Ring ($R$)

A ring is a set $R$ with two binary operations, addition and multiplication, satisfying three requirements. Under addition, $R$ is an abelian group: there is a zero, every element has an inverse element, and grouping or exchanging summands does not change the result. Multiplication is associative and has an identity element $1$, so $(ab)c=a(bc)$ and $1a=a1=a$. Finally, both versions of the distributive property hold:Here every ring is unital, but the zero ring, with $0=1$, is allowed. Commutativity of multiplication is an additional condition defining a commutative ring. These conventions agree with ^[ref].

The integers form a ring with their usual operations. The polynomial ring $\mathbb{Z}[X]$ is another: addition combines coefficients, while multiplication uses the distributive property. For example,The same rules justify symbolic calculations in both rings; this is why one definition can organize apparently different kinds of arithmetic. In every ring, $a0=0a=0$, and $a(-b)=-(ab)$. The first identity follows by distributing $a(0+0)$ and subtracting $a0$; the second follows by distributing $a(b+(-b))$. Thus familiar sign rules are consequences of the structure.

Order still matters in some rings. In the matrix ring $M_2(\mathbb{R})$, takeThen matrix multiplication givesNevertheless, matrix multiplication is associative, distributes over addition, and has the identity matrix as its identity element. This is a valid ring. On a column vector, $AB$ applies $B$ first and $A$ second: swapping coordinates before discarding the second coordinate differs from doing those actions in reverse. The failure of commutativity records a meaningful difference in function composition.

An associator ring is a ring in which equality of the principal ideals generated by two elements forces them to be associate elements: $(a)=(b)$ implies $a=ub$ for a unit in a ring $u$. The reverse implication always holds, since $a=ub$ and $b=u^{-1}a$ give both ideal inclusions. The property separates equality of generated ideals from equality of generators up to an invertible factor.

Every integral domain is an associator ring. If $a\ne 0$ and $(a)=(b)$, write $a=rb$ and $b=sa$. Then $(1-rs)a=0$, so cancellation gives $rs=1$ and $a=rb$ with $r$ a unit. The zero case is immediate. The converse fails: $\mathbb{Z}/4\mathbb{Z}$ has generator classes $\{0\}$, $\{2\}$ and $\{1,3\}$ for its three ideals, and $1,3$ differ by a unit, although $2$ is a nonzero zero divisor.

Finite direct product of rings preserve this property. Equality of principal ideals is coordinatewise, and a tuple of units is a unit. Every quotient ring of a principal ideal domain is also an associator ring: first, modulo a prime power $p^k$, every nonzero class is a power $p^e$, $e<k$, times a unit, and distinct powers generate distinct ideals. Next factor a nonzero nonunit modulus into prime powers and apply the Chinese remainder theorem. The zero modulus gives the original integral domain, while a unit modulus gives the zero ring.

A rng has an abelian group under addition, an associative multiplication, and the distributive property; a multiplicative identity element is not required. Every ring under this wiki's unital convention is a rng. A nonunital ring may happen to have an identity: the terminology removes an axiom rather than imposing the absence of an identity. For example, the even integers with their usual operations have no multiplicative identity.

Up to an operation-preserving bijection, there are two rngs with two elements. Write the elements as $0,u$. The additive abelian group must satisfy $u+u=0$. The distributive property forces all products involving zero to vanish, leaving only the choice $u^2=0$ or $u^2=u$. The first gives a zero-multiplication rng; the second gives the prime field ${\mathbb{F}}_2$, with identity $u$. Both satisfy the axioms, and they are not isomorphic because an isomorphism must send the unique nonzero element to the unique nonzero element and preserve its square. In particular, there is just one unital ring with two elements.

Any abelian group becomes a nonunital ring by declaring every product zero. Associativity and the distributive property hold because both sides are zero. When the group has more than one element, no multiplicative identity element exists: an identity would have to satisfy $ex=x$, whereas every such product is zero. This differs from the zero ring, which has just one element.

The characteristic of a rng is the least positive integer annihilating every element of its additive abelian group, or zero if no such integer exists. In a unital ring this equals the characteristic of a ring: $n1=0$ implies $nr=(n1)r=0$, and annihilating every element in particular annihilates $1$. In a finite rng it is the exponent of a finite group of the additive group.

A property of rings is a ring isomorphism invariant when any two rings related by a ring isomorphism either both have the property or both lack it. Being an integral domain is such a property: a ring isomorphism preserves nonzero elements, products, and zero.

