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For a field $F$ of characteristic $p$, the exact sequence $0\to {\mathbb{F}}_p\to F^{\mathrm{sep}}\stackrel{z\mapsto z^p-z}{\to }{F}^{\mathrm{sep}}\to 0$, together with additive normal-basis acyclicity, gives $H^2(\mathrm{Gal}(F^{\mathrm{sep}}/F),{\mathbb{F}}_p)=0$. Consequently every central finite embedding problem with group homomorphism kernel $C_p$ has a continuous weak solution: its pulled-back two-cocycle is a coboundary, whose correcting one-cochain supplies the lift. Over ${\mathbb{F}}_p((t))$, independent Artin–Schreier classes $t^{-m}$, with $m>0$ prime to $p$, allow the lift to be twisted by characters so that inertia maps onto the group homomorphism kernel as well as onto the original quotient. Induction through central order-$p$ quotients therefore realizes every finite p-group as the Galois group of a totally ramified extension with residue field ${\mathbb{F}}_p$.

A field can be ordered compatibly with addition and multiplication if and only if it is a formally real field. This ordering criterion concerns real algebra, and is different from Artin–Schreier theory of cyclic extensions in positive characteristic.

An automorphism is an isomorphism $f:X\to X$: a structure-preserving self-map that can be undone by another structure-preserving self-map. Equivalently, it is an invertible endomorphism. Its inverse satisfieswhere $1_X$ is the identity morphism. Both maps must be morphisms in the chosen category. For ordinary groups, rings, and vector spaces over a fixed field, a bijective morphism automatically has a compatible inverse. For topological spaces, an automorphism is a homeomorphism from the space to itself: a continuous bijection needs a continuous inverse as well. See Riehl, Category Theory in Context, Definition 1.1.10 and Example 1.1.11.

For example, $T(x,y)=(x+y,y)$ is an automorphism of the vector space ${\mathbb{R}}^2$ over the real numbers, with inverse $T^{-1}(u,v)=(u-v,v)$. Both are linear maps. The linear endomorphism $P(x,y)=(x,0)$ from endomorphism preserves addition and scalar multiplication too, but discards a coordinate and cannot be undone. Preserving the structure and preserving enough information for reversal are separate requirements.

The automorphisms of a fixed object in a locally small category form a group under composition in a category: the identity morphism is the identity element, and the inverse element of a composite reverses the order of its factors. This makes reversible internal changes calculable. For a group, the resulting automorphism group records which relabelings preserve all products; the examples in group automorphism show how restrictive that condition can be.

The bar resolution of an associative algebra $A$ over a field has degree-$n$ term $A^{\otimes (n+2)}$ as an $A^e$-module, with the outer factors carrying the left and right actions. Its boundary multiplies each consecutive pair with alternating signs, and its augmentation is multiplication $A\otimes A\to A$. It is exact: the $k$-linear contracting homotopy inserts $1$ at the left. Applying ${\mathrm{Hom}}_{A^e}(-,M)$ gives the Hochschild cochain complex of the bimodule $M$.

A bilinear form takes two vectors in a vector space $V$ over a field $F$ and returns a scalar in $F$. Its defining property is separate linearity: holding either input fixed gives a linear functional of the other. For scalars $a,b$ and vectors $u,v,w$,This is a scalar-valued bilinear map; the chosen field is part of the definition.^[ref]

Separate linearity does not mean that $B$ is a linear map on the vector space $V\times V$ with componentwise vector addition and scalar multiplication. Over the real numbers, $B(2v,2w)=4B(v,w)$, whereas a linear map on the pair would give $2B(v,w)$. The simplest example is $B(x,y)=xy$ on the one-dimensional vector space $\mathbb{R}$: fixing $y=3$ gives $x\mapsto 3x$, but doubling both $x$ and $y$ quadruples their product.

A bilinear form measures how two inputs interact. The dot product measures alignment; a symmetric bilinear form can describe a quadratic form; an alternating bilinear form can measure signed area. A nondegenerate bilinear form turns each vector into a unique linear functional and so connects geometry with the dual space. The same short definition therefore supports lengths, bilinear orthogonal complements, and symplectic vector spaces, depending on the additional properties imposed.

Over the complex numbers, a bilinear form is linear in both inputs without conjugating either scalar. A Hermitian form instead is a sesquilinear form and conjugates when its inputs are exchanged. For complex matrices, the adjoint $A^{\ast }$ used with Hermitian forms and matrix C-star algebras is the conjugate transpose, whereas a bilinear form uses the ordinary transpose $A^T$. For example, $B(z,w)=zw$ has $B(i,i)=-1$, while the Hermitian form $H(z,w)=z\overline{w}$ has $H(i,i)=1$.^[ref]

Over a field of characteristic three, the group algebra $A=k{S}_3$ has a single block of an Artinian algebra. Put $r=(123)$, $s=(12)$ and $a=r-1$. Then $a^3=0$, $sas=-a+a^2$, and $J(A)=aA$: the latter is a nilpotent ideal, with semisimple ring quotient $k{C}_2$. The center of an associative algebra is $k1\oplus k{a}^2\oplus k{a}^{2}s$, whose last two summands form a square-zero ideal. A central idempotent $d1+n$ satisfies $d^2=d$ and $(2d-1)n=0$, hence is zero or one. Thus no nontrivial central block decomposition exists.

