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A Hahn series over a field $k$ with exponents in an ordered abelian group $\Gamma$ is a formal sum ${\sum }_{\gamma }{a}_{\gamma }{t}^{\gamma }$ whose nonzero exponents form a well-ordered set. Such series form the Hahn series field.

For a projective scheme $X$ over a field $k$, the Hilbert functor assigns to a $k$-scheme $T$ the set of closed subschemes $W\subseteq X{\times }_{k}T$ which are of finite presentation, proper and flat over $T$. Pullback of families makes it a contravariant functor. After choosing a very ample line bundle, requiring a fixed fibre Hilbert polynomial $P$ gives the subfunctor represented by ${\mathrm{Hilb}}^P(X)$. The unrestricted functor is represented by the disjoint union of these Hilbert schemes.

A field $K$ is Hilbertian if each finite family of polynomials irreducible over $K(T)$ and separable in its root variable remains irreducible of the same degree at infinitely many $T=a\in K$, after any finitely many prescribed bad parameter values are excluded. Equivalently, every finite separable extension of $K(T)$ has infinitely many unramified $K$-rational places whose residue extension retains its full degree.

For a finitely generated graded module over a standard graded algebra over a field, with finite-dimensional graded pieces, the Hilbert function $h(n)=\dim M_n$ is eventually a polynomial in $n$. Its generating series is the Hilbert series. Thus the polynomial describes eventual graded growth, while the series records all graded pieces.

The Hochschild cohomological dimension of an associative algebra over a field is $\mathrm{Dim}(A)={\mathrm{pd}}_{A\otimes A^{\mathrm{op}}}A$. Equivalently, $H{H}^n(A,M)$ vanishes above that dimension for every bimodule $M$. A bound by one implies $H{H}^2(A,A)=0$ and hence formal rigidity from vanishing second Hochschild cohomology.

For an endomorphism of a finite-dimensional bounded chain complex over a field, the alternating chain trace equals the alternating trace on homology. Split each chain space into boundaries, representatives for homology, and a complement mapping isomorphically to the preceding boundaries. The chain-map relation makes the traces on the first and third summands cancel in successive degrees, leaving the homology traces. Over a field the induced cohomology maps are dual and have the same traces. This identity turns the absence of diagonal simplex contributions into a vanishing Lefschetz number.

Let $A$ be a commutative algebra over a field $k$, with a homomorphism $A\to k$ restricting to the identity on $k$, and let $I$ be its kernel. If $A$ is generated by finitely many elements, subtracting their images in $k$ gives finitely many generators in $I$. Every polynomial in those generators, modulo $I^2$, is a linear combination of their classes. Thus $I/I^2$, the algebraic cotangent space at this point, is finite-dimensional. For $A=k+xk[x,y]$ and $I=xk[x,y]$, the classes $x{y}^j$, $j\ge 0$, form an infinite basis modulo $I^2=x^{2}k[x,y]$. This is a subalgebra of a polynomial ring which is not a finitely generated algebra.

For a prime number $p$, a nonprincipal ultrafilter on $\mathbb{N}$ givesThe Łoś theorem makes $K$ a field of characteristic $p$. For every natural number $m$, almost all factors have at least $m$ elements, so $K$ is infinite.

Integral cohomology is cohomology with integer coefficients, $H^{\ast }(X;\mathbb{Z})$. Its cup product makes it a graded cohomology ring, retaining torsion and integral information that can disappear after changing coefficients to a field.

An integral domain is a nonzero commutative ring without zero divisors. Equivalently, a product of two nonzero elements is always nonzero. The zero ring is excluded, and $\mathbb{Z}/6\mathbb{Z}$ fails because $2\cdot 3=0$. Every field is an integral domain, but the integers show that the converse fails: $2$ has no multiplicative inverse in $\mathbb{Z}$. See MIT's Basic Properties of Rings, definitions 15.9–15.10 and example 15.11.

The practical benefit is multiplicative cancellation. In the integers, $2x=6$ has at most one solution and indeed $x=3$ works. Yet $2x=1$ has no integer solution. Thus multiplicative cancellation guarantees uniqueness when a solution exists; it does not guarantee division by every nonzero element. Passing to the fraction field $\mathbb{Q}$ supplies the missing solution $1/2$.