A principal ideal ring is a ring whose left ideals and right ideals each have one generator. For a commutative ring, this says that every ideal is a principal ideal. The examples here use commutative rings. Unlike a principal ideal domain, it may have zero divisors: $\mathbb{Z}/4\mathbb{Z}$ has this property, with ideals $(0)$, $(2)$, and $(1)$. A finite direct product of rings that are principal ideal rings is again a principal ideal ring. Indeed, the coordinate idempotent elements show that every ideal is a product of coordinate ideals; a tuple of their generators generates the product.

The matrix ring consists of all $n\times n$ matrices over a ring $R$, using entrywise addition and matrix multiplication. Its identity element is the identity matrix. Even when $R$ is a commutative ring, this ring need not have commutative multiplication.

The ring with a single element, which is both its additive and multiplicative identity. If a unital ring has $0=1$, then every element $r$ equals $r1=r0=0$, so it is this ring. Under the usual definition it is excluded from integral domains and fields. The quotient ring $\mathbb{Z}/(1)$ is an example.

A commutative ring of prime characteristic $p$ is perfect in the Frobenius sense when its Frobenius endomorphism is an automorphism. It is necessarily reduced: if $a$ is nilpotent, some $p$-power iterate is zero, and injectivity gives $a=0$. This use of perfect is distinct from the homological notion of a perfect ring. The Absolute Frobenius morphism of a scheme is an isomorphism exactly when all its local rings are perfect in this sense, because it is the identity on points and is an isomorphism on every stalk.

The opposite ring has the same additive group and identity as a ring $R$, with multiplication $a{\cdot }_{\mathrm{op}}b=ba$. Left modules over it are right modules over $R$. Forming the opposite exchanges left and right chain conditions and reverses the order of composition when interpreting endomorphisms as right multiplications. For an associative algebra with its scalar action retained, this construction gives the opposite algebra.

A subset closed under addition, additive inverses and multiplication, with the induced operations. Under the unital convention used here it also contains the identity of $R$. A subring of an integral domain is again an integral domain. The ring of invariants is one example: fixed elements remain fixed under all ring operations.

Every subring $R$ of the rational numbers containing $\mathbb{Z}$ is a principal ideal domain. If $a/b\in R$ is reduced, Bézout's identity gives integers $u,v$ with $ua+vb=1$, hence $1/b=u(a/b)+v\in R$. Thus every reduced denominator occurring in $R$ is a unit.

For an ideal $I$, write $I\cap \mathbb{Z}=d\mathbb{Z}$. Then $dR\subseteq I$. If $a/b\in I$ is reduced, $a=b(a/b)\in I\cap \mathbb{Z}$, so $a=dc$ for some integer $c$. Since $1/b\in R$, $a/b=d(c/b)\in dR$. Therefore $I=dR$, including the zero ideal. This is a concrete instance of extending and contracting ideals under localization of a ring.

A topological ring is a ring equipped with a topology making addition, additive inversion and multiplication continuous functions. An adic topology is obtained from powers of an ideal.

A pro-$p$ ring is a topological ring with continuous ring operations whose underlying additive group is a pro-p group. Standard local analytic examples are complete Noetherian local rings with finite residue field of characteristic $p$, such as ${\mathbb{Z}}_p$ and ${\mathbb{F}}_p[[t]]$. The topology is essential to the convergence of power series used for analytic group operations.

For a two-sided ideal $I$ of a ring $R$, the powers $I^n$ form a neighborhood base at zero for the adic topology. For an $R$-module $M$, use $I^{n}M$ on the left or $M{I}^n$ on the right. Convergence means that for every $n$ the differences eventually belong to the corresponding submodule. The formal power series ring $k[[t]]$ and formal power series module $M[[t]]$ are complete for $I=(t)$; convergence fixes each finite list of coefficients eventually. This is formal convergence, without an analytic condition on the size of coefficients.

For a complete Noetherian ring with ideal of definition $I$, completed localization is the completion of the localization of a ring $A_f$ in the induced adic topology. It gives the sections of the formal spectrum over $D(f)$. For $A=k[t][[q]]$, its completed localization at $t$ is $k[t,t^{-1}][[q]]$. Ordinary localization allows a bounded pole order in $t$ across all coefficients; the latter ring also contains ${\sum }_{n\ge 0}{q}^n{t}^{-n}$, whose pole orders are unbounded.

