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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.

An affine scheme is a scheme isomorphic to $\mathrm{Spec}A$ for some commutative ring $A$. Homomorphisms $A\to B$ correspond contravariantly to morphisms $\mathrm{Spec}B\to \mathrm{Spec}A$.

Restricting the functor of points to affine schemes gives a fully faithful functor from schemes to covariant set-valued functors on commutative rings. Given a natural transformation, apply it to the inclusions of an affine open cover of $X$. Naturality on affine open subschemes of overlaps makes the resulting maps to $Y$ agree, so they glue. For an arbitrary map from an affine scheme to $X$, cover its inverse images of these charts by affine open subschemes and apply naturality again. This proves that the glued map induces the prescribed natural transformation, and uniqueness follows from the original cover. In particular, the restricted functor of points determines the scheme up to an isomorphism of schemes.

An algebraic structure is a set equipped with specified algebraic operations satisfying selected axioms. An operation's domain is part of the structure: a binary operation on $S$ takes two elements of $S$ and produces another element of $S$. Groups specify one such operation, rings specify addition and multiplication, and vector spaces specify vector addition and scalar multiplication by a separate field.

Different choices of operations can give the same underlying set different structures. The integers form a group under addition, since every integer has an additive inverse. They do not form a group under multiplication: for example, $2$ has no multiplicative inverse among the integers. Even the nonzero integers fail that inverse requirement. In contrast, nonzero rational numbers form a group under multiplication.

Finite multiplication tables make such distinctions visible. Arithmetic modulo $5$ gives a field; arithmetic modulo $6$ gives a commutative ring with zero divisors. In the former, multiplying by any nonzero element permutes all residue classes. In the latter, multiplying by $2$ reaches only $0,2,4$, and $2\cdot 3=0$ even though both factors are nonzero.

The goal is to extract reusable properties rather than memorize isolated tables. A homomorphism preserves the specified operations, and an isomorphism shows that two structures have identical operation patterns after relabeling.

An algebra over a commutative ring $R$ is a ring $A$ with a specified unital ring homomorphism $R\to A$ whose image commutes with every element of $A$. Scalar multiplication is $r\cdot a=\phi (r)a$. For commutative $A$, the commutation requirement is automatic.

These two finite-sum transformations are inverses over any commutative ring. Substitute one sum into the other and use $\left(\genfrac{}{}{0ex}{}{n}{k}\right)\left(\genfrac{}{}{0ex}{}{k}{j}\right)=\left(\genfrac{}{}{0ex}{}{n}{j}\right)\left(\genfrac{}{}{0ex}{}{n-j}{k-j}\right)$. The inner alternating sum is $(1-1{)}^{n-j}$ by the binomial theorem, so only $j=n$ survives. In a Mahler expansion, $b_n$ is the nth finite difference of the sampled values $a_k$.

Every square matrix satisfies its own characteristic polynomial. More explicitly, for an $n$-by-$n$ matrix $A$ over a commutative ring with identity, formThenHere $0$ is the zero matrix, and the constant term multiplies the identity matrix. This is a polynomial evaluated at a matrix; it is stronger than saying that each eigenvalue is a root of a polynomial. Neither diagonalizability nor distinct eigenvalues is required.^[ref]

The practical payoff is that arbitrarily high matrix powers can be reduced to a combination of $I,A,\dots ,A^{n-1}$. These powers arise when repeatedly applying a linear map, solving a linear recurrence relation, or evolving a finite state space model. The characteristic polynomial turns that repeated action into a fixed-order scalar recurrence. The minimal polynomial may reduce the number of required powers further.

For a homomorphism of commutative rings $R\to A$ and two $A$-modules, the tensor product of modules over $A$ is the quotient of the tensor product of modules over $R$ imposing all relations $am\otimes n=m\otimes an$. The quotient map is $A$-linear when $A$ acts through the first factor on the source. This expresses the extra balancing imposed by change of rings.

For pairwise comaximal ideals in a commutative ring, reduction has the displayed ring isomorphism. The kernel is their intersection, equal to their product. Choose $e_i\in {\prod }_{j\ne i}{I}_j$ with $e_i\equiv 1(\mathrm{mod}{I}_i)$; then ${\sum }_i{a}_i{e}_i$ lifts any prescribed residue tuple. The equality between intersection and product follows from $1=u+v$, $u\in I$, $v\in J$, which writes each $x\in I\cap J$ as $xu+xv\in IJ$.

A comonoid in the monoidal category of modules over a commutative ring, with tensor ${\otimes }_k$ and unit $k$.