The polynomial ring $\mathbb{Z}[X]$ is also an integral domain. For example, the product of $2X+1$ and $3X-4$ has leading coefficient $6$ and cannot vanish. Generally the leading coefficient of a product is the product of the two leading coefficients; an integral domain prevents that coefficient from vanishing. This connects integer arithmetic to polynomial arithmetic through one reusable principle.

Historically, this connection grew far beyond elementary examples. Keith Conrad identifies the ring of integers of a number field and a ring of polynomial functions on a smooth plane curve as early important examples of Dedekind domains. This shared structure connects number theory with algebraic geometry; see Conrad's Ideal Factorization, section 6.

If a field $L$ is integral over a subdomain $A$, every nonzero $a\in A$ has an inverse in $L$ satisfying a monic equation over $A$. Multiplying that equation by $a^{r-1}$ expresses $a^{-1}$ as a polynomial in $a$ with coefficients in $A$. Hence $a^{-1}\in A$, and $A$ is itself a field. The embedding and integrality assumptions are essential; a fraction field is not generally integral over its source domain.

The inverse Galois problem asks which finite groups occur as Galois groups over a given field, with the classical problem taking the base field to be $\mathbb{Q}$. A regular realization over $K(T)$ and Galois-preserving Hilbert specialization solve an instance over a Hilbertian field. Realization over an algebraically closed constant field's rational function field alone does not supply descent to $\mathbb{Q}$.

An $n$-by-$n$ square matrix $A$ over a field is invertible when it has a two-sided matrix inverse. Equivalently, its linear map $x\mapsto Ax$ is a bijection: every output comes from exactly one input. For a square matrix, the following conditions are equivalent:Here matrix rank $n$ means both full column rank and full row rank. The square hypothesis matters: a rectangular matrix can have full column rank without reaching every output, or full row rank without determining every input.^[ref]

The equivalences explain why invertibility is a question about retaining information. A trivial kernel makes the linear map injective; the rank-nullity theorem then gives matrix rank $n$, so its columns form a basis and the linear map is also surjective. Assigning to each output its unique input gives a linear map: the preimage of $\alpha b+\beta c$ is $\alpha A^{-1}b+\beta A^{-1}c$. The determinant is nonzero precisely when the columns are linearly independent. Thus algebraic solvability and independent directions describe the same phenomenon; over the real numbers, nonzero oriented volume supplies its geometric interpretation.

An irreducible group representation is nonzero and has no nonzero proper invariant subspace. These are the pieces that cannot be decomposed further by taking a smaller symmetry-preserved vector space.

The real two-dimensional rotation through $120$ degrees illustrates the role of the coefficient field. A one-dimensional real invariant subspace would supply a real eigenvector, but the characteristic polynomial ${\lambda }^2+\lambda +1$ has no real root. The representation is therefore irreducible over the real numbers. Over the complex numbers, that polynomial has distinct roots of unity $e^{2\pi i/3}$ and $e^{-2\pi i/3}$, giving two invariant eigenlines. Irreducibility over the real numbers does not imply absolute irreducibility.

Stable invertible matrices, with stabilization $g\mapsto \mathrm{diag}(g,1)$, form the stable general linear group. Its elementary subgroup is its commutator subgroup, and the displayed quotient is its abelianization. Elementary matrices lift along any surjective ring map, even when arbitrary invertible matrices do not lift. Over a field, determinant identifies $K_1$ with its multiplicative unit group.

The Koszul complex on $f_1,\dots ,f_r$ has term $K_i={\bigwedge }^i{R}^r$ and differential $\partial (e_{j_1}\wedge \cdots \wedge e_{j_i})={\sum }_a(-1{)}^{a-1}{f}_{j_a}{e}_{j_1}\wedge \cdots \wedge \widehat{e_{j_a}}\wedge \cdots \wedge e_{j_i}$. Its degree-zero homology is $R/(f_1,\dots ,f_r)$. For a regular sequence it is exact in positive degrees, hence a Koszul resolution of that quotient. The variables of a polynomial ring form a regular sequence, so this resolves the coefficient field in length equal to the number of variables.