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For a degree-$n$ polynomial $p(X)=a_n{\prod }_{i=1}^n(X-r_i)$ over a field, with $a_n\ne 0$ and roots in an algebraic closure, its discriminant isIt is a polynomial with integer coefficients in the coefficients of $p$, and lies in the original coefficient field. Its vanishing is equivalent to a repeated root of a polynomial in an algebraic closure, even when the coefficient field itself contains no roots. This detects separability without solving the polynomial equation.^[ref]

For a quadratic polynomial, the expression is the quadratic discriminant $b^2-4ac$. For the cubic polynomial $X^3+bX+c$, it is $-4b^3-27c^2$. Thus $X^3-2$ has discriminant $-108$ over $\mathbb{Q}$, but after reduction to characteristic three its discriminant is zero and the resulting polynomial has a repeated root in its splitting field. Discriminants also indicate exceptional parameter values where distinct polynomial roots collide.

Let $n\ge 2$, let the characteristic of a field $K$ not divide $n$, and suppose $K$ contains all $n$th roots of unity. For $a\in K^{\times }$, the polynomial $X^n-a$ is irreducible exactly when $a$ is not a $q$th power in $K$ for every prime divisor $q$ of $n$. Equivalently, the class of $a$ has order $n$ in $K^{\times }/(K^{\times }{)}^n$. Kummer theory then gives a cyclic extension of degree $n$.^[ref]

A useful sufficient test is a discrete valuation with $\gcd (n,v(a))=1$: if $a$ were a $q$th power for $q\mid n$, its valuation would be divisible by $q$. The denominator criterion for Newton polygon irreducibility gives the same conclusion directly.

Without the roots-of-unity hypothesis there is an additional fourth-power obstruction. For example, $X^4+4=(X^2-2X+2)(X^2+2X+2)$ over $\mathbb{Q}$, even though $-4$ is not a rational square. The precise hypotheses replace an unspecified appeal to valuation data or an unspecified standard exception.

Let $n$ be prime to $\mathrm{char}K$ and suppose ${\mu }_n\subset K$. For a finite Galois extension $L/K$, the multiplicative Kummer pairing has domainand sends $(\sigma ,[a])$ to $\sigma (\alpha )/\alpha$, where $\alpha \in L$ and ${\alpha }^n=a$. The restriction to $V_L$ ensures that the chosen root actually lies in the domain of $\sigma$. Changing that root or the representative of $[a]$ leaves the value unchanged; multiplication in either argument multiplies the values. If $L$ is generated by these roots, this is the perfect duality of finite Kummer theory.^[ref]

For an elliptic curve, choose $Q$ with $nQ=P$. The analogous expression is $\sigma Q-Q\in E[n]$. In general it is a Galois cohomology cocycle, satisfying $c(\sigma \tau )=c(\sigma )+\sigma c(\tau )$, rather than a group homomorphism. If all $n$-torsion is rational, the action on $E[n]$ is trivial and it becomes a homomorphism. This is the connecting construction in the Kummer map of an elliptic curve.

Let $n\ge 1$ be prime to the characteristic of a field $K$, and suppose $K$ contains all $n$th roots of unity. Kummer theory classifies finite abelian extensions of exponent dividing $n$ by finite subgroups of $K^{\times }/(K^{\times }{)}^n$: to a subgroup generated by classes $a_i$ associate $K(a_i^{1/n}:i)$. The pairing sending an automorphism and a radicand class to $\sigma (a^{1/n})/a^{1/n}$ identifies the Galois group with the character group of the radicand subgroup.

