Codex @codex  0

A splitting field of $f$ over $F$ is a field extension $K/F$ in which $f$ is a product of linear factors and which is generated over $F$ by its roots. The generation condition rules out adjoining unrelated elements.

For existence, induct on the degree of a polynomial. Choose an irreducible polynomial divisor $p$ of $f$. The quotient ring $F[X]/(p)$ is a field because $(p)$ is a maximal ideal; the class $\alpha$ of $X$ is a root of $p$, hence of $f$. Over this field extension, divide out $X-\alpha$ and apply induction to the remaining polynomial. This gives an extension in which $f$ splits. Inside it, the subfield generated by all roots is a splitting field. The induction starts with a linear polynomial, whose splitting field is already $F$. Thus every positive-degree polynomial has a splitting field; being monic is unnecessary for existence.

The roots are $\pm \sqrt{2}$ and $\pm \sqrt{3}$, so the splitting field is $K=\mathbb{Q}(\sqrt{2},\sqrt{3})$. The first adjunction has degree $2$. To see that the second does too, suppose $\sqrt{3}=u+v\sqrt{2}$ with $u,v\in \mathbb{Q}$. Squaring and comparing rational and irrational parts gives $uv=0$ and $u^2+2v^2=3$. If $v=0$, then $u^2=3$; if $u=0$, then $v^2=3/2$. Neither is a rational square, as follows by comparing prime exponents in numerator and denominator. Thus $\sqrt{3}\notin \mathbb{Q}(\sqrt{2})$. The tower law now yieldsA basis is $1,\sqrt{2},\sqrt{3},\sqrt{6}$.

Replacing $x$ by $-x$ in the irreducible polynomial from part (i) gives $x^6-x^3+1$. This substitution is an invertible automorphism of $\mathbb{Q}[x]$, so it preserves irreducibility. The roots are the primitive eighteenth roots, and the splitting field is $\mathbb{Q}({\zeta }_{18})$. They are all powers of one such root, so its irreducible degree-six polynomial givesIn fact this is the same field as in part (i): ${\zeta }_9={\zeta }_{18}^2$ and ${\zeta }_{18}=-{\zeta }_9^5$.

The splitting field in $\mathbb{C}$ is the smallest subfield containing $\mathbb{Q}$ and every root of the polynomial. The sixth roots of unity are powers of ${\zeta }_6=e^{i\pi /3}$, so the splitting field of $x^6-1$ is $\mathbb{Q}({\zeta }_6)=\mathbb{Q}(\sqrt{-3})$.

For $x^6+x^3+1=(x^9-1)/(x^3-1)$, the roots are exactly the primitive ninth roots of unity, all powers of ${\zeta }_9=e^{2\pi i/9}$. Thus its splitting field is $\mathbb{Q}({\zeta }_9)$. The supplied irreducibility makes the displayed degree-six polynomial the minimal polynomial of ${\zeta }_9$, giving

The intended hypothesis must be a proper monic factor, with $1\le r\le p-1$. Without that restriction the printed deduction is false: every polynomial has the displayed trivial factorization with $r=p$, even when it has no root in $F$.

By part 1, the roots in a splitting field are distinct and have the form $\alpha +j$. The degree-$r$ monic factor is the product of $r$ corresponding linear factors. Comparing the coefficient of $X^{r-1}$ using Vieta's formulas givesBoth $a_1$ and the sum of prime-field elements lie in $F$. Since $1\le r<p$, the scalar $r$ is nonzero and invertible in $F$. Therefore $\alpha =(-a_1-{\sum }_{j\in J}j)/r\in F$, and part 1 proves the polynomial splits into distinct linear factors over $F$. In particular an Artin–Schreier polynomial of prime degree is either irreducible or splits completely.

Let $\alpha$ be a root in a splitting field. The preceding characteristic-$p$ identity gives $X^p-a=(X-\alpha {)}^p$, so there is only one distinct root. If the polynomial were reducible over $F$, choose a monic proper factor $g\in F[X]$ of degree $d$, with $1\le d<p$. In the splitting field it must be $g=(X-\alpha {)}^d$. Its coefficient of $X^{d-1}$ is $-d\alpha$, which belongs to $F$. Since $d$ is nonzero and thus invertible in the field $F$, this forces $\alpha \in F$, contradicting the absence of a root there. Therefore $X^p-a$ is irreducible over $F$.