For a monic degree-$n$ polynomial $g$ over a commutative ring $R$, let $A=R[X]/(g)$ and let $\lambda$ take the coefficient of $X^{n-1}$ in the reduced representative. The pairing $(a,b)\mapsto \lambda (ab)$ is unimodular on $1,X,\dots ,X^{n-1}$: reversing one basis order gives a triangular matrix with diagonal ones. When $A$ embeds in a separable field extension, Lagrange interpolation identifies $\lambda (a)$ with $\mathrm{Tr}(a/g^{\prime }(x))$. This yields the derivative formula for the inverse different.

Use the signature of unital commutative rings, its equational axioms, and the coherent sequents $0=1⊢\perp$ and $\top {⊢}_x\mathrm{Inv}(x)\vee \mathrm{Inv}(1-x)$, where $\mathrm{Inv}(x)$ abbreviates $\exists y(xy=1)$. In a set-based model the nonunits form its unique maximal ideal; this is the usual nonzero local ring. In a topos the disjunction expresses local existence of a unit witness. Its classifying topos is defined by the Zariski coverage on finitely presented rings.

For a commutative ring $R$, remove the section $[1:0:0]$ from ${\mathbb{P}}_R^2$. The two remaining standard charts have coordinate rings $R[a,b]$ and $R[a/b,b^{-1}]$ inside the Laurent polynomial ring $R[a,b,b^{-1}]$. Their intersection consists of constants: their monomial exponent regions are $i\ge 0,j\ge 0$ and $i\ge 0,j\le -i$, respectively. The Čech cohomology in degree one is free over $R$ on the missing monomials $a^i{b}^{-k}$ with $1\le k<i$. Equivalently its basis consists of the classes of $x_0^{m+n}/(x_1^m{x}_2^n)$ for $m,n\ge 1$. Degree zero is $R$ and every degree at least two is zero. The charts and their intersection are affine schemes, so this computes sheaf cohomology. If $R$ is nonzero, the first sheaf cohomology is nonzero and this scheme is not an affine scheme.

Commutative algebra studies commutative rings, ideals, modules, and the spectrum of a ring. Its basic question is which familiar rules of integer and polynomial arithmetic survive in a new setting. The distinction between units, zero divisors, and integral domains makes that question concrete.

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.

Two elements of a commutative ring are congruent modulo an ideal if their difference lies in that ideal, equivalently if their images in the quotient ring agree. This is an equivalence relation compatible with addition and multiplication. For $R={\mathbb{Z}}_p[[T]]$ and $I=p^{k}R$, it means every coefficient difference is divisible by $p^k$.

For a commutative ring $R$ and an ideal $I$, the maps $J\mapsto J/I$ and $H\mapsto q^{-1}(H)$ give inverse inclusion-preserving bijections between ideals $J\supseteq I$ of $R$ and ideals $H$ of the quotient ring $R/I$, where $q:R\to R/I$ is the quotient map. The preimage is an ideal containing the kernel $I$. Conversely, $q^{-1}(q(J))=J$ for $J\supseteq I$, and surjectivity of $q$ gives $q(q^{-1}(H))=H$. These identities prove the claim.

A module is cyclic if one element $m$ generates it: every element is $rm$ for some scalar $r\in R$. The generating element need not give unique coefficients. The module homomorphism $R\to M$, $r\mapsto rm$, is surjective. Over a commutative ring, its kernel is the annihilator ${\mathrm{Ann}}_R(m)$, givingThus cyclic modules are exactly the quotient modules $R/I$ for ideals $I$ in $R$. For a left module over a noncommutative ring, use a left ideal and interpret the quotient as a module, with no claim that it is a quotient ring. See John A. Beachy's Abstract Algebra Online, section 10.1.

The module $\mathbb{Z}/6\mathbb{Z}$ is generated by $\overline{1}$, and also by $\overline{5}$, whose successive multiples are $\overline{0},\overline{5},\overline{4},\overline{3},\overline{2},\overline{1}$. The element $\overline{2}$ generates only $\{\overline{0},\overline{2},\overline{4}\}$, a submodule with three elements. A cyclic module can be infinite and free, as $\mathbb{Z}=\mathbb{Z}\cdot 1$ shows, or finite and torsion, as $\mathbb{Z}/6\mathbb{Z}$ shows. One generator counts directions of generation, not the number of elements and not freedom from relations.

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.

For $n\ge 1$ and $a,b$ in a commutative ring, multiplying out the displayed polynomial factorization cancels all intermediate terms, leaving $a^n-b^n$. The case $n=2$ is the difference of squares. Unlike division by $a-b$, this identity also applies when $a=b$.

For real $1\le b\le a$, every summand lies between one and $a^{n-1}$. Consequently $n(a-b)\le a^n-b^n\le n(a-b)a^{n-1}$. This turns a one-step discrepancy into a bound after repeated multiplication, for example when estimating the global error of a one-step method on $y^{\prime }=y$.