posted 6 Jan 2022

Unique Factorization in k[x]k[x]

Notes Number Theory

Unique Factorization in k[x]k[x]

§ Polynomial Ring

The unique factorization theorem can be generalized into other rings, not just the integers.

Polynomial Ring

The ring k[x]k[x] are polynomials with coefficients in a field k.k.

§ Divisibility in k[x]k[x]

  • If f,g∈k[x],f, g \in k[ x ], then we say f∣gf | g iff ∃h∈k[x],g=fh.\exists h \in k[ x ], g = fh.
  • If deg⁡f\deg f denotes the degree of f,f, then deg⁡fg=deg⁡f+deg⁡g.\deg fg = \deg f + \deg g.
  • deg⁡f=0\deg f = 0 iff ff is a nonzero constant.
  • f∣gf | g and g∣fg | f iff f=cgf = cg where cc is a nonzero constant.

Units

The only polynomials that divide all others are the nonzero constants, i.e. the units of k[x]k[ x ]

Irreductible Polynomials

A nonconstant polynomial pp is said to be irreductible iff q∣pq | p implies qq is either a constant or a constant times p.p.

These are analogous to the prime numbers in Z.\mathbb{Z}.

§ Factorization

Every nonconstant polynomial is the product of irreductible polynomials.

It is clear that polynomials with degree 11 are irreductible. Assuming that all polynomials with degree-bound nn are the product of irreductible polynomials, and that deg⁡f=n.\deg f = n. If ff is irreductible, we are done. Otherwise, f=gh,f = gh, where q≤deg⁡g,deg⁡h<n.q \leq \deg g, \deg h \lt n. By the induction hypothesis gg and hh are themselves products of irreductible polynomials. Therefore, ff is also the product of irreductible polynomials as f=gh.f = gh.

Monic Polynomial

A polynomial ff is monic if its leading coefficient is 1.

Every polynomial, save for zero, is a constant times a monic polynomial.

Order in k[x]k[ x ]

Let pp be a monic irreductible polynomial. We define ordpf\text{ord}_p f to be the integer aa defined by the property that pa∣fp^a | f but that pa+1∤f.p^{a+1} \nmid f. Such an integer must exist since the degree of powers of pp is monotonically increasing. Notice that ordpf=0  ⟺  p∤f.\text{ord}_p f = 0 \iff p \nmid f.

Sps. f∈k[x],f \in k[ x ], then

f=c∏ppa(p),f = c \prod_p p^{a(p)},

where the product is over all the monic irreductible polynomials, and cc is a constant.

The constant cc and the exponents a(p)a(p) are uniquely determined by f,f, and a(p)=ordpf.a(p) = \text{ord}_p f.

§ Greatest Common Divisor

Let f,g∈k[x].f, g \in k[ x ]. If g≠0,g \neq 0, then there exists h,k∈k[x]h, k \in k[ x ] such that f=hg+r,f = hg + r, where r=0r = 0 or r≠0r \neq 0 and deg⁡r<deg⁡g.\deg r \lt \deg g.

If g∣f,g | f, then h=f/gh = f / g and r=0.r = 0. Otherwise, let r=f−hgr = f - hg be the polynomial of least degree among all polynomials of the form f−lgf - lg with l∈k[x].l \in k[ x ]. If deg⁡r≥deg⁡g,\deg r \geq \deg g, then let the leading term of rr be axdax^d and that of gg be bxm.bx^m. Then r−ab−lxd−mg=f−(h+ab−ldd−m)gr - ab^{-l}x^{d-m}g = f - (h + ab^{-l}d^{d - m})g has degree less than r.r. Contradiction.

Ideals on k[x]k[ x ]

If f1,f2,…,fn∈k[x],f_1, f_2, \dots, f_n \in k[ x ], then (f1,f2,…,fn)(f_1, f_2, \dots, f_n) is the set of all polynomials of the form f1h1+f2h2+⋯+fnhn,f_1h_1 + f_2h_2 + \dots + f_nh_n, where h1,h2,…,hn∈k[x].h_1, h_2, \dots, h_n \in k[ x ].