// Target: ringed-space.bigb

The adic completion is the inverse limit of the quotient modules $M/I^{n}M$. An element is a compatible family of residue classes at every order. For $R=k[t]$ and $M=V{\otimes }_{k}k[t]$, this gives the formal power series module $V[[t]]$. The canonical map from $M$ need not be injective in general: its kernel is ${\bigcap }_n{I}^{n}M$. It is injective for this polynomial example because a nonzero polynomial has finite degree. Successive changes of coordinates in formal rigidity from vanishing second Hochschild cohomology converge in this completion because the order-$r$ change is the identity modulo $t^r$.

If $I=(a_1,\dots ,a_s)$, evaluation $X_i\mapsto a_i$ gives a ring map to the $I$-adic completion, with each series evaluated modulo $I^r$ using only terms of total degree below $r$. It is surjective: successively represent the residual class in $I^r/I^{r+1}$ by a homogeneous degree-$r$ polynomial in the $a_i$. If $R$ is a commutative Noetherian ring, Noetherianity of formal power series rings therefore makes its completion Noetherian.

For any ideal $I$ and surjective module map $f:M\to N$, compatible representatives $n_r$ modulo $I^{r}N$ can be lifted recursively. Having chosen $f(m_r)=n_r$, lift $n_{r+1}-n_r\in I^{r}N$ to ${\delta }_r\in I^{r}M$ and put $m_{r+1}=m_r+{\delta }_r$. These define the required element of the adic completion of a module. No finite-generation hypothesis on the modules is required.

A ring is von Neumann regular if every element $a$ has an element $b$ with $a=aba$. In a commutative ring, this is $a=a^{2}b$. An arbitrary direct product of rings whose factors are fields has this property by taking inverses in each nonzero coordinate. Every localization at a prime ideal of a commutative von Neumann regular ring is a field.

The direct product of a family of rings consists of all tuples of their elements, with coordinatewise addition and multiplication. Unlike a direct sum, tuples may have infinitely many nonzero coordinates.

For a commutative ring $k$ and a set $X$, the set $k^X$ of all functions $X\to k$ is a commutative ring under pointwise addition and multiplication. It is the direct product of rings with one copy of $k$ for each $x\in X$.

If $k$ is a field and $X$ is finite, every ideal is generated by some of the singleton indicator functions $e_x$. Indeed, whenever $f$ in the ideal has $f(x)\ne 0$, multiplication by $e_x/f(x)$ puts $e_x$ in the ideal. Every function in that ideal is a finite sum of its coordinate multiples of these $e_x$.

If $X$ is infinite and $k$ is a nonzero commutative ring, the functions with finite support form an ideal without a finite generating set of an ideal. A finite collection of such functions vanishes outside one finite subset, as does the ideal it generates, whereas a singleton indicator function outside that subset belongs to the ideal of all finitely supported functions. Thus, for a field $k$, $k^X$ is a Noetherian ring exactly when $X$ is finite.

For a field $k$, the shift on the direct product of rings ${\prod }_{n\ge 0}k$ preserves coordinatewise addition, multiplication and the unit. It is a surjective ring endomorphism because an arbitrary first coordinate can be prepended to a desired output. Its kernel of a ring homomorphism consists of the sequences supported in coordinate zero, so it is not injective. The ring is not Noetherian: the ideals generated by the first $n+1$ coordinate idempotents form a strictly ascending chain.

A quotient ring of a finite direct product of rings whose factors are fields is a product of a subset of those factors. By ideals of a finite product of rings, write $J={\prod }_j{J}_j$. Since a field has only the zero ideal and the whole field as ideals, every $J_j$ is one of these. The ring homomorphism $(x_j)\mapsto (x_j+J_j)$ is surjective with kernel of a ring homomorphism $J$, so the first isomorphism theorem for rings gives $({\prod }_j{K}_j)/J\cong {\prod }_j(K_j/J_j)$. Discard the zero ring factors. If no factors survive, the result is the zero ring.

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 tuple in a direct product of rings is a unit exactly when each coordinate is a unit. Indeed, multiplying by an inverse tuple gives an inverse in each coordinate; conversely, the coordinate inverses form an inverse tuple. This identifies the unit group with the direct product of groups of the individual unit groups, for finite or infinite families of unital rings.

The characteristic of a ring $R$ is the least positive integer $n$ satisfying $n\cdot 1_R=0$, or zero if no such integer exists. A nonzero ring of positive characteristic has prime characteristic when it is an integral domain.

