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A group representation over a field is absolutely irreducible if it remains irreducible after every field extension. For a finite-dimensional representation it is enough to check an algebraic closure. Irreducibility over the original field alone is weaker: the two-dimensional real rotation representation of the cyclic group of order three splits into two lines over the complex numbers.

Fix a number field $K$, a positive integer $d$, and a finite set $S$ of its finite places. There are finitely many field extensions $L/K$ of degree at most $d$ unramified outside $S$, inside a fixed algebraic closure. To see this, view them as extensions of the rational numbers of bounded degree. Their ramified rational primes belong to a fixed finite set. At each of these primes, the different exponent is bounded in terms of the local degree and ramification index, giving a bound for the absolute field discriminant. The Hermite–Minkowski theorem gives finitely many fields up to isomorphism, and each has only finitely many embeddings into the fixed closure. This useful form of arithmetic compactness is stated and proved for $K=\mathbb{Q}$ in Andrew Sutherland's number-theory notes, Theorem 14.25; the preceding reduction gives the general case.

If $\alpha$ and $\beta$ are separable algebraic elements, put $K=F(\alpha )$ and $L=K(\beta )$. The minimal polynomial of $\beta$ over $K$ divides its separable polynomial over $F$, and is therefore separable. The extension count for field embeddings gives $[K:F]$ embeddings of $K$ into an algebraic closure, each with $[L:K]$ extensions to $L$. Hence $L$ has $[L:F]$ embeddings. For $E=F(\alpha +\beta )$, each embedding of $E$ has at most $[L:E]$ extensions, so there are at least $[E:F]$ embeddings of $E$ and hence exactly that number. Since $E$ is a simple field extension, its minimal polynomial has $[E:F]$ distinct roots. Thus $\alpha +\beta$ is a separable algebraic element.

The completion ${\mathbb{C}}_p$ of an algebraic closure of ${\mathbb{Q}}_p$ is an algebraically closed field. Approximate a polynomial over ${\mathbb{C}}_p$ by one over ${\overline{\mathbb{Q}}}_p$, use continuity of roots over a non-Archimedean field to obtain a nearby algebraic root, and then use Krasner's lemma to show that the original root already belongs to ${\mathbb{C}}_p$.

The algebraic conjugates of an algebraic element over a field $K$ are the polynomial roots of its algebraic minimal polynomial over $K$ in an algebraic closure.

The algebraic conjugates of an algebraic number are the polynomial roots of its algebraic minimal polynomial, equivalently its images under field embeddings of its number field into an algebraic closure.

A division point above $P$ is a point $Q$ over an algebraic closure satisfying $[N]Q=P$, for a specified nonzero integer $N$. Two such points differ by an $N$-torsion point of an elliptic curve. If the characteristic of a field does not divide $N$, there are $N^2$ distinct division points; in other characteristics the scheme fibre still has length $N^2$, but its geometric points can be fewer. The unramified division torsors at good primes control their coordinate fields over local fields.

For a finite field extension $L/K$, multiplication by $\alpha \in L$ is a $K$-linear map $m_{\alpha }:L\to L$. Its determinant is the field norm:This does not depend on the chosen basis. If $L/K$ is a separable field extension, thenwhere the product runs over all $[L:K]$ field embeddings into an algebraic closure of $K$. Embeddings are counted even when several give the same value on $\alpha$: for $r\in K$, the result is $r^{[L:K]}$. This avoids ambiguity in the phrase “product of algebraic conjugates”.^[ref]

For $L=\mathbb{Q}(\sqrt{2})$ over $K=\mathbb{Q}$, write $\alpha =a+b\sqrt{2}$. In the basis $1,\sqrt{2}$, multiplication by $\alpha$ has columns $(a,b)$ and $(2b,a)$, henceThe determinant calculation and the product of the two field embeddings give the same answer. In particular,A field norm is an arithmetic invariant; it is not a nonnegative norm measuring length, and need not be positive on nonzero elements.

