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Algebraic topology translates questions about topological spaces and continuous maps into questions about groups and other structures from algebra. A topological invariant compresses the enormous amount of data in a space into something one can calculate. For example, the fundamental group records which based loops can be continuously deformed into one another. A filled disk has trivial fundamental group, whereas a circle has fundamental group $\mathbb{Z}$: an integer remembers how many times a loop goes around.

This compression also respects maps. By functoriality of the fundamental group, a continuous map preserving the chosen basepoints induces a group homomorphism; successive maps induce successive group homomorphisms. Thus geometry supplies arithmetic constraints. A proposed homotopy equivalence must induce an isomorphism, so incompatible fundamental groups rule it out before anyone attempts to construct the deformation.

The beauty is that a whole annulus can retain exactly the same homotopy type as its inner circle. Its thickness disappears under a deformation retraction, while the obstruction to shrinking a loop survives. Equal fundamental groups alone do not establish a homeomorphism or a homotopy equivalence between arbitrary topological spaces.

This turns a visual question into a reusable test. Suppose a continuous map sends the inner circle of an annulus once around another circle. Could it extend over a filled disk spanning that boundary? Such an extension would supply a homotopy shrinking the image loop to a point. Its winding number would have to change from one to zero, which homotopy invariance of winding number forbids. Algebra rules out the extension even when no particular formula for it has been proposed.

An algebra isomorphism is a bijective linear map preserving multiplication and identity between two algebras over the same field. In classification of a square-zero extension of an algebra, an equivalence must additionally induce the identity on the prescribed quotient and kernel.

Al-Khwarizmi wrote on algebra and treated quadratic equations using both algebraic procedures and geometry.

An almost commutative algebra over a field is an algebra with an exhaustive increasing nonnegative multiplicative filtration whose associated graded ring is commutative and finitely generated as an algebra over the field. A common presentation uses finitely many degree-one generators, scalar degree-zero part, and commutators that lower degree. The finite-generation requirement is essential: an infinitely generated commutative polynomial ring has commutative associated graded algebra but need not be Noetherian.

Original-edition title and metadata from New York Public Library. Used as evidence for the term universal arithmetic; the title alone does not define every contemporary meaning of algebra.

Online source: old-digitalcollections.nypl.org/collections/arithmetica-universalis

For a finite-dimensional algebra, the Auslander–Reiten quiver has one vertex for each isomorphism class of finite-dimensional indecomposable modules, and its arrows record irreducible module homomorphisms. A homomorphism is irreducible when it is neither a split injection nor a split surjection and every factorization forces the first map to split injectively or the second to split surjectively.

A block of an associative algebra with a finite decomposition of its identity is a two-sided direct summand $Ae$ corresponding to a primitive central idempotent $e$. Its identity is $e$, and it cannot be split into two nonzero algebras by central idempotents. For a group algebra of a finite group, this agrees with a block of a group algebra. Orthogonal primitive central idempotents give $A={⨁}_{i}A{e}_i$.

Cauchy's functional equation asks for functions $f:\mathbb{R}\to \mathbb{R}$ satisfyingThe identity must hold for every pair of real numbers, and a solution is an additive function. It says that combining two inputs before applying $f$ gives the same result as combining their outputs.^[ref]

For example, $f(x)=3x$ satisfies the identity because $3(x+y)=3x+3y$. At $x=2/3$ and $y=5/6$ the two sides giveFor positive widths, a rectangle of height three makes this visible: cutting its width into $x$ and $y$ partitions its area into $3x$ and $3y$. The same identity also holds for negative inputs, where $3x$ is a signed output rather than an ordinary area.

Every function $f(x)=cx$ with a fixed real number $c$ solves Cauchy's functional equation. The surprising question is whether these are all the solutions. Rational homogeneity of an additive function forces $f(q)=qf(1)$ for every rational number $q$, but reaching an irrational number requires an additional argument. Continuity at just one point or being a bounded function on an interval of positive length supplies that argument; without such a condition, discontinuous additive functions exist.^[ref]

This makes Cauchy's functional equation a useful test of the boundary between algebra and analysis: a rule about addition fixes all rational multiples, while a tiny amount of continuity fixes the entire real line. For example, the continuous exponential functional equation reduces to an additive function after taking a logarithm, so this result explains why its nonzero solutions are exponential functions.