In a finite unital ring $R$, some positive multiple of $1$ is zero, because two of the finitely many successive multiples coincide. If $n=\mathrm{char}R$, the permutation $r\mapsto r+1$ has every cycle of a permutation of length exactly $n$: returning to $r$ after $k$ steps means $k1=0$, whose least positive solution is $n$. These disjoint cycles partition the finite set $R$, so $n$ divides its cardinality. The zero ring has cardinality and characteristic both one.

The positive characteristic of a ring which is an integral domain is a prime number. The characteristic cannot be one because $1\ne 0$. If its least positive value were $n=ab$ with $1<a,b<n$, minimality would make $a1$ and $b1$ nonzero, while $(a1)(b1)=n1=0$ would give zero divisors.

The analogous result holds for a nonzero commutative nonunital ring without zero divisors, using the characteristic of a rng. If $n=ab$ were composite and minimal, choose $r$ with $ar\ne 0$. For every $s$, $(ar)(bs)=n(rs)=0$, so cancellation forces $bs=0$ for every $s$, contradicting minimality of $n$.

In a nonzero unital ring with finite odd unit group, the unit $-1$ has order dividing two. By Lagrange's theorem, that order also divides $|U(R)|$, so $-1=1$ and the characteristic of a ring is two. Consequently $\overline{a}\mapsto a{1}_R$ defines a unital ring homomorphism ${\mathbb{F}}_2\to R$. This map still exists for the zero ring, whose characteristic is one under the least-positive-integer convention.

A division ring is a nonzero ring in which every nonzero element has a multiplicative inverse. Its multiplication need not be commutative.

A nonzero ring is simple when its only two-sided ideals are zero and the whole ring.

A semisimple ring is a ring that is semisimple as a module over itself. Equivalently, every module over it is semisimple, so every short exact sequence of modules splits.

The group ring $R[G]$ consists of finite formal sums ${\sum }_{g\in G}{r}_{g}g$ with coefficients in a ring $R$. Addition is coefficientwise and multiplication extends the group operation and multiplication in $R$ distributively. A group algebra is the case in which $R$ is a field.

If $A$ is a left Noetherian ring and a right Noetherian ring, and $G$ is a poly-(cyclic or finite) group, then $A[G]$ satisfies both chain conditions. Induct along the subnormal series. A finite factor gives a finite free extension on each side, handled by finite-module extension preserves Noetherianity. An infinite cyclic factor gives a skew Laurent polynomial ring, with the twisting automorphism induced by conjugation by a lift of the cyclic generator. The noncommutative Hilbert basis theorem and clearing negative powers prove the required chain conditions at that step. In particular $\mathbb{Z}[G]$ is Noetherian.

The augmentation homomorphism $\epsilon :R[G]\to R$ sends ${\sum }_g{r}_{g}g$ to ${\sum }_g{r}_g$. Its kernel is the augmentation ideal.

In the ordinary group algebra of the additive p-adic integers, let $I$ be the augmentation ideal. The kernel of the map to the group algebra of ${\mathbb{Z}}_p/p^r{\mathbb{Z}}_p$ is generated by $[p^{r}h]-1=([h]-1{)}^{p^r}$, so it is contained in $I^{p^r}$. The quotient is the cyclic-group algebra ${\mathbb{F}}_p[T]/(T^{p^r})$, so its augmentation ideal has zero $p^r$th power and the reverse inclusion holds. Every finite degree of the associated graded ring can be read from such a quotient, giving ${\mathrm{gr}}_I{\mathbb{F}}_p[{\mathbb{Z}}_p]\cong {\mathbb{F}}_p[X]$. The completion is ${\mathbb{F}}_p[[T]]$, but completion is not needed for the graded calculation.

In characteristic $p$, the augmentation ideal $I$ of the group algebra of a finite $p$-group is nilpotent. Choose a central element $z$ of order $p$ and put $t=z-1$, so $t^p=0$. The quotient by $t$ is the group algebra of the smaller quotient group. Induction gives $I^m\subseteq tkP$ for some $m$, hence $I^{mp}=0$. For a normal $p$-subgroup $P$ of $G$, conjugation stability gives $(IkG{)}^m=I^{m}kG$, proving nilpotence of the kernel of $kG\to k(G/P)$.

