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

**Definition**: The*content*of $f\in \mathbb{Z}\left[x\right]$ is the greatest common divisor of its coefficients.**Definition**: A polynomial $f\in \mathbb{Z}\left[x\right]$ is*primitive*if its content is $1$.**Lemma 3.10.1**: If $f,g\in \mathbb{Z}\left[x\right]$ are primitive, then so is $fg$.**Theorem 3.10.1 (Gauss Lemma)**: If $f\in \mathbb{Z}\left[x\right]$ is primitive and can be factored as $f=gh$ with $g,h\in \mathbb{Q}\left[x\right]$, then it can be factored as $f={g}^{\mathrm{\prime}}{h}^{\mathrm{\prime}}$ with ${g}^{\mathrm{\prime}},{h}^{\mathrm{\prime}}\in \mathbb{Z}\left[x\right]$.**Corollary**: If $f\in \mathbb{Z}\left[x\right]$ 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 $f\in \mathbb{Z}\left[x\right]$, expressed as $f\left(x\right)={a}_{0}+{a}_{1}x+\cdots +{a}_{n}{x}^{n}$. If a prime $p$ exists such that $p\mid {a}_{0},p\mid {a}_{1},\dots p\mid {a}_{n-1}$, but $p\nmid {a}_{n}$ and ${p}^{2}\nmid {a}_{0}$, then $f$ is irreducible over $\mathbb{Q}$.

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.

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 $f\in R\left[x\right]$ is primitive and factors in $F\left[x\right]$, then it factors in $R\left[x\right]$.

Suppose there exist $p,q\in F\left[x\right]$ such that $f=pq$. Then by clearing “denominators”, we can write $p\left(x\right)=[a,b]{p}^{\mathrm{\prime}}\left(x\right),\phantom{\rule{1em}{0ex}}q\left(x\right)=[c,d]{q}^{\mathrm{\prime}}\left(x\right)$

with $a,b,c,d\in R$ and ${p}^{\mathrm{\prime}},{q}^{\mathrm{\prime}}\in R\left[x\right]$ primitive. Then, we must have $bd\cdot f\left(x\right)=ac\cdot {p}^{\mathrm{\prime}}\left(x\right){q}^{\mathrm{\prime}}\left(x\right)\mathrm{.}$

Because $f,{p}^{\mathrm{\prime}},{q}^{\mathrm{\prime}}$ are all primitive, and primitive polynomials are closed under multiplication, we must have $bd=ac$. Therefore $f\left(x\right)=[ac,bd]{p}^{\mathrm{\prime}}\left(x\right){q}^{\mathrm{\prime}}\left(x\right)={p}^{\mathrm{\prime}}\left(x\right){q}^{\mathrm{\prime}}\left(x\right)$

is an expression of $f$ as a product of polynomials in $R\left[x\right]$.

Write the coefficients of the polynomial as ${a}_{0}=-p$, ${a}_{n}=1$ and ${a}_{i}=0$ for $i=1,\dots ,n-1$. The result is immediate from the Eisenstein criterion because $p\mid {a}_{i}$ for $i<n$ while ${p}^{2}\nmid {a}_{0}$ and $p\nmid {a}_{n}$.

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 $f\in F\left[x\right]$ is irreducible, we can compose it with another $g\in F\left[x\right]$ and consider the irreducibility of $f\circ g\in F\left[x\right]$. If $f$ is reducible, say $f\left(x\right)=p\left(x\right)q\left(x\right)$, then so too is $f\circ g$ because $f\left(g\right(x\left)\right)=p\left(g\right(x\left)\right)q\left(g\right(x\left)\right)$. Hence the contrapositive statement: if $f\left(g\right(x\left)\right)$ is irreducible, this must mean that $f\left(x\right)$ itself is irreducible for otherwise it would have furnished us with a factorization of $f\circ g$. Needless to say, it is often difficult to determine reducibility of a polynomial, and there are instances when such an $f\circ g$ is easier to study than $f$.