Then, (f1,f2,…,fn)(f_1, f_2, \dots, f_n) is the ideal generated by f1,f2,…,fn.f_1, f_2, \dots, f_n.

Given f,g∈k[x],f, g \in k[ x ], there exists d∈k[x]d \in k[ x ] such that (f,g)=(d).(f, g) = (d).

Sps. d∈(f,g)d \in (f, g) and it is an element of least degree.

(⇒\Rightarrow) Clearly, (d)⊆(f,g).(d) \subseteq (f, g).

(⇐\Leftarrow) Let c∈(f,g).c \in (f, g). If d∤c,d \nmid c, then there exists polynomials h,rh, r such that c=hd+rc = hd + r with deg⁡r<deg⁡d.\deg r \lt \deg d. Since c,d∈(f,g),c, d \in (f, g), we get r=c−hd⊆(f,g).r = c - hd \subseteq (f, g). Contradiction, rr has degree less than d.d. Therefore, d∣cd | c and c∈(d).c \in (d).

Greatest Common Divisor

Sps. f,g∈k[x].f, g \in k[ x ]. Then d∈k[x]d \in k [ x ] is said to be the greatest common divisor of ff and gg iff d∣fd | f and d∣gd | g and every other common divisor of ff and gg divides d.d.

The greatest common divisor of two polynomials is determined up to multiplication by a constant. If we require the gcd to be monic, then it is uniquely determined which is what we mean when we say the greatest common divisor.

Let f,g∈k[x].f, g \in k[ x ]. By Lemma 3, there exists d∈k[x]d \in k[ x ] such that (f,g)=(d).(f, g) = (d). Then, dd is a gcd of ff and g.g.

Since f∈(d)f \in (d) and g∈(d),g \in (d), we have d∣fd | f and d∣g.d | g. Sps. that h∣fh | f and h∣g.h | g. Then hh divides every polynomial of the form fl+gmfl + gm with l,m∈k[x],l, m \in k[ x ], particularly h∣d.h | d.

§ Coprimality

Coprime

f,g∈k[x]f, g \in k[ x ] are said to be coprime (or relatively prime) iff the common divisors of ff and gg are constants, i.e. (f,g)=(1).(f, g) = (1).

Proposition 2

If ff and gg coprime, we have that (f,g)=(1)(f, g) = (1) so there are polynomials l,ml, m such that lf+mg=1.lf + mg = 1. Thus, lfh+mgh=h.lfh + mgh = h. Since ff divides the left hand side, f∣h.f | h.

Corollary 1

If pp is an irreductible polynomial and p∣fg,p | fg, then p∣fp | f or p∣g.p | g.

Since pp is irreductible, either (p,f)=(p)(p, f) = (p) or (p,f)=(1).(p, f) = (1). In the former, p∣f,p | f, q.e.d.

In the latter case, p,fp, f coprime, and p∣fp | f follows from proposition 2.

Corollary 2

If pp is a monic irreductible polynomial and f,g∈k[x],f, g \in k[ x ], we have ordpfg=ordpf+ordpg.\text{ord}_p fg = \text{ord}_p f + \text{ord}_p g.

Refer to proof to corollary 2 for prop. 1.

§ Unique Factorization Proof

Finally, we have the tools to prove the unique factorization of a polynomial.

If we apply ordq\text{ord}_q to both sides of

f=c∏ppa(p).f = c\prod_p p^{a(p)}.

We get

ordq(f)=ordq(c)+∑pa(p) ordq(p).\text{ord}_q (f) = \text{ord}_q (c) + \sum_p a(p)\,\text{ord}_q (p).

Clearly q∤cq \nmid c as cc is a constant, so ordq(c)=0.\text{ord}_q (c) = 0. Then ordq(p)=0\text{ord}_q (p) = 0 if q≠pq \neq p and 11 otherwise. Finally, we get that ordq(f)=a(q).\text{ord}_q (f) = a(q). This shows that the exponents are uniquely determined by ff and so is c.c.