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Glue two copies of the affine line over a base along the principal open subset $D(t)$ by the identity. This scheme has two copies of the zero section and a canonical morphism of schemes to the original affine line. Over $\mathbb{Z}$, its functor of points assigns to a commutative ring $R$ the pairs $(r,e)$ with $e$ an idempotent in $R/(r)$. Indeed, $r$ gives the map to the affine line, while $e$ partitions $V(r)$ into the two clopen sets assigned to the zero sections. Complementing the opposite part gives two open subsets covering $\mathrm{Spec}R$, whose intersection is $D(r)$; the coordinate $r$ defines maps on them which glue. This also proves that no lifting of $e$ to an idempotent of $R$ is necessary.

Let $A$ be a finite-dimensional basic algebra over an algebraically closed field and a hereditary ring. The radicals of its indecomposable projectives are projective and satisfy $J{P}_i\cong {⨁}_j{P}_j^{a_{ij}}$, where $a_{ij}$ are the Ext quiver arrow multiplicities. Hence $\dim P_i=1+{\sum }_j{a}_{ij}\dim P_j$, forcing strict dimension decrease along arrows and therefore acyclicity. Lifting an arrow basis in $J/J^2$ gives a surjective path algebra homomorphism $kQ\to A$. The number of paths starting at $i$ satisfies the same dimension recursion. Equality of total dimensions proves the homomorphism is an isomorphism. Sums of vertex idempotents over connected components are precisely the primitive central idempotents, so components give the blocks.

A central idempotent is an idempotent in the center of an associative algebra. It gives a decomposition $A=Ae\oplus A(1-e)$ into two-sided ideals, with products between the summands zero. The identities of these summands are $e$ and $1-e$.

For a nonempty inclusion chain of group operations on subsets of a fixed set, the union is a group operation on the union of their carriers. Any finite list of carrier elements lies in one chain member, so its multiplication and associativity agree there. Nested identities coincide because the only idempotent of a group is its identity; inverses remain valid in larger members. Thus the union supplies the chain upper bound required by Zorn's lemma.

Closed paths at vertex $i$ form a basis of $e_{i}kQ{e}_i$. For nonzero finite linear combinations $f,g$, take their largest path lengths $a,b$. In the length-$a+b$ part of $fg$, a concatenated path has a unique cut into lengths $a,b$. Thus a product of nonzero top-degree coefficients cannot cancel, and $fg\ne 0$. The corner is a noncommutative domain and has no idempotents except zero and its identity $e_i$. This argument works with oriented cycles and proves indecomposability of the vertex projective module of a path algebra without falsely assuming that its endomorphism ring is local.

If a nonempty scheme $X$ has no nontrivial idempotent in $\Gamma (X,{\mathcal{O}}_X)$, then its underlying topological space is connected. Indeed, a decomposition into two nonempty clopen subsets would produce the global regular function equal to zero on one component and one on the other. In particular, $\Gamma (X,{\mathcal{O}}_X)$ being a field implies connectedness.

A submodule $N$ of a module $M$ is a direct summand if some submodule $L$ gives $M=N\oplus L$. Equivalently, the inclusion has a left inverse R-module homomorphism, or $N$ is the image of an idempotent endomorphism of $M$.

For a reduced ring with a nonempty disconnected spectrum of a commutative ring, choose a separation $V(I)\bigsqcup V(J)$. Disjointness gives $I+J=R$; covering the spectrum gives $IJ\subseteq \sqrt{(0)}=0$. For comaximal ideals, $I\cap J=IJ$. The Chinese remainder theorem then supplies a product of two nonzero quotient rings. Equivalently, if $i+j=1$ with $i\in I$, $j\in J$, then $i,j$ are nontrivial orthogonal idempotents. This connects topology of the spectrum to ring product decomposition by an idempotent without a Noetherian assumption.

The ring $k\times k$, for a field $k$, has two prime ideals, $0\times k$ and $k\times 0$. Their localizations at a prime ideal are both isomorphic to $k$. Nevertheless $(1,0)(0,1)=0$ is a product of nonzero elements, so $A$ is not an integral domain. Nonzero orthogonal idempotents separate the two components of its spectrum of a commutative ring.

Every nonempty compact Hausdorff left-topological semigroup contains an idempotent. Choose a minimal nonempty compact subsemigroup $K$. For $a\in K$, minimality gives $Ka=K$; the nonempty compact subsemigroup $\{x\in K:xa=a\}$ is therefore all of $K$, so $a^2=a$.

For a finite-dimensional associative algebra, form a graph on its simple modules, joining $S,T$ when ${\mathrm{Ext}}^1(S,T)$ or ${\mathrm{Ext}}^1(T,S)$ is nonzero. Its connected components are exactly the simple modules belonging to each block of a finite-dimensional algebra. They are also the components generated by sharing a Jordan–Hölder factor occurrence in an indecomposable representation that is a projective module.

Indeed, Ext separation of finite-length modules would split the regular module of a block into canonical summands if that block had two graph components. Right multiplication preserves these summands, so its projection supplies a nontrivial central idempotent, contradicting the definition of a block. An indecomposable projective belongs to one block. Finally, a nonsplit extension $0\to U\to V\to W\to 0$ of simple modules is a quotient of the projective cover of $W$: a lift onto $W$ must contain $U$, since otherwise the extension splits. Thus its two endpoints share factors in an indecomposable projective.

