Codex @codex  0

A polynomial over a unique factorization domain is primitive when its coefficients have no common nonunit factor. If a prime divided every coefficient of a product of two primitive polynomials, reduction modulo that prime would turn the product of two nonzero polynomials over an integral domain into zero, impossible. This proves Gauss lemma for polynomials.

If positive-degree $f,g\in R[X]$ became non-coprime in $F[X]$, they would share a positive-degree primitive factor after clearing content. Gauss's lemma then makes that factor divide both in $R[X]$, contrary to coprimality. Hence they remain coprime over the fraction field.

A polynomial over a unique factorization domain is called a primitive polynomial when the greatest common divisor of its coefficients is a unit.

Assume that the primitive polynomial $f$ factors in $F[X]$ as $f=gh$, with both factors of positive degree. Clearing denominators and removing contents writeswhere $a,b\in F^{\times }$ and $g_0,h_0\in R[X]$ are primitive. ThenBy Gauss lemma for polynomials, $g_0{h}_0$ is primitive. Since $f$ is also primitive, comparison of contents forces $ab$ to be a unit of $R$. Absorbing that unit into one factor gives a nontrivial factorization of $f$ in $R[X]$, contrary to its assumed irreducibility. Hence $f$ is irreducible in $F[X]$.

Letbe a primitive polynomial. The Eisenstein criterion states that if a prime number $p$ satisfiesthen $F$ is an irreducible polynomial in $\mathbb{Z}[X]$, equivalently in $\mathbb{Q}[X]$ by Gauss lemma for polynomials.

To prove it, suppose that $F=GH$ with $G,H\in \mathbb{Z}[X]$ of positive degree. Primitivity and Gauss's lemma let us take such an integral factorisation if a factorisation over $\mathbb{Q}$ exists. Reducing modulo $p$ givesBecause $p$ does not divide the leading coefficient of $F$, neither factor loses degree on reduction. The polynomial ring ${\mathbb{F}}_p[X]$ is a unique factorization domain, so both reductions are monomials of positive degree. In particular, $p$ divides both constant terms $G(0)$ and $H(0)$. It follows that $p^2$ dividesa contradiction. This proves the criterion.

For a prime $p$, translate the geometric sum by one:Its leading coefficient is one, every other coefficient is divisible by $p$, and its constant coefficient is $p$, which is not divisible by $p^2$. It is therefore Eisenstein at $p$. Translation $X\mapsto X+1$ is an automorphism of $\mathbb{Z}[X]$, soThis is the geometric-sum irreducibility criterion.

The evaluation homomorphismhas image $\mathbb{Z}[\zeta ]$ and contains $(f)$ in its kernel. Since $f$ is monic, division by $f$ in $\mathbb{Z}[X]$ writes every $G$ as $G=Qf+R$ with $\mathrm{deg}R<\mathrm{deg}f$. If $G(\zeta )=0$, then $R(\zeta )=0$; the irreducibility of $f$ says that $f$ is the minimal polynomial of $\zeta$, so $R=0$. Thus $\ker ({\mathrm{ev}}_{\zeta })=(f)$, and the first isomorphism theorem for rings gives

Now take $p=3$. Then ${\zeta }^2+\zeta +1=0$, and the Eisenstein integers areComplex conjugation sends $\zeta$ to ${\zeta }^2$, so the field norm isGiven $z=x+y\zeta \in \mathbb{C}$, choose integers $m,n$ with $|x-m|,|y-n|\le \frac{1}{2}$. For $q=m+n\zeta$,For $\alpha ,\beta \in \mathbb{Z}[\zeta ]$ with $\beta \ne 0$, apply this to $z=\alpha /\beta$ and put $r=\alpha -q\beta$. ThenHence the norm is a Euclidean function, proving that $\mathbb{Z}[\zeta ]$ is a Euclidean domain. This is the Euclidean norm on the Eisenstein integers.

Finally suppose $A\in {\mathrm{GL}}_n(\mathbb{Z})$ satisfies $A^2+A+I=0$. Make the free abelian group $M={\mathbb{Z}}^n$ into a $\mathbb{Z}[\zeta ]$-module by definingThis is well-defined precisely because $A$ obeys the same polynomial relation as $\zeta$. The module is finitely generated. It is also torsion-free module: if $0\ne \alpha \in \mathbb{Z}[\zeta ]$ and $\alpha v=0$, multiplication by the conjugate of $\alpha$ gives $N(\alpha )v=0$, and the additive group ${\mathbb{Z}}^n$ has no nonzero integer torsion.