The field norm is multiplicative, $N(\alpha \beta )=N(\alpha )N(\beta )$, because $m_{\alpha \beta }=m_{\alpha }{m}_{\beta }$ and determinants are multiplicative. It vanishes exactly at zero. For a number field over $\mathbb{Q}$, it takes algebraic integers to integers, and an algebraic integer is a unit exactly when its field norm is $1$ or $-1$.^[ref][ref]

This makes the field norm a useful test for divisibility. If nonzero $\alpha ,\beta \in {\mathcal{O}}_L$ and $\alpha$ divides $\beta$, then the integer $N(\alpha )$ divides $N(\beta )$. For example, $3+\sqrt{2}$ has field norm $7$, so it cannot divide $2$, whose field norm is $4$. It does divide $7$, with quotient $3-\sqrt{2}$. This necessary condition can reject a proposed factor without solving for its quotient.

The complex algebraic numbers form a field, the algebraic closure of $\mathbb{Q}$ inside $\mathbb{C}$. To see closure under the field operations, if $\alpha ,\beta$ are algebraic numbers, the tower law makes $\mathbb{Q}(\alpha ,\beta )$ a finite-dimensional field extension of $\mathbb{Q}$. Every element of this extension is an algebraic element, because its successive powers are linearly dependent. In particular it contains $\alpha \pm \beta$, $\alpha \beta$ and $\alpha /\beta$ when $\beta \ne 0$.

It is an infinite-degree algebraic extension. A primitive $p$th root of unity has minimal polynomial of an algebraic element $1+X+\cdots +X^{p-1}$ of degree $p-1$ for every prime $p$, so the degrees of its finite subextensions are unbounded. Finally a complex root of a polynomial whose coefficients are algebraic numbers is itself algebraic over their finite-degree coefficient field and hence over $\mathbb{Q}$, again by the tower law. Thus every such polynomial splits in this field, establishing the algebraic closure assertion.

For a polynomial with nonzero constant coefficient over a complete discretely valued field, plot coefficient exponents against their valuations. A lower Newton polygon segment of slope $s$ and horizontal length $r$ accounts for exactly $r$ roots of valuation $-s$, counted with multiplicity in an algebraic closure. Zero coefficients are omitted. Reflecting the horizontal axis changes the sign of every slope, so slope conventions must be fixed before reading root valuations.

For an affine plane algebraic curve $F(x,y)=0$, a point over an algebraic closure is nonsingular when the displayed condition holds. Its Zariski tangent space has dimension one, and a local coordinate can be chosen in the direction of a nonzero partial derivative. A smooth algebraic curve has all points nonsingular. Over a complete local field, the Hensel lemma lifts a nonsingular residue point by solving for a coordinate with unit partial derivative.

For a reduced plane algebraic curve, an ordinary double point is a point whose lowest nonzero local homogeneous term has degree $2$ and factors into two distinct linear factors over an algebraic closure. The two factors give distinct tangent directions. In characteristic not $2$, $x^2+y^2+\text{higher terms}$ is an example.

Proceed by induction on $n=\dim V$. The zero-dimensional case is immediate. Over $\mathbb{C}$, the fundamental theorem of algebra supplies a root $\lambda$ of the characteristic polynomial. Since $\det (T-\lambda I)=0$, there is a nonzero vector $v_1$ with $T{v}_1=\lambda v_1$. The line $W=\mathbb{C}{v}_1$ is invariant, so $T$ induces a linear map $\overline{T}$ on $V/W$. By induction, choose an upper triangular basis for that quotient and lift its vectors to $v_2,\dots ,v_n$. Then $T{v}_j\in \mathrm{span}(v_1,\dots ,v_j)$ for every $j$. Thus $v_1,\dots ,v_n$ is the required basis and $\boxed{T\text{ is triangularizable}}$.

The same conclusion holds over $F$ when the characteristic polynomial splits over $F$. Indeed an eigenvalue then lies in $F$, and the block matrix arising from the invariant line gives ${\chi }_T(X)=(X-\lambda ){\chi }_{\overline{T}}(X)$. The quotient characteristic polynomial therefore also splits, so the induction works over $F$ without requiring algebraic closure.