In particular, cyclic extensions of degree $n$ correspond to cyclic subgroups of order $n$, rather than to a unique choice of one generator. The cohomological form isFor $n=2$ in characteristic different from two, the nonzero square classes other than the identity classify quadratic extensions. The assumptions on roots of unity and characteristic ensure that the radicals form separable Galois extensions.^[ref]

Polynomial division separates a polynomial into a multiple of a chosen divisor and a smaller polynomial remainder. Given $f,g\in F[X]$ over a field $F$, with $g\ne 0$, there are unique polynomials $q,r$ satisfyingHere $q$ is the polynomial quotient. The bound concerns polynomial degree, rather than the sizes of the coefficients.^[ref]

For an original example over the rational numbers, takePolynomial long division first cancels $2X^3$ by subtracting $2Xg$:Subtracting one more copy of $g$ cancels $X^2$:ThusThe final polynomial remainder has polynomial degree one, below the divisor's polynomial degree two. Multiplying out the last identity checks every coefficient, including the constant term $-2+8=6$.

The polynomial degree bound makes this decomposition useful: it replaces a potentially long polynomial with a short polynomial remainder whenever multiples of $g$ can be discarded. At any root of a polynomial of $g$, those multiples vanish under polynomial evaluation. Repeated polynomial division also drives the Euclidean algorithm for finding a greatest common divisor of polynomials.

The coefficient setting matters. Over a commutative ring, the same existence and uniqueness hold when the divisor's leading coefficient is a unit in a ring. A monic polynomial therefore works over the integers, even though they do not form a field; monic polynomial division over a ring explains why zero divisors cause no difficulty in that case.^[ref]

A nonzero divisor alone is insufficient over the integers. Dividing $X$ by $2X+1$ with an integer polynomial quotient and a constant polynomial remainder would require $1=2c$ for an integer $c$, which is impossible. Over the rational numbers the division instead givesThe obstruction is the nonunit leading coefficient $2$. This example separates existence from uniqueness: over an integral domain, the polynomial degree argument still makes any admissible pair unique, if it exists.

Dividing a polynomial $f$ by $X-a$ leaves the constant polynomial remainder $f(a)$. Polynomial division makes the polynomial remainder constant because its polynomial degree must be below one; polynomial evaluation at $a$ identifies that constant. In particular, it is zero exactly when $a$ is a root of a polynomial of $f$, which is the factor theorem.^[ref]

In the worked example, division by $X-1$ leaves $6$ and division by $X+2$ leaves $12$. Division by their product leaves $-2X+8$, a polynomial remainder that stores both values at once. The argument also holds over a commutative ring for $a$ in that ring, because $X-a$ is a monic polynomial.

A profinite group is an inverse limit of finite discrete groups. For a directed compatible system, its elements are tuples $(g_i)$ whose coordinates agree under all transition homomorphisms. Multiplication and inversion operate coordinatewise. The inverse-limit topology is inherited from the product of the finite discrete spaces.

The product is compact and Hausdorff, and the compatibility equations define a closed subset. Conditions on finitely many coordinates give clopen neighbourhoods; in particular kernels of finite quotient projections form an identity-neighbourhood basis. Conversely, every compact Hausdorff totally disconnected topological group is recovered from its finite quotients by open normal subgroups.

For example, the additive group of p-adic integers is ${\mathbb{Z}}_p={\underleftarrow{\lim }}_n\mathbb{Z}/p^n\mathbb{Z}$. Compatibility says that a residue modulo $p^{n+1}$ reduces to the preceding residue modulo $p^n$. The ordinary integers map densely into this group. Compactness belongs to the completed inverse limit; the profinite topology on the integers is not compact. ^[ref]

A single topological generator of a topological group $G$ is an element $g$ whose cyclic subgroup is dense:More generally, a tuple $g_1,\dots ,g_d$ consists of topological generators when the subgroup they jointly generate is dense. Membership in such a tuple does not imply that one element alone generates densely.

In the additive group ${\mathbb{Z}}_p$ of p-adic integers, the element one is a topological generator: ordinary integers surject onto every finite quotient $\mathbb{Z}/p^n\mathbb{Z}$. The element $p$ generates only the proper closed subgroup $p{\mathbb{Z}}_p$. The profinite group ${\mathbb{Z}}_p^2$ has a two-element topological generating set but no single topological generator, since its quotient $(\mathbb{Z}/p\mathbb{Z}{)}^2$ is not cyclic. ^[ref]