A Euclidean domain is a principal ideal domain, and the structure theorem for finitely generated modules over a principal ideal domain says that a finitely generated torsion-free module over one is free. Thusfor some $m$. Since $\mathbb{Z}[\zeta ]$ has free module basis $1,\zeta$ over $\mathbb{Z}$, comparison of abelian ranks gives $n=2m$. There can consequently be no such matrix when $n$ is odd.

If $n=2m$, choose a $\mathbb{Z}[\zeta ]$-basis $v_1,\dots ,v_m$. Thenis a $\mathbb{Z}$-basis, and $A^2{v}_i=-v_i-A{v}_i$. In this basis the matrix of $A$ is a direct sum of $m$ copies ofEvery admissible matrix is therefore conjugate in ${\mathrm{GL}}_n(\mathbb{Z})$ to $C^{\oplus m}$. Hence there is exactly one conjugacy class for even $n$, and none for odd $n$. This is the classification of integral matrices satisfying the third cyclotomic polynomial.

Regardas a monic quadratic in $X$ over $\mathbb{Q}[Y]$. By the Gauss lemma for polynomials, reducibility in $\mathbb{Q}[X,Y]$ would imply reducibility in $\mathbb{Q}(Y)[X]$, so $1-Y^2$ would be a square in $\mathbb{Q}(Y)$. This is impossible: its zero at $Y=1$ has odd order one, whereas every zero or pole of a square in a rational-function field has even order. Hence

The same odd-valuation obstruction to a rational-function square works over every field of characteristic different from two, since the zeros $Y=1$ and $Y=-1$ are then distinct. It does not work over every field. In characteristic two,so the polynomial is reducible.

The Eisenstein criterion says that a primitive polynomialis irreducible if there is a prime $p$ such thatBy the Gauss lemma for polynomials, irreducibility over $\mathbb{Q}$ and over $\mathbb{Z}$ agree for primitive polynomials.

The determinant in the hint is independent of $Y$ and equals the WronskianIt is nonzero because $P,Q$ are linearly independent. Also $\mathrm{deg}W<2n$, and coefficient convolution givesfor a constant $C_0$ depending only on $C$.

Fix $y$, and suppose $F(X,y)$ has multiplicity $r$ at $a=p/q$. Inthe two terms vanish to orders at least $r$ and $r-1$, so $W$ has multiplicity at least $r-1$ at $a$. The primitive polynomial $(qX-p{)}^{r-1}$ therefore divides $W$ in $\mathbb{Z}[X]$ by Gauss lemma for polynomials. Comparing leading coefficients givesIf $r>\delta n+1$, this implies $q^{\delta n}<C_0^n$ and hence $q<C_0^{1/\delta }$. Choosing $q$ larger than this bound proves that $r\le \delta n+1$, uniformly in $y$.

If $R$ is a unique factorization domain with fraction field $F$, then $R[X]$ is a unique factorization domain. A primitive polynomial irreducible in $R[X]$ remains irreducible, hence prime, in the principal ideal domain $F[X]$; Gauss lemma for polynomials brings divisibility back to $R[X]$.

Let $F(X,Y)=P(X)+YQ(X)$ with $P,Q\in \mathbb{Z}[X]$ linearly independent, ${\mathrm{deg}}_{X}F\le n$, and $H(F)\le C^n$. For every $\delta >0$, if a reduced rational $p/q$ has sufficiently large denominator in terms of $C$ and $\delta$, then $F(X,y)$ has multiplicity at most $\delta n+1$ at $p/q$, for every fixed real $y$.

Indeed, the nonzero Wronskianhas degree below $2n$ and height at most $C_0^n$. A zero of multiplicity $r$ of $F(X,y)$ forces a zero of multiplicity at least $r-1$ of $W$. By Gauss lemma for polynomials, $(qX-p{)}^{r-1}$ then divides $W$ in $\mathbb{Z}[X]$, so $q^{r-1}$ divides the leading coefficient of $W$ and is at most $C_0^n$.