Proceed by induction on $n=\dim V$. The zero-dimensional case is immediate. Over $\mathbb{C}$, the fundamental theorem of algebra supplies a root $\lambda$ of the characteristic polynomial. Since $\det (T-\lambda I)=0$, there is a nonzero vector $v_1$ with $T{v}_1=\lambda v_1$. The line $W=\mathbb{C}{v}_1$ is invariant, so $T$ induces a linear map $\overline{T}$ on $V/W$. By induction, choose an upper triangular basis for that quotient and lift its vectors to $v_2,\dots ,v_n$. Then $T{v}_j\in \mathrm{span}(v_1,\dots ,v_j)$ for every $j$. Thus $v_1,\dots ,v_n$ is the required basis and $\boxed{T\text{ is triangularizable}}$.

The same conclusion holds over $F$ when the characteristic polynomial splits over $F$. Indeed an eigenvalue then lies in $F$, and the block matrix arising from the invariant line gives ${\chi }_T(X)=(X-\lambda ){\chi }_{\overline{T}}(X)$. The quotient characteristic polynomial therefore also splits, so the induction works over $F$ without requiring algebraic closure.

The eigenvalues in $K$ are precisely the roots in $K$ of the companion polynomial $d$; over an algebraic closure take all its roots. Thus the form determines their root set, but an explicit list generally requires factoring or solving $d$ and cannot be read off by inspection. This is the relevant distinction in the qualification “with no further calculations”. Over a general field the polynomial may have no roots in $K$ at all, and its diagonal entries are not its eigenvalues. If “determinable” means only uniquely specified rather than immediately readable, all the information is of course fixed by the matrix, including this root set.

A Hilbertian field is a field over which generic irreducibility can be retained by specializing a parameter. More precisely, for polynomials $f_i(T,X)$ irreducible over $K(T)$ and separable in $X$, and a nonzero exclusion polynomial $h(T)$, the Hilbert subsetis infinite. Denominators, leading coefficients and discriminants may be incorporated into $h$. The infinite condition permits simultaneous avoidance of any finitely many bad places. Separability is part of the definition, particularly in positive characteristic.

The field-theoretic interpretation is useful in constructive Galois theory. If $L/K(T)$ is finite separable and generated by a root of $f$, specialization at a good $a$ is the residue extension at a place over $T-a$. Irreducibility of $f(a,X)$ means its residue degree is the entire degree $[L:K(T)]$. Thus a Hilbertian field has infinitely many unramified rational places which are completely inert in each prescribed finite separable extension. For a finite family, one can work in a primitive-element presentation of their separable compositum; preserving its full degree preserves the degrees of all the subextensions.

Hilbert's irreducibility theorem says that $\mathbb{Q}$ is Hilbertian, and gives the corresponding assertion for number fields, including infinitely many integral specializations in its classical form. Rational function fields $k(U)$ over arbitrary fields $k$ are also Hilbertian. In particular, global function fields are Hilbertian just as number fields are. These standard field classes and the specialization viewpoint are discussed in Haran and Jarden's introduction to Hilbertian fields. The theorem does not say every parameter is good: $X^2-T$ is generically irreducible, but a square parameter makes it split. For simultaneous irreducibility, one excludes more exceptional parameters, not the entire field.

Hilbertianity survives finite extensions. For the separable case, let $K$ be Hilbertian, $M/K$ finite separable and $f\in M(T)[X]$ irreducible and separable. A primitive element of $M(T)(\alpha )$ over $K(T)$ has degree $[M:K]\mathrm{deg}f$. Apply Hilbertianity to its minimal polynomial and exclude the presentation's denominators. The resulting residue field contains the constant copy of $M$ and has the same total degree over $K$, so its degree over $M$ is $\mathrm{deg}f$. Therefore $f(a,X)$ remains irreducible over $M$, even for infinitely many $a$ chosen in $K$. This argument also handles finitely many polynomials by a compositum. Purely inseparable finite extensions are treated by separating the separable part and transporting its rational places through the unique purely inseparable extension; the finite-extension stability statement remains valid.

