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A field has characteristic zero when no positive integer multiple of its identity vanishes. Equivalently, its prime field is isomorphic to $\mathbb{Q}$. Every irreducible polynomial over such a field is a separable polynomial.

A splitting field of $f$ over $K$ is an extension generated over $K$ by all its roots in which $f$ splits into linear factors. A nonconstant irreducible polynomial is separable polynomial when its roots in a splitting field are distinct, equivalently when it is coprime to its formal derivative. For a general polynomial, the convention compatible with the printed assertion is that all its irreducible factors are separable; repeated occurrences of the same factor are allowed under this convention.

For a finite field of characteristic $p$, the finite-field Frobenius automorphism $x\mapsto x^p$ is injective and hence surjective. If an irreducible polynomial $g$ had zero derivative, every nonzero exponent would be divisible by $p$. Taking $p$th roots of its coefficients would give $g=h^p$, contradicting irreducibility. Thus $g^{\prime }\ne 0$, and irreducibility gives $\gcd (g,g^{\prime })=1$. Every irreducible factor is therefore separable: finite fields are perfect.

If instead one defines a separable polynomial to mean that the entire polynomial has no repeated roots, the printed claim for every $f$ is false: $(t-1{)}^2$ is a counterexample over any finite field. The proof above establishes the precise intended perfect-field assertion; a polynomial has entirely distinct roots if and only if it is also squarefree.

A splitting field of a nonzero polynomial $f$ over $K$ is an extension in which $f$ splits into linear factors and which is generated over $K$ by all its roots. For a constant nonzero polynomial the splitting field is $K$; the zero polynomial does not have a root-generated splitting field in this standard definition.

Given two splitting fields $L,L^{\prime }$, enumerate the roots generating $L$ and extend a $K$-embedding one root at a time. If an embedding is defined on $K({\alpha }_1,\dots ,{\alpha }_j)$, the minimal polynomial of the next root divides $f$ over that field. Its image under the embedding therefore divides $f$ over the corresponding subfield of $L^{\prime }$, and has a root in $L^{\prime }$. The quotient construction $K(\alpha )\cong K[t]/(m_{\alpha })$ extends the embedding to that root. The polynomial need not be a separable polynomial.

At the end this gives a $K$-embedding $L\to L^{\prime }$. It sends the distinct roots of $f$ injectively to its roots, hence permutes the finite set of distinct roots and contains every root in its image. Since these generate $L^{\prime }$, the embedding is onto. Thus the splitting field is unique up to a (K)-isomorphism.

Write $|F|=p^n$, where $n=[F:{\mathbb{F}}_p]$ since $F$ is a finite-dimensional vector space over its prime field. Every nonzero element satisfies $x^{p^n-1}=1$ by Lagrange theorem, so $F$ is exactly the splitting field of $X^{p^n}-X$ over ${\mathbb{F}}_p$. The derivative is $-1$, making it a separable polynomial. Thus the extension is a normal field extension and a separable field extension, hence a Galois extension.

The finite-field Frobenius automorphism $\sigma (x)=x^p$ fixes ${\mathbb{F}}_p$ and has ${\sigma }^n=1$. If ${\sigma }^d=1$ with $0<d<n$, all $p^n$ elements would be roots of the degree-$p^d$ polynomial $X^{p^d}-X$, impossible. It has order $n$, equal to the degree of the Galois extension. Therefore

Because $K$ has characteristic $p$, the prime field ${\mathbb{F}}_p$ lies in $K$, andfor every $c\in {\mathbb{F}}_p$. These are all $p$ roots. Moreover $f^{\prime }(t)=-1$, so $f$ is a separable polynomial.

Let $m(t)$ be the minimal polynomial of an algebraic element of $\alpha$ over $K$. Translation by $c\in {\mathbb{F}}_p$ sends an irreducible factor $g(t)$ of $f(t)$ to the irreducible factor $g(t-c)$. The additive group of ${\mathbb{F}}_p$ therefore acts on the irreducible factors, and every orbit has size $1$ or $p$. An orbit of size $p$ would already contribute $p\mathrm{deg}m$ to the degree-$p$ polynomial $f$, forcing $\mathrm{deg}m=1$. Then $\alpha \in K$, all roots $\alpha +c$ lie in $K$, and $L=K$, contrary to the hypothesis.

