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The splitting field of $f\in K[t]$ is the smallest extension $L/K$ in which $f$ is a product of linear factors; equivalently, $L$ is generated over $K$ by all its roots of a polynomial. To prove existence and uniqueness of splitting fields, adjoin a root of an irreducible polynomial dividing $f$ and repeat until $f$ splits. If $L$ and $L^{\prime }$ are two resulting fields, an embedding between the fields generated so far extends by sending each newly adjoined root to a root of its transformed minimal polynomial. Repetition gives a $K$-embedding $L\to L^{\prime }$, whose image contains every root of $f$ and is therefore all of $L^{\prime }$. Thus the splitting field exists and is unique up to $K$-isomorphism.

Now write $K={\mathbb{F}}_q$, and let the distinct irreducible factors of $f$ have degrees $d_1,\dots ,d_r$. The irreducible factors of a finite-field Frobenius polynomial show that a root of the degree-$d_i$ factor lies in ${\mathbb{F}}_{q^n}$ exactly when $d_i\mid n$. Hence the splitting field over a finite field isEvery finite field is a perfect field, so $L/K$ is separable; it is normal because it is a splitting field. It is consequently a Galois extension. More explicitly, the Galois group of a finite field extension is the cyclic group of order $m$ generated by the finite-field Frobenius automorphism $x\mapsto x^q$.

Finally, let $N_2(n)$ denote the number of monic irreducible polynomials over a finite field of degree $n$ over ${\mathbb{F}}_2$. Sinceand $\mu (1)=1$, $\mu (3)=-1$, $\mu (9)=0$, we obtain

The cyclotomic polynomial of the primitive $p$th root $\omega$ isBy the irreducibility of cyclotomic polynomials, this is the minimal polynomial of $\omega$ over $\mathbb{Q}$. The prime cyclotomic field degree therefore gives

TakeThen $AB=B\ne 0$ and $(AB{)}^2=0$, whereas $BA=0$. Hence their minimal polynomials are

For every nonnegative integer $k$,Consequently, if $m_{AB}(AB)=0$, thenBy the divisibility characterization of the minimal polynomial,Interchanging $A$ and $B$ also gives $m_{AB}(t)\mid t,m_{BA}(t)$. Therefore the Minimal polynomials of AB and BA can differ only in the exponent of the factor $t$, and that exponent differs by at most one.

If $AB$ is diagonalizable, its minimal polynomial has no repeated roots. The preceding divisibility says that every nonzero Jordan block of $BA$ has size one and every zero-eigenvalue block has size at most two. Squaring $BA$ preserves the one-dimensional nonzero blocks and sends every size-at-most-two nilpotent block to zero. Thus the Jordan normal form of $(BA{)}^2$ is diagonal, so

Let $L$ be a splitting field of $f(X)=X^m-1$ over $k$. Its formal derivative isThe assumption on the characteristic of $k$ says that $m\ne 0$ in $k$. Any root $\alpha$ of $f$ is nonzero, and hence $f^{\prime }(\alpha )=m{\alpha }^{m-1}\ne 0$. Thus $f$ and $f^{\prime }$ have no common root, so $f$ is a separable polynomial. Since it has degree $m$ and splits over $L$, it has exactly $m$ distinct roots. This is the separability of roots of unity.

Fix a primitive root of unity ${\zeta }_m=e^{2\pi i/m}$. Define the $m$th cyclotomic polynomial byIt is monic and has degree given by the Euler totient function $\phi (m)$. Partitioning all $m$th roots of unity according to their exact multiplicative order gives the cyclotomic factorizationWe now prove by mathematical induction that ${\Phi }_m\in \mathbb{Z}[X]$. The base case is ${\Phi }_1=X-1$. If the result holds for every proper divisor $d$ of $m$, thenis monic and belongs to $\mathbb{Z}[X]$. Polynomial division of $X^m-1$ by the monic integer polynomial $G$ produces a quotient and remainder in $\mathbb{Z}[X]$. The factorization over $\mathbb{C}$ says that the remainder is zero and the quotient is ${\Phi }_m$, proving the claim.