The property is not inherited by arbitrary algebraic extensions. An algebraic closure is not Hilbertian: in characteristic zero, $X^2-T$ is irreducible over $K(T)$ but every specialization has a root in $K$. In characteristic $p$, use $X^p-X-T$ instead; a hypothetical rational-function root would give its right side a pole of order divisible by $p$, contradicting its simple pole at infinity. All constant specializations split over the algebraically closed field.

Finite fields also fail. For $K={\mathbb{F}}_q$ of characteristic $p$, considerIts right-hand additive datum has a pole of order $q+1$, not divisible by $p$, and cannot be $b^p-b$ for a rational function $b$. The polynomial is therefore irreducible over $K(T)$: its roots are $\alpha +c$ for $c\in {\mathbb{F}}_p$, so its splitting-field group has prime order unless a root already belongs to the base field. At every $a\in K$, however, $a^q-a=0$ and the specialization is $X^p-X$, which splits. This gives a concrete failure of the required specialization property.

The main constructive application is Galois-preserving Hilbert specialization. Suppose $E/K(T)$ is finite Galois with group $G$, and use a primitive element of the entire splitting field, not merely a generator of one root field. A good specialization retaining degree $|G|$ has residue group embedded in $G$ by unramified reduction. Since its order is still $|G|$, it is exactly $G$. Preserving only the irreducibility of the original degree-$n$ polynomial would prove transitivity, which is weaker than preserving its full group. Resolvents or the primitive generator eliminate that ambiguity.

Consequently a regular realization of $G$ over $K(T)$, with constant field $K$, yields realizations over the Hilbertian field $K$. This is the bridge from branch-cycle constructions to the Inverse Galois problem over number fields. A realization over $\mathbb{C}(T)$ alone cannot be specialized to $\mathbb{Q}$; descent of its coefficients and constants is an additional step. Rational rigidity can supply that descent under its appropriate hypotheses, while ordinary complex rigidity answers only the geometric uniqueness question.

For example, the family $X^n-X-T$ has full symmetric geometric monodromy in characteristic zero. It is irreducible because it is linear in $T$. Its derivative has $n-1$ distinct critical roots satisfying $n{X}^{n-1}=1$, and their distinct critical values are $-(n-1)X/n$. Each simple critical point gives a transposition. These finite branch cycles generate the monodromy, since the infinity loop is the inverse product of the finite loops. Their transitive connected-edge action gives $S_n$. The arithmetic group is also $S_n$, so the constant field is $\mathbb{Q}$. Hilbert specialization then produces symmetric Galois groups over $\mathbb{Q}$. For $n=2$ the same conclusion follows directly from the square-root family.

Finally, the good parameters form a large arithmetic set without being every point or generally a Zariski-open set. Reducibility can occur at infinitely many exceptional points, as the square example shows. The language of thin sets packages those exceptions as images of generically finite separable covers of degree greater than one, together with finitely many points. Hilbertianity asserts that such finitely many exceptional constructions cannot exhaust the rational line. It is an existence principle with powerful constructive consequences, not by itself an algorithm that certifies a particular chosen parameter.

The PDF has $q$-power exponents; the polynomial is a linearized polynomial over a finite field, not the degree-$d$ polynomial produced by the converted TeX. Let $V$ be its root set in an algebraic closure. Because $(x+y{)}^{q^i}=x^{q^i}+y^{q^i}$ and $c^{q^i}=c$ for $c\in {\mathbb{F}}_q$,Thus $V$ is an ${\mathbb{F}}_q$-vector space. Its derivative is $a_0\ne 0$, and its degree is $q^d$, so it has exactly $q^d$ distinct roots. A finite vector space of that size has dimension $d$.

Every splitting-field automorphism fixes ${\mathbb{F}}_q$ and carries roots to roots. Hence it acts linearly on $V$. The action is faithful because the splitting field is generated by the roots. Choosing a basis identifies this injection withDifferent bases conjugate the matrix subgroup. The linear action itself is canonical, while a particular matrix realization is basis-dependent.

