Topics in Algebra, Chapter 3.10

This page covers section 3.10 (“Polynomials over the Rational Field”).

Topics covered: 3.10

  • Definition: The content of fZ[x] is the greatest common divisor of its coefficients.

  • Definition: A polynomial fZ[x] is primitive if its content is 1.

  • Lemma 3.10.1: If f,gZ[x] are primitive, then so is fg.

  • Theorem 3.10.1 (Gauss Lemma): If fZ[x] is primitive and can be factored as f=gh with g,hQ[x], then it can be factored as f=gh with g,hZ[x].

  • Corollary: If fZ[x] is monic and factors as a product of rational polynomials, then it factors as a product of monic integer polynomials.

  • Theorem 3.10.2 (Eisenstein Criterion): Let fZ[x], expressed as f(x)=a0+a1x++anxn. If a prime p exists such that pa0,pa1,pan1, but pan and p2a0, then f is irreducible over Q.

The problems below are paraphrased from/inspired by those given in Topics in Algebra by Herstein. The solutions are my own unless otherwise noted. I will generally try, in my solutions, to stick to the development in the text. This means that problems will not be solved using ideas and theorems presented further on in the book.

Herstein 3.10.1: Let R be a Euclidean ring and let F be the field of fractions of R. Prove that Gauss’s lemma holds when R replaces Z and F replaces Q.

This is almost word for word the proof of theorem 3.10.1. We can generalize to the extent stated here, noting that R being Euclidean gives us a notion of greatest common divisor, of polynomial content and of a polynomial being primitive. The statement of Gauss’s lemma is that if fR[x] is primitive and factors in F[x], then it factors in R[x].

Suppose there exist p,qF[x] such that f=pq. Then by clearing “denominators”, we can write p(x)=[a,b]p(x),q(x)=[c,d]q(x)

with a,b,c,dR and p,qR[x] primitive. Then, we must have bdf(x)=acp(x)q(x).

Because f,p,q are all primitive, and primitive polynomials are closed under multiplication, we must have bd=ac. Therefore f(x)=[ac,bd]p(x)q(x)=p(x)q(x)

is an expression of f as a product of polynomials in R[x].

Herstein 3.10.2: Let pZ be prime. Show that xnp is irreducible over Q.

Write the coefficients of the polynomial as a0=p, an=1 and ai=0 for i=1,,n1. The result is immediate from the Eisenstein criterion because pai for i<n while p2a0 and pan.

Herstein 3.10.3: Let pZ be prime. Show that 1+x++xp1 is irreducible over Q.

It is useful to realize that polynomials can sometimes be transformed in ways that respect their irreducibility. For instance, if we are interested in showing that fF[x] is irreducible, we can compose it with another gF[x] and consider the irreducibility of fgF[x]. If f is reducible, say f(x)=p(x)q(x), then so too is fg because f(g(x))=p(g(x))q(g(x)). Hence the contrapositive statement: if f(g(x)) is irreducible, this must mean that f(x) itself is irreducible for otherwise it would have furnished us with a factorization of fg. Needless to say, it is often difficult to determine reducibility of a polynomial, and there are instances when such an fg is easier to study than f.

In the present case, let f(x)=1+x++xp1. Herstein hints to consider 1+(x+1)+(x+1)2++(x+1)p1

which is the composition fg of f with g(x)=x+1. Expanding with the binomial theorem, f(g(x))=m=0p1(x+1)m=m=0p1n=0m(mn)xn=n=0p1xnm=np1(mn),

where we reorder the sums in the last step. The sum over m can be done in closed form (see below) and so we have f(g(x))=n=0p1(pn+1)xn.

Now the Eisenstein criterion may be used on fg: the coefficients of xn for n=0,,p2 are all divisible by p lemma 2 of my chapter 1 solutions, while the coefficient of x0 is not divisible by p2 and the coefficient of xp1 is not divisible by p. Then fg is irreducible over Q and, by the comments made above, f is irreducible over Q.

To show that m=np1(mn)=(pn+1),

for n>0, we proceed by induction on p (in the context of the above problem, p is prime, but for this purpose it is an arbitrary positive integer). For p=1 it is true in a trivial sense, where we interpret (ab) to be zero if a<b. Now supposing the result for p, we would like to show it for p+1: m=np(mn)=m=np1(mn)+(pn)=(pn+1)+(pn).

As is well known, (pn+1)+(pn)=p!(n+1)!(pn1)![n+1pn+1]=(p+1n+1).

This proves the result.

Herstein 3.10.4 (Rational Root Theorem): Suppose m,nZ are coprime and (xmn)p where pZ[x]. Letting p(x)=arxr++a0, prove that ma0 and nar.

First, let the content of p be denoted c, and write p(x)=cpˉ(x) where pˉZ[x] is primitive. We would like to show that (nxm)pˉ. We note that (xmn)cpˉ implies (xmn)pˉ because a degree one polynomial is irreducible and thus divides one of the two factors (and c has degree zero). Then there exists qQ[x] such that pˉ(x)=(xmn)q(x).

As in a proof of Gauss’s lemma, we proceed to pull out a factor of 1n from (xmn) and pull out all denominators and common factors from the coefficients in q. We can then write pˉ(x)=K(nxm)(αr1xr1++α1x+α0)

where KQ has accumulated all of the unwanted factors, and αiZ for i=1,,r1 are coprime. The content of the left hand side is 1, while the content of the right hand side is K, because (n,m)=1 and all common factors from q were pulled out so that the content of the α-polynomial is also 1. Therefore K=1. Multiplying out some terms of pˉ, we see that pˉ(x)=nαr1xr+mα0.

Then ar=cnαr1 so that nar, and a0=cmα0 so that ma0.

Herstein 3.10.5: Let aQ be such that xa divides a monic polynomial pZ[x]. Prove that aZ.

We can put this scenario into the language of problem 3.10.4, writing a=m/n with m,n coprime integers. Then we have that n must divide the coefficient of the highest order term, which is 1 for a monic polynomial. There are only two choices for n: ±1. Thus a=±mZ.