It remains to prove irreducibility of cyclotomic polynomials. Let $f\in \mathbb{Z}[X]$ be a monic irreducible polynomial dividing ${\Phi }_m$, let $\epsilon$ be a root of $f$, and write ${\Phi }_m=fg$. We claim that ${\epsilon }^p$ is a root of $f$ for every prime number $p\nmid m$. Otherwise ${\epsilon }^p$ is a root of $g$, so $\epsilon$ is a root of $g(X^p)$. Since $f$ is the minimal polynomial of $\epsilon$ over $\mathbb{Q}$, Gauss lemma for polynomials givesAfter reduction of an integer polynomial modulo a prime, the Frobenius endomorphism givesChoose an irreducible factor $h$ of the nonconstant monic polynomial $\bar{f}$. Then $h\mid \bar{g}$, so $h^2\mid \bar{f}\bar{g}={\bar{\Phi }}_m$. The cyclotomic factorization would then make $h^2$ divide $X^m-1$ over ${\mathbb{F}}_p$. This is impossible because its derivative $m{X}^{m-1}$ is coprime to it when $p\nmid m$.

Thus $f({\epsilon }^p)=0$ for every prime $p\nmid m$. Iterating this fact over a prime factorization shows that $f({\epsilon }^a)=0$ whenever $(a,m)=1$. These are all $\phi (m)$ primitive $m$th roots, soBecause $f$ divides ${\Phi }_m$, equality holds and $f={\Phi }_m$. Hence every cyclotomic polynomial is irreducible over $\mathbb{Q}$.

Finally, direct multiplication givesAlsoApplying the cyclotomic factorization to the two factors on the right givesand thereforeSince $X^2-X+1={\Phi }_6(X)$, cancellation yieldsThe thirtieth cyclotomic polynomial is irreducible over $\mathbb{Q}$ by the theorem just proved, so $F$ is irreducible.

The structure theorem for finitely generated modules over a principal ideal domain applies because every Euclidean domain is a principal ideal domain. It says that a finitely generated $R$-module is isomorphic towhere the nonzero nonunits $d_i$ are unique up to multiplication by units. They are the invariant factors.

For $V_{\alpha }$, multiplication by $X$ is the linear map $\alpha$. Since $V$ is finite-dimensional over $F$, $V_{\alpha }$ is a finitely generated torsion module over the polynomial ring $F[X]$, so it has no free summand. Choosing each invariant factor $a_i$ to be monic givesThe basis $1,X,\dots ,X^{\mathrm{deg}{a}_i-1}$ of each cyclic summand makes multiplication by $X$ a companion matrix. Concatenating these bases therefore puts $\alpha$ in rational canonical form.

On $F[X]/(a_i)$, a polynomial annihilates multiplication by $X$ exactly when it is divisible by $a_i$. It follows that the minimal polynomial and characteristic polynomial areThe second identity follows blockwise from the characteristic polynomial of a companion matrix. Since every $a_i$ divides $a_k$, the product ${\chi }_{\alpha }$ annihilates every cyclic summand. Thus ${\chi }_{\alpha }(\alpha )=0$, which is the Cayley-Hamilton theorem.

For the displayed matrix, the given generators of $\ker \theta$ are the columns ofElementary row and column operations over $\mathbb{R}[X]$ give the Smith normal formConsequently the nonunit invariant factors areand, as a check,

Gauss lemma for polynomials states that the product of two primitive polynomials over a unique factorization domain is primitive. In particular, a primitive polynomial in $\mathbb{Z}[X]$ is reducible over $\mathbb{Q}$ exactly when it is reducible over $\mathbb{Z}$.

Eisenstein criterion says that a primitive polynomialis irreducible over $\mathbb{Q}$ if there is a prime $p$ such that $p\nmid a_n$, $p\mid a_j$ for every $j<n$, and $p^2\nmid a_0$. To prove it, suppose $F=GH$ with both factors of positive degree. After removing contents, Gauss's lemma lets us take $G,H\in \mathbb{Z}[X]$ primitive. Reduction modulo $p$ givesso both reduced factors are monomials of positive degree. Their constant terms are therefore divisible by $p$, making $p^2$ divide $a_0=G(0)H(0)$, a contradiction.

