This page covers section 3.10 (“Polynomials over the Rational Field”).
Definition: The content of is the greatest common divisor of its coefficients.
Definition: A polynomial is primitive if its content is .
Lemma 3.10.1: If are primitive, then so is .
Theorem 3.10.1 (Gauss Lemma): If is primitive and can be factored as with , then it can be factored as with .
Corollary: If 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 , expressed as . If a prime exists such that , but and , then is irreducible over .
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.
This is almost word for word the proof of theorem 3.10.1. We can generalize to the extent stated here, noting that 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 is primitive and factors in , then it factors in .
Suppose there exist such that . Then by clearing “denominators”, we can write
with and primitive. Then, we must have
Because are all primitive, and primitive polynomials are closed under multiplication, we must have . Therefore
is an expression of as a product of polynomials in .
Write the coefficients of the polynomial as , and for . The result is immediate from the Eisenstein criterion because for while and .
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 is irreducible, we can compose it with another and consider the irreducibility of . If is reducible, say , then so too is because . Hence the contrapositive statement: if is irreducible, this must mean that itself is irreducible for otherwise it would have furnished us with a factorization of . Needless to say, it is often difficult to determine reducibility of a polynomial, and there are instances when such an is easier to study than .
In the present case, let . Herstein hints to consider
which is the composition of with . Expanding with the binomial theorem,
where we reorder the sums in the last step. The sum over can be done in closed form (see below) and so we have
Now the Eisenstein criterion may be used on : the coefficients of for are all divisible by lemma 2 of my chapter 1 solutions, while the coefficient of is not divisible by and the coefficient of is not divisible by . Then is irreducible over and, by the comments made above, is irreducible over .
To show that
for , we proceed by induction on (in the context of the above problem, is prime, but for this purpose it is an arbitrary positive integer). For it is true in a trivial sense, where we interpret to be zero if . Now supposing the result for , we would like to show it for :
As is well known,
This proves the result.
First, let the content of be denoted , and write where is primitive. We would like to show that . We note that implies because a degree one polynomial is irreducible and thus divides one of the two factors (and has degree zero). Then there exists such that
As in a proof of Gauss’s lemma, we proceed to pull out a factor of from and pull out all denominators and common factors from the coefficients in . We can then write
where has accumulated all of the unwanted factors, and for are coprime. The content of the left hand side is , while the content of the right hand side is , because and all common factors from were pulled out so that the content of the -polynomial is also . Therefore . Multiplying out some terms of , we see that
Then so that , and so that .
We can put this scenario into the language of problem 3.10.4, writing with coprime integers. Then we have that must divide the coefficient of the highest order term, which is for a monic polynomial. There are only two choices for : . Thus .