Let $k$ be an algebraically closed field and let a block of a group algebra $B$ have a central defect group of a block $D$. Its quotient by $J(kD)B$ is a defect-zero block, hence a full matrix algebra over $k$. The algebra $B$ is free over the central local ring $kD$. Lift a primitive idempotent from this matrix quotient; its corner is free of rank one over $kD$, and is therefore $kD$. Lifting matrix units gives $B\cong {\mathrm{Mat}}_r(kD)$.

For an associative algebra $A$ over a field $k$, a derivation is a $k$-linear map $D:A\to A$ satisfying $D(ab)=D(a)b+aD(b)$. It is the case $M=A$ of a derivation into a bimodule; the order of factors matters when $A$ is noncommutative. Substituting $a=b=1$ gives $D(1)=0$. Such maps form a Lie algebra under the commutator $[D,E]=D\circ E-E\circ D$.

More generally, for a commutative ring $R$, a derivation from a commutative $R$-algebra $B$ into a $B$-module $M$ is an $R$-linear map $D:B\to M$ with $D(r)=0$ for $r\in R$ and $D(b{b}^{\prime })=bD(b^{\prime })+b^{\prime }D(b)$. The second term has this order because the algebra is commutative. Derivations of smooth functions are vector fields provide a geometric example.

A division algebra over a field $k$ is a division ring equipped with a unital algebra structure over $k$, with $k$ mapping into its center of an associative algebra. Every nonzero element is invertible; it need not be finite-dimensional over $k$.

A unital subalgebra $A\subseteq \mathrm{End}(V)$ closed under the adjoint operator, on a finite-dimensional real or complex inner product space, equals its double commutant of an operator algebra. Here is a proof that does not require a classification of algebras. For $N=\dim V$, choose a basis $(v_1,\dots ,v_N)$ and let $W=\{(a{v}_1,\dots ,a{v}_N):a\in A\}\subseteq V^N$. The diagonal action of $A$ preserves $W$ and $W^{\perp }$, so the orthogonal projection onto $W$ commutes with that action. Each matrix entry of this projection belongs to $A^{\prime }$. Hence the diagonal action of $b\in A^{\prime \prime }$ commutes with the projection and preserves $W$. Since $1\in A$, the tuple $(v_1,\dots ,v_N)$ belongs to $W$, and therefore $(b{v}_1,\dots ,b{v}_N)=(a{v}_1,\dots ,a{v}_N)$ for one $a\in A$. This implies $b=a$. Without the unital hypothesis the conclusion is $A^{\prime \prime }=A+\mathbb{F}1$.

An equation asserts that two mathematical expressions have the same value. Solving it means finding all allowed assignments to its variables that make that assertion true. It converts a relationship into a constraint: a measurement, a geometric intersection, or an equilibrium model can leave an unknown value to determine.

The allowed values matter. The equation $2x=1$ has solution set $\{1/2\}$ over the rational numbers, but the empty set over the integers. Over the real numbers, $x^2=1$ has two solutions, $-1$ and $1$. A mathematical identity, such as $(x+1{)}^2=x^2+2x+1$, holds for every permitted real $x$. The familiar equality sign therefore does not by itself say whether one is checking a particular value, solving for an unknown, or stating an identity.

For a worked linear equation over the real numbers,The first step subtracts three from both sides; the second divides both sides by the nonzero number two. Reversing the steps recovers the original equation, so these are equivalent equations with the same solution set $\{3\}$. Checking the value gives $2\cdot 3+3=9$. The role of reversible operations and the classification by solution sets are described in College Algebra 2e, §2.2: Solving Linear Equations in One Variable.

When the two sides define real-valued functions $f$ and $g$ on a common domain of a function, solutions are exactly the input values at which their function graphs intersect. Equivalently, they are the zeros of a function $f-g$. This gives a useful connection between symbolic algebra and geometry: rearranging an equation can simplify the comparison without changing the inputs where equality holds.