Thus $m(t-c)=m(t)$ for every $c\in {\mathbb{F}}_p$. If $1\le d=\mathrm{deg}m<p$, comparison of the coefficient of $t^{d-1}$ in $m(t-c)-m(t)$ for $c\ne 0$ gives $-dc$ times the nonzero leading coefficient, a contradiction. Hence $\mathrm{deg}m=p$, and since $m$ is a monic divisor of the monic degree-$p$ polynomial $f$, we have $m=f$. Thus $f$ is irreducible over $K$.

All roots $\alpha +c$ already belong to $K(\alpha )$, so the splitting field isAs the splitting field of a separable polynomial, $L/K$ is a Galois extension. For each $c\in {\mathbb{F}}_p$, irreducibility gives a distinct $K$-automorphismThese exhaust the $p=[L:K]$ automorphisms. This is an Artin–Schreier extension, and

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.

Write $q=p^r$, where $p=\mathrm{char}K$. Forthe derivative isso $h$ is a separable polynomial and all its roots in its splitting field $L$ are distinct.

Let $R$ be the set of these roots. The Frobenius endomorphism and the identitiesshow that $R$ is closed under addition and multiplication. It is also closed under additive inverses and, for $x\ne 0$, inversion. Thus $R$ is a finite field. Since $h$ splits into $q^n$ distinct linear factors, $|R|=q^n$. The field generated by all roots is therefore $R$ itself, so $L=R$ andHence the polynomial characterization of a finite field gives

Now let $f\in K[X]$ be irreducible of degree $n$, and let $\alpha$ be a root in an algebraic closure. The minimal polynomial of an algebraic element givesso $K(\alpha )$ has $q^n$ elements. Its nonzero elements form a finite group under multiplication, of order $q^n-1$; hence ${\alpha }^{q^n}=\alpha$. Therefore $\alpha$ is a root of $X^{q^n}-X$, and its minimal polynomial $f$ divides every polynomial over $K$ that vanishes at $\alpha$. Thus the irreducible factors of a finite-field Frobenius polynomial satisfy

If a monic polynomial $f$ has roots ${\alpha }_1,\dots ,{\alpha }_n$ in a splitting field, its polynomial discriminant isIt is a symmetric polynomial in the roots, so it belongs to the base field; it is nonzero exactly when $f$ is a separable polynomial.

Putthe Vandermonde determinant. A root permutation $\sigma$ sends $D$ to $\mathrm{sgn}(\sigma )D$. Hence the Galois group of a polynomial is contained in the alternating group $A_n$ exactly when every Galois automorphism fixes $D$, which by the fixed-field property is equivalent to $D\in K$. This implies that $\mathrm{disc}(f)=D^2$ is a square in $K$. Conversely, if $\mathrm{disc}(f)=d^2$ for $d\in K$, then $D^2=d^2$, so $D=\pm d\in K$. The Galois group therefore fixes $D$ and consists of even permutations. This proves the discriminant criterion for an alternating Galois group.

Now letThe rational root theorem shows that none of $\pm 1,\pm 2$ is a root, so this cubic is irreducible over $\mathbb{Q}$. The discriminant of a depressed cubic isThis is not a square in $\mathbb{Q}$, so the Galois group of an irreducible cubic is $S_3$ over $\mathbb{Q}$.

Over $K=\mathbb{Q}(\sqrt{-19})$, the discriminant is the squareThe cubic remains irreducible: a root in the degree-two extension $K/\mathbb{Q}$ would generate over $\mathbb{Q}$ both a degree-three field, by irreducibility, and a subfield of a degree-two field, contradicting the tower law for field extensions. Its Galois group over $K$ is consequently $A_3\cong C_3$.

In characteristic two, the formal derivative in positive characteristic ofisA polynomial is separable exactly when it and its derivative have no common root. If $r$ is a root of $f^{\prime }$, thenThus $f$ and $f^{\prime }$ have a common root exactly when $b=0$. Therefore $f$ is a separable polynomial if and only if $b\ne 0$.

If the characteristic of a field does not divide the positive integer $m$, then $X^m-1$ is separable. Its formal derivative is $m{X}^{m-1}$, and a common root would have to be both nonzero and zero.