An algebraic integer is a complex number satisfying a monic polynomial in $\mathbb{Z}[X]$. Let $\alpha$ be one and let $m_{\alpha }\in \mathbb{Q}[X]$ be its monic minimal polynomial. The usual proof using polynomial content shows that $m_{\alpha }\in \mathbb{Z}[X]$. The setis the kernel of the polynomial-evaluation ring homomorphism, hence an ideal. Every $f\in I_{\alpha }$ is divisible by $m_{\alpha }$ in $\mathbb{Q}[X]$. Because $m_{\alpha }$ is monic, polynomial division of $f$ by $m_{\alpha }$ takes place in $\mathbb{Z}[X]$, and the remainder must vanish by minimality. Thereforewhere $m_{\alpha }$ is monic and irreducible.

Fortranslation givesThis is Eisenstein at $p=3$, so $f(X+1)$ and therefore $f(X)$ are irreducible over $\mathbb{Q}$. HenceIrreducibility and Gauss's lemma also show that $(f)$ is a prime ideal of $\mathbb{Z}[X]$, soIt is not a field. The class of $2$ is nonzero but not a unit: reducing further modulo $2$ gives a nonzero quotientin which the image of $2$ is zero, whereas a unit must remain a unit under every unital quotient map. Thus

A direct determinant calculation givesMoreover,has rank one, so its kernel has dimension two, while $(A-I{)}^2=0$ and $A\ne I$. Hence the minimal polynomial isThere are two Jordan blocks, because the eigenspace has dimension two, and the largest has size two, because the minimal polynomial has exponent two. Thus

The minimal polynomial $m_{\alpha }$ is the unique monic polynomial of least positive degree such that $m_{\alpha }(\alpha )=0$. The Cayley-Hamilton theorem says that $\alpha$ satisfies its characteristic polynomial, so such polynomials exist. Dividing any two monic candidates of least degree shows uniqueness.

If $m_{\alpha }(x)=x^m$, then ${\alpha }^m=0$ but ${\alpha }^{m-1}\ne 0$. The least $r$ for which $({\alpha }^3{)}^r=0$ is the least integer satisfying $3r\ge m$. Hence

SetFor every $i$,Consequently $q({\alpha }^3)$ contains $m_{\alpha }(\alpha )$ as a polynomial factor and is zero. By the divisibility property of the minimal polynomial,

If the minimal polynomial of $\gamma$ has degree $d$, then
${\mathbb{F}}_p(\gamma )$ has $p^d$ elements and is an intermediate field of
$L/{\mathbb{F}}_p$. The tower law for field extensions givesso $d\mid n$.

The multiplicative group of a finite field is cyclic. Choose a generator
$\gamma$ of $L^{\times }$, of order $p^n-1$. If $\gamma$ lay in a proper intermediate field of degree $d<n$, its order would divide $p^d-1<p^n-1$, a contradiction. Thus
${\mathbb{F}}_p(\gamma )=L$ and its minimal polynomial has degree $n$.

For arbitrary $r\ge 1$, let $E$ be the splitting field of
$X^{p^r}-X$ over ${\mathbb{F}}_p$. Its roots form a field: the Frobenius endomorphism shows they are closed under addition, subtraction, multiplication, and inversion. The derivative is $-1$, so there are exactly $p^r$ distinct roots. Thus this root field has order $p^r$. Applying the preceding generator argument supplies an element whose minimal polynomial over ${\mathbb{F}}_p$ has degree $r$, proving that an irreducible polynomial of every positive degree exists.

An element $\alpha$ is separable over $K$ when its minimal polynomial has distinct roots in a splitting field. A polynomial has a repeated root exactly when it has a common root with its formal derivative. Hence
$\gcd (f,f^{\prime })=1$ implies that $f$ has no repeated root and $\alpha$ is separable.

Let $f$ be the minimal polynomial of $\beta$. Repeatedly factor through the Frobenius untilwhere $g^{\prime }\ne 0$. Irreducibility of $f$ makes $g$ irreducible, and
$\alpha ={\beta }^{p^h}$ has separable minimal polynomial $g$. PutThen $M/K$ is separable andEvery $\gamma \in L=K(\beta )$ is a polynomial in $\beta$ of degree below $[L:K]$. The freshman's-dream identity givesso $L/M$ is purely inseparable.

Finally, any intermediate field satisfying the stated conditions consists of elements separable over $K$, so it lies in the maximal separable subextension just constructed. Conversely, the condition that all relevant $p^h$th powers lie in that field puts ${\beta }^{p^h}$, and hence $M$, inside it. The two inclusions prove uniqueness.