For a point blowup of a smooth algebraic surface $\psi :\tilde{X}\to X$, where $X$ is a smooth projective surface over an algebraically closed field, multiplication by the exceptional section to power $m$ identifies $H^0(\tilde{X},m{\psi }^{\ast }{K}_X)$ with $H^0(\tilde{X},m{K}_{\tilde{X}})$. Every effective pluricanonical representative must contain $mE$, since after subtracting $j<m$ copies its intersection with $E$ is $-m+j<0$. The projection formula for sheaves and ${\psi }_{\ast }\mathcal{O}=\mathcal{O}$ identify the remaining sections with those on $X$. No effectivity of $K_X$ itself is assumed.

Over an algebraically closed field, the blowup of a smooth algebraic surface at a closed point replaces the point by an exceptional smooth rational curve $E\cong {\mathbb{P}}^1$, with $E^2=-1$. Locally it is the blowup of the affine plane at the origin. It is a proper birational morphism, an isomorphism off the exceptional curve, and the blown-up surface is again smooth and projective when the original is projective.

For a point blowup of a smooth algebraic surface $\psi :\tilde{X}\to X$ over an algebraically closed field, with exceptional curve $E$, $K_{\tilde{X}}={\psi }^{\ast }{K}_X+E$ after choosing compatible representatives. In local coordinates $y=xu$, the differential $dx\wedge dy=xdx\wedge du$ has one additional zero along $E$. Thus the formula is valid in every characteristic.

The center of an indecomposable finite-dimensional block of a group algebra is a commutative Artinian local ring. Its quotient by its Jacobson radical is a field, giving the block's central character. Over a splitting field for finite group representations the residue field is the coefficient field; without this assumption one retains the residue field rather than pretending every central character is scalar-valued over $k$.

A central simple algebra over a field $k$ is a finite-dimensional associative $k$-algebra that is a simple ring and whose center is exactly $k$.

The characteristic of a field $F$ is the least positive integer $n$ for which $n\cdot 1_F=0$, or zero if no such integer exists.

In a field of characteristic two, $1+1=0$. The usual splitting of a tensor product into its symmetric square and exterior square by $(I\pm \tau )/2$ is unavailable.

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.

An elliptic curve with complex multiplication is called a CM elliptic curve. Distinguish geometric endomorphisms from endomorphisms defined over the base field. In characteristic zero, adjoining the imaginary quadratic field selected by the tangent action makes all geometric endomorphisms defined over the base. Good split primes have ordinary reduction; good inert primes have supersingular reduction.

If $k$ is a field and $H^{\ast }(F;k)$ has finite total dimension, the cohomology cross product gives the displayed graded-ring isomorphism for arbitrary spaces. Over a field, every chain complex splits into its homology with zero differential and contractible summands. The Eilenberg–Zilber theorem thus replaces the fiber chain complex by a bounded finite-dimensional complex; dualizing its tensor product with $C_{\ast }(X;k)$ gives a finite direct sum of shifted cochain complexes of $X$. Its cohomology is the displayed tensor product. Finiteness prevents the infinite-product obstruction that can occur for two infinite discrete spaces. The standard product formula and its finiteness hypothesis are discussed in pi.math.cornell.edu/~hatcher/AT/ATch3.pdf, Theorem 3.15.

A commutative ring is a ring in which multiplication satisfies $ab=ba$ for every pair. The integers and the polynomial ring $\mathbb{Z}[X]$ are examples. So is $\mathbb{Z}/n\mathbb{Z}$, the quotient ring whose elements are residue classes in modular arithmetic. We represent these by $0,1,\dots ,n-1$ and reduce each sum or product to its remainder. All following calculations modulo $6$ use this convention.

In $\mathbb{Z}/6\mathbb{Z}$, $2\cdot 3=0$, although neither factor is zero. Thus $2$ and $3$ are zero divisors: nonzero elements whose product with another nonzero element can vanish. Also $2\cdot 1=2\cdot 4=2$. Removing the common factor $2$ would falsely assert $1=4$. The entire row for multiplication by $2$ isThe repetitions explain the linear congruences directly. The equation $2x=2$ has exactly the two solutions $x=1,4$, whereas $2x=1$ has none. A common factor can erase information, and a nonzero coefficient does not guarantee that an equation has a solution.

A unit in a ring is an element $u$ with a multiplicative inverse, meaning $uv=vu=1$ for some $v$. In $\mathbb{Z}/6\mathbb{Z}$, the units are exactly $1$ and $5$, since $5^2=1$. The other nonzero elements, $2,3,4$, are zero divisors; for $4$, use $4\cdot 3=0$. In a nonzero commutative ring, a unit cannot be a zero divisor: multiplying $ub=0$ by its multiplicative inverse forces $b=0$. But a nonzero element need not be a unit or a zero divisor: the integer $2$ is neither in $\mathbb{Z}$.

By contrast, in $\mathbb{Z}/5\mathbb{Z}$, every row for a nonzero multiplier is a permutation of $0,1,2,3,4$. Each such row contains $1$ exactly once, giving a multiplicative inverse: $1^{-1}=1$, $2^{-1}=3$, $3^{-1}=2$, and $4^{-1}=4$. Hence this commutative ring is a field. For example, $2x=1$ now has the unique solution $x=3$. Reading the tables reveals which algebraic steps the arithmetic permits.

The commutator of two elements of an associative algebra is $[A,B]=AB-BA$. Matrix Lie algebras use the commutator as their Lie bracket.