Put $N=(q^d-1)/(q-1)$, so $N>1$ and $N\equiv 1$ in the characteristic. For a nonzero root $x$, division by $x$ givesThe projective point ${\mathbb{F}}_{q}x$ is therefore represented by $s=x^{q-1}$, satisfyingTwo nonzero roots give the same $s$ exactly when their ratio lies in ${\mathbb{F}}_q^{\times }$. Thus the $N$ projective points are precisely the roots of $P_t$. The polynomial $s^N+s+t$ is irreducible in ${\mathbb{F}}_q[s,t]$, since it is linear in $t$ and its quotient ring is ${\mathbb{F}}_q[s]$. Gauss's lemma makes it irreducible over ${\mathbb{F}}_q(t)$, so the action on projective points is transitive.

Fix one such coordinate $s$, so $t=-s^N-s$. The other coordinates satisfyWe prove its irreducibility even over ${\overline{\mathbb{F}}}_q(s)$. Make the birational change $z=Y/s$, $h=1/s$. Multiplication by $h^{N-1}$ transforms the equation intoChoose a nontrivial $N$th root of unity $\zeta$ over the algebraic closure. Since the characteristic does not divide $N$, $B$ has a simple zero at $\zeta$. Viewed as a polynomial in $h$ over ${\overline{\mathbb{F}}}_q[z]$, $h^{N-1}+B(z)$ is Eisenstein at the prime $z-\zeta$: the leading coefficient is one, every intermediate coefficient is zero, and the constant term is divisible by that prime exactly once. It is therefore irreducible.

The birational substitution preserves irreducibility after inverting $s$ or $h$. No factor supported solely on $s=0$ is lost, because $R(Y,0)=Y^{N-1}+1$ is nonzero. Gauss's lemma then proves that $R(Y,s)$ is irreducible over ${\overline{\mathbb{F}}}_q(s)$ and hence over ${\mathbb{F}}_q(s)$. Its roots are distinct: a repeated root would have $Y^{N-1}+1=0$ as well as $Y^N+Y+t=0$, forcing $t=0$, impossible generically.

The stabilizer of the chosen projective point is $\mathrm{Gal}(E/{\mathbb{F}}_q(s))$, and the irreducibility of $R$ makes it transitive on all the remaining projective points. This proves projective transitivity of a trinomial linearized polynomial, namely two-transitivity on the one-dimensional subspaces. The supplied group-theoretic result now applies to the linear subgroup from part (a), givingThe Eisenstein argument remains valid even though $N-1$ is divisible by the characteristic; it concerns irreducibility in $h$, not separability of that projection. Separability of the projective polynomial has been checked independently.

Write $d=2n$ with $n\ge 2$, and let $E$ be the splitting field. The root set of $X^{q^{2n}}+X^{q^2}+tX$ is an $n$-dimensional vector space over ${\mathbb{F}}_{q^2}$, by the argument of part (a) with $q$ replaced by $q^2$.

The constants ${\mathbb{F}}_{q^2}$ really occur inside $E$, rather than only in the ambient algebraic closure. Choose a nonzero root $\alpha$. For every $c\in {\mathbb{F}}_{q^2}$, $c\alpha$ is also a root, so both elements lie in $E$ andLet $K={\mathbb{F}}_{q^2}(t)$. Since $K/{\mathbb{F}}_q(t)$ is Galois of degree two,This index statement does not require proving that $E$ has no still larger constant field.

As an extension of $K$, $E$ is the splitting field of the same polynomial. Applying part (b) with base constant size $q^2$ and dimension $n$ givesThe special linear group is the kernel of determinant in the general linear group, so every element of $H$ normalizes it. Therefore $H$ is the required normal index-two subgroup normalizing $SL(d/2,q^2)$.

In fact the full group is semilinear over ${\mathbb{F}}_{q^2}$: an element outside $H$ acts on constants as $c\mapsto c^q$ and satisfies $g(cv)=c^{q}g(v)$. Field automorphisms preserve determinant one, so this also explains the normalizer behavior geometrically. This is the quadratic constant extension of a q-squared linearized splitting field, not the polynomial of part (b) with its $X^q$ term unchanged.