The Dedekind factorization theorem says that, when a rational prime $p$ does not divide the index of an order $\mathbb{Z}[\theta ]\subseteq {\mathcal{O}}_K$, a factorizationof the minimal polynomial modulo $p$ gives

Let $K=\mathbb{Q}(\sqrt{d})$ with square-free $d$, and let $p$ be odd. The resulting splitting of rational primes in a quadratic field is:

Finally suppose $d>0$ and
$\alpha =A/C+(B/C)\sqrt{d}$ has norm $-1$, with integers
$A,B,C$ chosen coprime. ThenIf an odd $p\equiv 3(\mathrm{mod}4)$ ramified, then $p\mid d$. Reduction modulo $p$ gives
$A^2\equiv -C^2(\mathrm{mod}p)$. Since $-1$ is not a square modulo such a prime, $p\mid A,C$. Dividing the equation's divisibility shows $p\mid B$ as well (use that square-free $d$ has $p$-adic valuation one), contradicting coprimality. Hence no prime $p\equiv 3(\mathrm{mod}4)$ ramifies.

The automorphisms $1,\sigma ,\dots ,{\sigma }^{n-1}$ are distinct and hence linearly independent as maps $L\to L$ by the linear independence of distinct field embeddings. Therefore the operatoris not the zero map. Choose $\beta \in L$ for which $\alpha =T(\beta )\ne 0$. Reindexing the sum and using ${\sigma }^n=1$ givesThis is the Lagrange resolvent eigenvector for a cyclic field automorphism.

Let $F=L^{\sigma }$ be the fixed field. Since $\sigma ({\alpha }^n)={\alpha }^n$, the element $\alpha$ is a root ofIts orbit under $⟨\sigma ⟩$ isand these $n$ elements are distinct because $\alpha \ne 0$ and $\zeta$ is a primitive root of unity. Hence the minimal polynomial has at least $n$ distinct roots. It also divides the displayed degree-$n$ polynomial, so

Because $\mathrm{char}K=0$, the finite extension $M/K$ is separable. By the primitive element theorem, write $M=K(\theta )$. The minimal polynomial of $\theta$ over $K$ has a root in the algebraically closed field $L$, so sending $\theta$ to that root defines a $K$-field embedding $\iota :M\hookrightarrow L$.

The cyclic group $⟨\sigma ⟩$ acts faithfully on $L$ after replacing $d$ by the order of $\sigma$. The Artin fixed-field theorem givesand says that $L/K$ is a finite Galois extension with cyclic Galois group $⟨\sigma ⟩$. The image $E=\iota (M)$ is an intermediate field. Every subgroup of a cyclic group is normal, so the normal subextension criterion shows that $E/K$ is Galois, and its Galois group is a quotient of $⟨\sigma ⟩$, hence cyclic. Transporting this structure through $\iota$ proves that $M/K$ is Galois with cyclic Galois group. This is the finite extensions of the fixed field of a finite-order automorphism of an algebraically closed field theorem.

Algebraic closedness is necessary. Take $L=\mathbb{Q}$ and $\sigma =\mathrm{id}$, so $K=L^{\sigma }=\mathbb{Q}$, and let $M=\mathbb{Q}(\sqrt[3]{2})$. The extension has degree three but is not normal, since the two nonreal roots of $X^3-2$ do not belong to $M$. It is therefore not Galois.

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.

A finite extension $L/K$ is normal when every irreducible polynomial in $K[T]$ having one root in $L$ splits completely over $L$.

Suppose first that $L/K$ is normal. Choose ${\alpha }_1,\dots ,{\alpha }_m$ with $L=K({\alpha }_1,\dots ,{\alpha }_m)$, and let $f_j$ be the minimal polynomial of ${\alpha }_j$ over $K$. Every $f_j$ splits in $L$ by normality, and $L$ is generated by their roots because it is already generated by the particular roots ${\alpha }_j$. Thus $L$ is the splitting field of $f_1\cdots f_m$.

Conversely, suppose that $L$ is the splitting field of $f\in K[T]$. Every $K$-embedding $\sigma :L\to \overline{K}$ permutes the roots of $f$, so $\sigma (L)=L$. By the embedding characterization of normality, $L/K$ is normal. This proves the finite normal extension as a splitting field criterion.