 1, r+s=nr + s = n. Modulo pp: fˉ=gˉhˉ=aˉnxn\bar{f} = \bar{g}\bar{h} = \bar{a}_n x^n in (Z/pZ)[x](\mathbb{Z}/p\mathbb{Z})[x]. Since Z/pZ[x]\mathbb{Z}/p\mathbb{Z}[x] is an integral domain, gˉ=bxr\bar{g} = bx^r And hˉ=cxs\bar{h} = cx^s for some b,c∈Z/pZb, c \in \mathbb{Z}/p\mathbb{Z}. In particular, the constant terms of gg and hh are both divisible by pp. But then p2p^2 divides a0a_0 Contradicting condition (3). ■\blacksquare

Problem. Show that x2+1x^2 + 1 is irreducible in R[x]\mathbb{R}[x] but reducible in C[x]\mathbb{C}[x].

Solution

Solution. In R[x]\mathbb{R}[x]: if x2+1=(x+a)(x+b)x^2 + 1 = (x + a)(x + b) with a,b∈Ra, b \in \mathbb{R} Then a+b=0a + b = 0 and ab=1ab = 1 Giving −a2=1-a^2 = 1 Which has no real solution. So x2+1x^2 + 1 is irreducible In R[x]\mathbb{R}[x].

In C[x]\mathbb{C}[x]: x2+1=(x+i)(x−i)x^2 + 1 = (x + i)(x - i). ■\blacksquare

Problem. Use the Euclidean algorithm to compute gcd⁡(x4+x3+x2+x+1,x3+1)\gcd(x^4 + x^3 + x^2 + x + 1, x^3 + 1) in Q[x]\mathbb{Q}[x].

Solution

Solution. Apply the division algorithm:

x4+x3+x2+x+1=x(x3+1)+(x2+x+1)x^4 + x^3 + x^2 + x + 1 = x(x^3 + 1) + (x^2 + x + 1) x3+1=(x−1)(x2+x+1)+2x^3 + 1 = (x - 1)(x^2 + x + 1) + 2

Since 22 is a non-zero constant (a unit in Q[x]\mathbb{Q}[x]), the polynomials are coprime: gcd⁡(x4+x3+x2+x+1,x3+1)=1\gcd(x^4 + x^3 + x^2 + x + 1, x^3 + 1) = 1. ■\blacksquare

Problem. Show that f(x)=x5−5x+3f(x) = x^5 - 5x + 3 is irreducible in Q[x]\mathbb{Q}[x].

Solution

Solution. By the rational root theorem, possible rational roots are ±1,±3\pm 1, \pm 3. f(1)=−1f(1) = -1, f(−1)=7f(-1) = 7, f(3)=216f(3) = 216, f(−3)=−269f(-3) = -269. No rational roots.

Since deg⁡(f)=5\deg(f) = 5 If ff is reducible, it must have an irreducible factor of degree 11 or 22. No degree-11 factor means no rational root. We check for degree-22 factors by reducing modulo 22: fˉ=x5+x+1\bar{f} = x^5 + x + 1 in F2[x]\mathbb{F}_2[x]. fˉ(0)=1\bar{f}(0) = 1, fˉ(1)=1\bar{f}(1) = 1 So no roots in F2\mathbb{F}_2. The only irreducible quadratic in F2[x]\mathbb{F}_2[x] is x2+x+1x^2 + x + 1. Division gives x5+x+1=(x2+x+1)(x3+x2)+1x^5 + x + 1 = (x^2+x+1)(x^3 + x^2) + 1 So x2+x+1x^2 + x + 1 does not divide fˉ\bar{f}.

Thus ff has no factor of degree 11 or 22 So ff is irreducible in Q[x]\mathbb{Q}[x]. ■\blacksquare

  • If F⊆KF \subseteq K is a field extension, then F[x]F[x] is a subring of K[x]K[x].
  • For a field FF, the polynomial ring F[x]F[x] is a Euclidean domain, hence a PID and a UFD.
  • If RR is a UFD, then R[x]R[x] is a UFD (Gauss’s lemma extends unique factorisation).
  • If RR is a PID, then R[x]R[x] is a UFD but generally not a PID.
  • The degree map deg⁡:F[x]∖{0}→N\deg : F[x] \setminus \{0\} \to \mathbb{N} is a Euclidean valuation.

Polynomial rings extend arithmetic to algebraic structures. The ring F[x] consists of formal sums of powers of x with coefficients in a field, and it inherits the familiar operations of addition and multiplication. The division algorithm allows you to divide one polynomial by another, producing a quotient and remainder, just as with integers. This makes polynomial rings Euclidean domains, where unique factorisation into irreducibles holds. Eisenstein’s criterion provides a practical test for irreducibility: if a prime divides all coefficients except the leading one, and its square does not divide the constant term, the polynomial is irreducible. Polynomial rings are fundamental to coding theory, cryptography, and algebraic geometry.

  • Assuming that irreducibility in R[x]R[x] implies irreducibility in S[x]S[x] for an extension S⊇RS \supseteq R. The polynomial x2+1x^2 + 1 is irreducible over Q\mathbb{Q} but reducible over C\mathbb{C}.
  • Forgetting that Eisenstein’s criterion requires the leading coefficient not to be divisible by pp. If p∣anp \mid a_n the criterion gives no information about reducibility.
  • Confusing the polynomial ring F[x]F[x] with the ring of formal power series F[[x]]F[[x]]. In F[[x]]F[[x]] every element is a unit except those with zero constant term.
  • Neglecting to check that a proposed factorisation respects the coefficient ring. Over Z[x]\mathbb{Z}[x], a factorisation into monic polynomials in Q[x]\mathbb{Q}[x] may not lie in Z[x]\mathbb{Z}[x].
  • Coding theory: BCH and Reed-Solomon codes are constructed using polynomial rings over finite fields Fq\mathbb{F}_q.
  • Cryptography: The ring Zp[x]/(f(x))\mathbb{Z}_p[x]/(f(x)) where ff is irreducible gives a finite field Fpn\mathbb{F}_{p^n}, used in elliptic curve cryptography.
  • Algebraic geometry: The coordinate ring of an affine variety is a quotient of a polynomial ring.
  • Signal processing: The discrete Fourier transform can be viewed as evaluating a polynomial at roots of unity.

10.7 Worked Example: Irreducibility via Substitution

Section titled “10.7 Worked Example: Irreducibility via Substitution”

Problem. Show that f(x)=x4+4x+2f(x) = x^4 + 4x + 2 is irreducible in Q[x]\mathbb{Q}[x].

Solution

By Eisenstein with p=2p = 2: 2∣0,4,22 \mid 0, 4, 2 but 2∤12 \nmid 1 (leading coefficient) and 4∤24 \nmid 2 (constant term). So ff is irreducible in Q[x]\mathbb{Q}[x]. ■\blacksquare