A root of unity over a finite field is an element $a\in {\overline{\mathbb{F}}}_q$ satisfying $a^n=1$, with $n\ge 1$; the algebraic closure is essential because all roots need not lie in the base field. Write $p=\mathrm{char}{\mathbb{F}}_q$ and $n=p^{a}m$, where $p\nmid m$. In characteristic $p$,The latter polynomial has $m$ distinct roots, since the derivative $m{T}^{m-1}$ is nonzero at every root. Those roots are closed under multiplication and inversion and contain $1$, so $E(n,q)$ is a finite multiplicative subgroup of a field. The proof in the preceding solution makes it a cyclic group. In particularThis also explains why the coprimality qualification is needed.

The multiplicative group of a finite field ${\mathbb{F}}_{q^s}^{\times }$ is cyclic of order $q^s-1$. It contains the entire group of $m$th roots precisely when $m\mid q^s-1$. For necessity, a generator of $E(n,q)$ has order $m$ and Lagrange's theorem applies. For sufficiency, a cyclic group whose order is divisible by $m$ contains $m$ roots of $T^m-1$; since there are exactly $m$ roots in the algebraic closure, these are all of them. Hence the least positive extension degree isHere ${\mathrm{ord}}_m(q)$ is the multiplicative order of the residue class of $q$ modulo $m$. If $n$ is coprime to $q$, replace $m$ by $n$. Existence of this order follows because $q$ is a unit modulo $m$.

A primitive $n$th root of unity over a finite field has multiplicative order exactly $n$. If $p\mid n$, no such element exists, since all roots have order dividing $m<n$. If $\gcd (n,q)=1$, choose a generator $\zeta$ of $E(n,q)$. The element ${\zeta }^j$ has order $n/\gcd (n,j)$, so it is primitive precisely when $\gcd (n,j)=1$. Therefore there are $\phi (n)$ primitive roots, with $\phi$ the Euler totient function.

If $\omega$ is primitive of order $n$, then $\omega \in {\mathbb{F}}_{q^{\ell }}$ exactly when ${\omega }^{q^{\ell }}=\omega$, equivalently $n\mid q^{\ell }-1$. The implication from that equality to membership follows because the roots of $T^{q^{\ell }}-T$ form ${\mathbb{F}}_{q^{\ell }}$, as proved in the preceding solution. Consequentlyfor $n>1$, and the degree is one for $n=1$. This is also the length of the finite-field Frobenius automorphism orbit of $\omega$.

Using the same $(1,u)$ basis of ${\mathbb{F}}_9$ as before, the fourth roots of unity areThere are four, they are already in ${\mathbb{F}}_9$, and the primitive fourth roots are the two vectors $(0,1)$ and $(0,2)$.

Krasner's lemma states: let $K$ be complete for a Non-Archimedean absolute value, let $\alpha$ be a separable algebraic element over $K$, and let $\beta$ be algebraic over $K$, all in a fixed algebraic closure with the extended field absolute value. Ifthen $K(\alpha )\subseteq K(\beta )$. If $\alpha \in K$, the conclusion is automatic and there is no conjugate-distance condition to check.

To prove it, put $F=K(\beta )$ and take a finite splitting field $N/F$ for the minimal polynomial of $\alpha$ over $F$. That is a separable polynomial, so $N/F$ is Galois. The finite extension $F$ is complete, and uniqueness of extensions of an field absolute value from a complete non-Archimedean field makes every $\sigma \in \mathrm{Gal}(N/F)$ an isometry. It fixes $\beta$, soIf $\sigma (\alpha )\ne \alpha$, it is another $K$-conjugate and this inequality contradicts the strict hypothesis. Every such automorphism therefore fixes $\alpha$. The Fundamental theorem of Galois theory gives $\alpha \in F$, provingOnly $\alpha$ needs to be a separable algebraic element: the proof works even when $\beta$ is inseparable, because the splitting field is taken over $K(\beta )$.