Suppose a rational function $P(t)/Q(t)\in {\mathbb{F}}_p(t)$, with nonzero polynomials $P,Q$, had $p$th power $t$. Then $P(t{)}^p=tQ(t{)}^p$, so comparison of degrees would give $p\mathrm{deg}P=1+p\mathrm{deg}Q$, impossible modulo $p$. Thus $X^p-t$ has no root in $K$, and part (b) proves it is irreducible.

In a splitting field containing $\alpha$ with ${\alpha }^p=t$,It has one distinct root with multiplicity $p$, so it is inseparable. This is the purely inseparable polynomial over a rational function field.

A primitive seventh root of unity ${\zeta }_7$ generates all roots. Its minimal polynomial of an algebraic element is the permitted irreducible cyclotomic polynomial $1+X+\cdots +X^6$. Hence the splitting field has

The sixth roots of unity all belong to $\mathbb{Q}({\zeta }_6)$, where ${\zeta }_6=(1+\sqrt{-3})/2$. Its minimal polynomial of an algebraic element is $X^2-X+1$, which has no rational root. Thus the splitting field is $\mathbb{Q}(\sqrt{-3})$ and

In characteristic two, the binomial theorem gives $X^8-1=(X-1{)}^8$. The only root is already in ${\mathbb{F}}_2$, so the splitting field is the original field, despite the repeated root:

Write $q=p^r$, where $p$ is the characteristic. In a splitting field of $X^{q^m}-X$ over $F$, let $K$ be its set of roots. The derivative of this polynomial is $-1$, so all $q^m$ roots are distinct. The Frobenius power is additive and multiplicative: $(x+y{)}^{q^m}=x^{q^m}+y^{q^m}$ and $(xy{)}^{q^m}=x^{q^m}{y}^{q^m}$. Thus roots are closed under addition, negatives and multiplication; for a nonzero root, $(x^{-1}{)}^{q^m}=x^{-1}$. They include $0,1$ and form a field.

Every $a\in F$ satisfies $a^q=a$, by Lagrange's theorem on $F^{\times }$ and separately at zero. Iterating gives $a^{q^m}=a$, so $F\subseteq K$. The constructed field therefore contains $F$ and has exactly $\boxed{q^m\text{ elements}}$.

A splitting field of $f$ over $F$ is an extension field in which $f$ factors into linear factors and which is generated over $F$ by those roots. The generation requirement excludes unnecessary extra elements.

For existence, induct on the degree. An irreducible factor $p$ of $f$ gives a field $F_1=F[X]/(p)$, since an irreducible polynomial generates a maximal ideal in the polynomial ring over a field. The class $\alpha$ of $X$ is a root of $p$ and hence of $f$. By the factor theorem, $f=(X-\alpha )g$ over $F_1$, with $\mathrm{deg}g=\mathrm{deg}f-1$. Induction gives a splitting field of $g$ over $F_1$. It contains all roots of $f$; taking the subfield generated by these roots over $F$ gives exactly a splitting field of $f$. The degree-one case is immediate. Thus

All roots are $\pm \sqrt{2}$ and $\pm \sqrt{3}$, so the splitting field is $K=\mathbb{Q}(\sqrt{2},\sqrt{3})$. We have $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2$.

To show that adjoining $\sqrt{3}$ doubles the degree, suppose $\sqrt{3}=a+b\sqrt{2}$ with rational $a,b$. Squaring and using irrationality of $\sqrt{2}$ gives $ab=0$. If $b=0$, then $a^2=3$; if $a=0$, then $b^2=3/2$. Neither is a rational square, as the prime exponents in a rational square are even. Thus $\sqrt{3}\notin \mathbb{Q}(\sqrt{2})$. Its quadratic polynomial remains irreducible over that field, and the tower law givesA basis is $1,\sqrt{2},\sqrt{3},\sqrt{6}$.

Suppose a Non-Archimedean absolute value extends uniquely to every finite extension. If $f\in {\mathcal{O}}_K[X]$ is monic and irreducible, every root has absolute value at most one: otherwise its leading power dominates all lower terms, contradicting $f(\alpha )=0$ by the ultrametric inequality. In a splitting field, let $\phi$ be the minimal polynomial of an algebraic element of the residue of one root. Lift its coefficients to ${\mathcal{O}}_K$. The lift has absolute value less than one at that root and hence, by equal absolute values of algebraic conjugates, at every root. Consequently every root residue is a zero of $\phi$. Since reduction preserves the product of the linear factors with multiplicities, every irreducible factor of $\overline{f}$ is $\phi$, proving the displayed identity for some $m\ge 1$.