In the present case, let $f\left(x\right)=1+x+\cdots +{x}^{p-1}$. Herstein hints to consider $1+(x+1)+(x+1{)}^{2}+\cdots +(x+1{)}^{p-1}$

which is the composition $f\circ g$ of $f$ with $g\left(x\right)=x+1$. Expanding with the binomial theorem, $f\left(g\right(x\left)\right)=\sum _{m=0}^{p-1}(x+1{)}^{m}=\sum _{m=0}^{p-1}\sum _{n=0}^{m}\left(\genfrac{}{}{0px}{}{m}{n}\right){x}^{n}=\sum _{n=0}^{p-1}{x}^{n}\sum _{m=n}^{p-1}\left(\genfrac{}{}{0px}{}{m}{n}\right),$

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 $f\left(g\right(x\left)\right)=\sum _{n=0}^{p-1}\left(\genfrac{}{}{0px}{}{p}{n+1}\right){x}^{n}\mathrm{.}$

Now the Eisenstein criterion may be used on $f\circ g$: the coefficients of ${x}^{n}$ for $n=0,\dots ,p-2$ are all divisible by $p$ lemma 2 of my chapter 1 solutions, while the coefficient of ${x}^{0}$ is not divisible by ${p}^{2}$ and the coefficient of ${x}^{p-1}$ is not divisible by $p$. Then $f\circ g$ is irreducible over $\mathbb{Q}$ and, by the comments made above, $f$ is irreducible over $\mathbb{Q}$.

To show that $\sum _{m=n}^{p-1}\left(\genfrac{}{}{0px}{}{m}{n}\right)=\left(\genfrac{}{}{0px}{}{p}{n+1}\right),$

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 $\left(\genfrac{}{}{0px}{}{a}{b}\right)$ to be zero if $a<b$. Now supposing the result for $p$, we would like to show it for $p+1$: $\sum _{m=n}^{p}\left(\genfrac{}{}{0px}{}{m}{n}\right)=\sum _{m=n}^{p-1}\left(\genfrac{}{}{0px}{}{m}{n}\right)+\left(\genfrac{}{}{0px}{}{p}{n}\right)=\left(\genfrac{}{}{0px}{}{p}{n+1}\right)+\left(\genfrac{}{}{0px}{}{p}{n}\right)\mathrm{.}$

As is well known, $\left(\genfrac{}{}{0px}{}{p}{n+1}\right)+\left(\genfrac{}{}{0px}{}{p}{n}\right)=\frac{p!}{(n+1)!(p-n-1)!}[\frac{n+1}{p-n}+1]=\left(\genfrac{}{}{0px}{}{p+1}{n+1}\right)\mathrm{.}$

This proves the result.

First, let the content of $p$ be denoted $c$, and write $p\left(x\right)=c\stackrel{\u02c9}{p}\left(x\right)$ where $\stackrel{\u02c9}{p}\in \mathbb{Z}\left[x\right]$ is primitive. We would like to show that $(nx-m)\mid \stackrel{\u02c9}{p}$. We note that $(x-\frac{m}{n})\mid c\stackrel{\u02c9}{p}$ implies $(x-\frac{m}{n})\mid \stackrel{\u02c9}{p}$ because a degree one polynomial is irreducible and thus divides one of the two factors (and $c$ has degree zero). Then there exists $q\in \mathbb{Q}\left[x\right]$ such that $\stackrel{\u02c9}{p}\left(x\right)=(x-\frac{m}{n})q\left(x\right)\mathrm{.}$

As in a proof of Gauss’s lemma, we proceed to pull out a factor of $\frac{1}{n}$ from $(x-\frac{m}{n})$ and pull out all denominators and common factors from the coefficients in $q$. We can then write $\stackrel{\u02c9}{p}\left(x\right)=K\cdot (nx-m)({\alpha}_{r-1}{x}^{r-1}+\cdots +{\alpha}_{1}x+{\alpha}_{0})$

where $K\in \mathbb{Q}$ has accumulated all of the unwanted factors, and ${\alpha}_{i}\in \mathbb{Z}$ for $i=1,\dots ,r-1$ 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 $\alpha $-polynomial is also $1$. Therefore $K=1$. Multiplying out some terms of $\stackrel{\u02c9}{p}$, we see that $\stackrel{\u02c9}{p}\left(x\right)=n{\alpha}_{r-1}{x}^{r}+\cdots -m{\alpha}_{0}\mathrm{.}$

Then ${a}_{r}=cn{\alpha}_{r-1}$ so that $n\mid {a}_{r}$, and ${a}_{0}=-cm{\alpha}_{0}$ so that $m\mid {a}_{0}$.

We can put this scenario into the language of problem 3.10.4, writing $a=m\mathrm{/}n$ 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 $n$: $\pm 1$. Thus $a=\pm m\in \mathbb{Z}$.