Not every operation is reversible on every domain of a function. Dividing $x(x-1)=0$ by $x$ gives $x-1=0$ only after restricting to $x\ne 0$; it loses the solution $x=0$ if that restriction is forgotten. Squaring both sides can instead introduce an extraneous solution, because distinct real numbers can have the same square. The source equation, its domain of a function, and the direction of each implication should travel together.

Two equations are equivalent on a specified domain of a function if they have the same solution set. Adding the same defined mathematical expression to both sides is reversible. Multiplying or dividing both sides by a mathematical expression is reversible only where that expression is nonzero and invertible in the arithmetic being used. In a field, every nonzero scalar is invertible; a general ring needs an additional check.

For example, $x^2=1$ and $(x-1)(x+1)=0$ are equivalent equations over the real numbers, by the distributive property. Concluding $x=1$ or $x=-1$ additionally uses the absence of zero divisors. The modular example in algebra shows why that extra inference can fail in a ring with zero divisors.

A finite-dimensional algebra has finite representation type when there are only finitely many isomorphism classes of finite-dimensional indecomposable modules. This counts module types, not the dimensions or the number of direct sums formed from them. For a block of a group algebra, finite representation type is equivalent to a cyclic defect group of a block.

A formal associative deformation is a unital $k[[t]]$-bilinear product continuous for the adic topology on $A[[t]]$ of the form $a\ast b=ab+{\sum }_{r\ge 1}{t}^r{\mu }_r(a,b)$. Each coefficient ${\mu }_r$ is a $k$-bilinear map $A\otimes A\to A$, and the whole product satisfies associativity. For an infinite-dimensional algebra, $A[[t]]$ is the completed tensor product ${\underleftarrow{\lim }}_{n}A\otimes k[t]/(t^n)$ and generally differs from the ordinary $A\otimes k[[t]]$.

In characteristic $p>0$, the displayed map on the cocenter of a finite-dimensional algebra is additive and Frobenius-semilinear. Expand a $p$th power into words: nonconstant cyclic-word orbits have length $p$ and cancel modulo commutators. Also $(ab{)}^p$ and $(ba{)}^p$ have the same cocenter class, so the map kills additive commutators and is well defined.

Over a perfect field with split semisimple quotient, the stable image of this map has dimension equal to the number of simple modules of the algebra. Indeed, the kernel of $C(A)\to C(A/J(A))$ has representatives in the nilpotent Jacobson radical, so an iterate kills it. On the semisimple algebra quotient, each matrix-block coordinate is a field trace, and $\mathrm{tr}(a^p)=\mathrm{tr}(a{)}^p$. For a split semisimple quotient, Frobenius is bijective on every coordinate. In a group algebra over an algebraically closed field, the stable image has the p-regular element conjugacy classes as its basis, proving the simple-module counting theorem.

For a nonzero finitely generated algebra over a field, choose a finite-dimensional generating vector subspace $V$ containing $1$. The dimension is $\mathrm{GKdim}A={\mathrm{limsup}}_{n\to \infty }\log ({\dim }_k{V}^n)/\log n$. Changing $V$ only rescales the filtration index by bounded factors. A nonzero finite-dimensional algebra has dimension zero; a polynomial ring in $r$ variables has dimension $r$. The zero algebra is often assigned dimension $-\infty$.

For a nonzero right module $M$ that is a finitely generated module, choose a finite-dimensional generating vector subspace $W$ and define ${\mathrm{GKdim}}_{A}M={\mathrm{limsup}}_{n\to \infty }\log ({\dim }_{k}W{V}^n)/\log n$, where $V$ generates the algebra and contains $1$. For a left module use $V^{n}W$. It is independent of these choices and measures polynomial-order growth of a generating filtration of a module.

With $s=(12)$, the idempotents $e_{\pm }=(1\pm s)/2$ split the right regular module into two three-dimensional projective modules. Their tops are respectively the trivial representation and sign representation. The nilpotent ideal $J(k{S}_3)$ ensures that any nonzero direct summand has nonzero top, so the one-dimensional tops make these projective modules indecomposable. Their successive radical series of a module factors are trivial, sign, trivial and sign, trivial, sign. This decomposition concerns right modules; the block of S3 in characteristic three remains a single algebra.