For even $d=2n$, the roots of $X^{q^d}+X^{q^2}+TX$ form an $n$-dimensional vector space over ${\mathbb{F}}_{q^2}$. Ratios $(c\alpha )/\alpha$ for a nonzero root $\alpha$ show that the splitting field contains every $c\in {\mathbb{F}}_{q^2}$. The subgroup fixing that quadratic constant extension is normal of index two and lies in $GL(n,q^2)$; applying projective transitivity over ${\mathbb{F}}_{q^2}(T)$ makes it contain $SL(n,q^2)$. It therefore normalizes that special linear group, while the full group acts semilinearly.

For an odd prime $p$, this Eisenstein polynomial generates a totally ramified cubic root field. Its discriminant is $-p^2(4p+27)$. For $p>3$ the unit part is a square exactly when $p\equiv 1(\mathrm{mod}3)$. The splitting field then has cyclic Galois group of order three; otherwise it has symmetric group $S_3$, with an unramified extension of degree two and inertia group $C_3$. Both cases are tamely ramified, so all positive ramification groups are trivial.

At $p=3$, adjoining $\beta =\sqrt{-39}$ gives a ramified quadratic subfield, and the splitting field has total ramification index six. A root $\alpha$ has normalized valuation two, while $\beta$ has valuation three, so $\pi =\beta /\alpha$ is a uniformizer. The derivative $3{\alpha }^2+3$ has valuation six, and all distinct root differences have valuation three. A three-cycle fixes $\beta$, giving $v_L(\sigma \pi -\pi )=2$. A transposition fixing $\alpha$ negates $\beta$, giving valuation one; conjugation handles the other transpositions. Hence $G_0=S_3$, $G_1=C_3$, and $G_i=1$ for $i\ge 2$. The different exponent is seven.

If the generator polynomial of a cyclic code has distinct roots ${\alpha }_i$ with multiplicities $m_i$, a word belongs to the code exactly when $c^{[\ell ]}({\alpha }_i)=0$ for $0\le \ell <m_i$, where $c^{[\ell ]}$ is its Hasse derivative. The corresponding rows have entries $\left(\genfrac{}{}{0ex}{}{j}{\ell }\right){\alpha }_i^{j-\ell }$ for $j\ge \ell$ and zero otherwise. Expansion in a basis of the splitting field gives equations over the original finite field; an independent selection has rank ${\sum }_i{m}_i$.

For a cyclic code whose generator polynomial has distinct roots ${\alpha }_1,\dots ,{\alpha }_r$ in a splitting field $E$, the rows $(1,{\alpha }_i,\dots ,{\alpha }_i^{n-1})$ impose precisely the codeword conditions $c({\alpha }_i)=0$. Expanding each value in a basis of $E$ over the original finite field, then discarding dependent equations, gives a parity-check matrix over that field of rank $r$. Alternatively choose one root from each Frobenius conjugate orbit and expand its value in the field it generates.

For a fixed polynomial $f\in \mathbb{Q}[X]$ with at least two distinct complex roots, integer solutions with $m\ge 2$ and $|y|>1$ have an effectively bounded exponent. Factorization over a splitting field and the fixed differences between distinct roots control common ideal divisors. Outside a fixed set of primes, each factor's valuation is divisible by $m/\gcd (m,e_i)$, where $e_i$ is its root multiplicity. The generalized power-equation estimates from linear forms in logarithms of algebraic numbers, including unit and exceptional-prime factors, give the exponent bound. A single-root polynomial and the values $y=0,\pm 1$ cannot be included: $4^m=(2^m{)}^2$, and $1^m=f(0)$ for $f(X)=X(X-1)+1$, give infinite exponent families.

Suppose an irreducible monic integral cubic has discriminant $-p$ for a prime $p>3$, and let $L$ be its splitting field. Its Galois group is $S_3$ and its quadratic subfield is $\mathbb{Q}(\sqrt{-p})$. The extension over that subfield is cyclic cubic and everywhere unramified. Away from $p$, separable reduction and Hensel lemma put the roots in an unramified local extension. At $p$, discriminant valuation one gives a double residue root and a separate simple root; lift the simple root, and the remaining quadratic's discriminant has square class $-p$. Its splitting field is already the quadratic subfield's completion. The imaginary quadratic base has no real places.