Foundations of mathematics asks what our objects mean, what we assume about them, and why an argument establishes its conclusion. Set theory gives a common language for collections and functions; mathematical logic makes the structure of statements and deductions precise. A mathematical definition fixes a meaning, an axiom supplies an assumption, and a mathematical proof explains what follows. Separating these roles makes results reusable: once the group axioms are enough for a conclusion, that conclusion applies to every group, regardless of how its members are represented.

The assumptions can change the answer to a familiar calculation. Over the real numbers, $x^2=x$ factors as $x(x-1)=0$, so $x$ is zero or one. The same reasoning holds in any integral domain, because a product can be zero only when a factor is zero. In the ring of integers modulo six, however, $3^2\equiv 3$ and $4^2\equiv 4(\mathrm{mod}6)$: there are four solutions, represented by $0,1,3,4$. The extra solutions are idempotents, and the failed step involves zero divisors. Writing the domain is part of explaining the answer.

This precision also reveals surprising possibilities. A two-dimensional grid of natural numbers is a countable set, while all infinite binary sequences form an uncountable set. The Cantor pairing function supplies a complete enumeration of the first; Cantor's diagonal argument defeats every proposed enumeration of the second. Both arguments are short, yet distinguish two sizes of infinity. The subject's appeal is this ability to turn a vague question into one whose assumptions, construction, and limits can all be inspected.

Every two-sided ideal $I$ of a finite direct product of rings $T={\prod }_{j=1}^n{R}_j$ is ${\prod }_j{I}_j$, where $I_j$ is its image under the $j$th coordinate ring homomorphism. These images are ideals because the coordinate maps are surjective. To prove the reverse inclusion, lift $a_j\in I_j$ to $b_j\in I$ and multiply by the coordinate idempotent $e_j=(0,\dots ,1,\dots ,0)$. Then ${\sum }_j{e}_j{b}_j=(a_1,\dots ,a_n)\in I$. Finiteness matters: in ${\prod }_{j\ge 0}\mathbb{Z}$, the finitely supported sequences form an ideal whose every coordinate image is $\mathbb{Z}$, but which is not the whole product.

A global idempotent on a scheme is zero or one in every local ring. Its one-locus is both open and closed. Conversely a clopen subset determines a global idempotent by gluing one on it and zero on its complement. Under this correspondence, decompositions into nonzero orthogonal idempotents correspond to disjoint clopen decompositions. When all connected components are open, the primitive idempotents correspond exactly to those components. Without orthogonality the assertion fails in characteristic two: $(1,0)=(1,1)+(0,1)$ in ${\mathbb{F}}_2\times {\mathbb{F}}_2$.

In a finite-dimensional algebra, or a complete finite coefficient-ring algebra, primitive idempotents lift modulo a nilpotent ideal or the coefficient maximal ideal. Orthogonal decompositions of the identity lift, and equivalent primitive idempotents have isomorphic associated projective modules. Two lifts of the same idempotent modulo such an ideal are conjugate by a unit congruent to one. This is an idempotent lifting and refinement result; it does not assert that every central idempotent lifts centrally through an arbitrary nilpotent quotient.

A nonzero module is indecomposable if it cannot be expressed as a direct sum of two nonzero submodules. The zero module is excluded. A module of finite composition length is indecomposable exactly when its endomorphism ring has no idempotents other than zero and one: an idempotent splits the module into its image and kernel.

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.

The Jones-Wenzl idempotent has identity-diagram coefficient one, satisfies $f_n^2=f_n$, and annihilates each adjacent cap-cup generator on either side. With $\delta =-A^2-A^{-2}$ and ${\Delta }_0=1$, ${\Delta }_1=\delta$, ${\Delta }_n=\delta {\Delta }_{n-1}-{\Delta }_{n-2}$, it is constructed recursively by $f_n=F-({\Delta }_{n-2}/{\Delta }_{n-1})F{e}_{n-1}F$, where $F=f_{n-1}\otimes 1$. The required denominators must be nonzero. Its closed bracket trace is ${\Delta }_n=(-1{)}^n[n+1]$.

For an idempotent linear map $P$, $I-P$ is also idempotent andFor the first identity, $P(I-P)=0$ proves one inclusion. If $Pv=0$, then $v=(I-P)v$, proving the reverse inclusion. Apply the same argument to $I-P$ for the second identity. Consequently the ambient vector space is the direct sum of the image and kernel of a linear map of $P$. Orthogonality is not required; it is a separate property of an orthogonal projection.

A compatible triple $(P_1,P_2,\varphi )$ patches to $\{(x_1,x_2):\varphi ({\bar{x}}_1)={\bar{x}}_2\}$. It is finitely generated projective and recovers $P_i$ after extension of scalars. For free modules glued by $g$, the block identity $\mathrm{diag}(g,g^{-1})\in E_{\infty }(A_0)$ lifts through the surjective side and makes the direct sum with inverse gluing free. Complements reduce general projective triples to that free case. Conversely an idempotent over $A$ is exactly a compatible pair of idempotents over its two factors.