The group ring of an infinite cyclic group is the Laurent polynomial ring $R=\mathbb{Z}[t,t^{-1}]$, with augmentation $\epsilon (t)=1$. Multiply any element in the kernel by a power of $t$ to obtain a polynomial vanishing at one. The factor theorem makes it divisible by $t-1$. Thus the augmentation ideal is $(t-1)R$. Since $R$ is an integral domain, multiplication by $t-1$ is an injective module map, so this ideal is a free $R$-module of rank one, even though its additive group has infinite rank.

A noncommutative ring is a ring whose multiplication is not required to be commutative.

An element $x$ of a ring is nilpotent when $x^n=0$ for some positive integer $n$. A field has no nonzero nilpotent elements.

An element of an associative algebra is square-zero if its square vanishes. Every such element is a nilpotent element. These elements are preserved by an algebra isomorphism, so their structure distinguishes algebras. Their set need not be a vector space: the matrix units $E_{12}$ and $E_{21}$ are square-zero, while their sum has square equal to the identity in $M_2(K)$. In a square-zero ideal, by contrast, every pairwise product vanishes and all its elements are square-zero. In $K[t]/(t^3)$ the square-zero elements are exactly the multiples of $t^2$, over every field: $(yt+z{t}^2{)}^2=y^2{t}^2$ forces $y=0$.

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.

A regular sequence in a commutative ring $A$ is a finite list such that $(f_1,\dots ,f_r)$ is proper and $f_j$ is a non-zero-divisor on $A/(f_1,\dots ,f_{j-1})$ for each $j$. In a graded ring with homogeneous $f_j$, injectivity can be checked on homogeneous elements because components of distinct degrees cannot cancel. It gives short exact sequences $0\to R_{j-1}(-\mathrm{deg}{f}_j)\to R_{j-1}\to R_j\to 0$ for the successive graded quotients.

The depth of a Noetherian local ring is the maximal length of an $R$-regular sequence in its maximal ideal. It is at most its Krull dimension. Quotienting by a nonzero divisor lowers depth by one. This invariant measures how many independent successive nonzero-divisor equations the local algebra supports.

The dual-number ring over a field $k$ consists of elements $a+b\epsilon$ with ${\epsilon }^2=0$. The element $\epsilon$ is a nonzero nilpotent element; its ideal is the ring's only prime ideal. Hence its affine scheme has the same one-point topology as $\mathrm{Spec}k$ while its structure sheaf differs.

For $A=k[\epsilon ]/({\epsilon }^2)$ and $k=A/(\epsilon )$, the kernel and image of multiplication by $\epsilon$ both equal $(\epsilon )$. Thus the displayed sequence is a free resolution. Applying ${\mathrm{Hom}}_A(-,k)$ makes every differential zero, so ${\mathrm{Ext}}_A^i(k,k)\cong k$ for every $i\ge 0$. The module $k$ therefore has infinite projective dimension.

The formal power series ring $R[[T]]$ consists of all expressions ${\sum }_{n\ge 0}{a}_n{T}^n$ with $a_n\in R$. Addition is coefficientwise, while multiplication uses the Cauchy product; no analytic convergence is required.

For a formal power series over a field of characteristic zero, the displayed coefficientwise operation chooses constant coefficient zero. Its formal derivative is the original formal power series: differentiating the term with exponent $n+1$ cancels its denominator. Conversely, formally integrating a formal derivative removes the original constant term. These are algebraic identities; analytic convergence is not required.

If the constant coefficient of $A$ is invertible, then for every formal power series $B$ there is a unique quotient $Q$ satisfying $AQ=B$. Comparing coefficients determines $Q$ recursively; multiplying $B$ by the formal power series reciprocal of $A$ computes the same quotient.

For a formal power series $A$ with zero constant term over a characteristic-zero field, its exponential is $\exp A={\sum }_{k\ge 0}{A}^k/k!$. Every coefficient is a finite sum because $A^k$ starts in degree at least $k$. In finite characteristic, truncated computations require the factorial denominators that occur to be invertible.

To solve $\log G=A$ with $A(0)=0$, start with $G=1$ and update $G\leftarrow G(1+A-\log G)$ at successively doubled truncation precision. The correction removes the current logarithmic error to first order. This is Newton iteration applied to the formal power series exponential, using a formal power series logarithm.

For a commutative Noetherian ring, the ideals of lowest possible coefficients of an ideal in $R[[X]]$ form an ascending chain. Choose lifts of finite generating sets through the stabilization degree. Successive coefficient cancellation expresses any series in the ideal using these finitely many lifts. The multiplier series converge coefficientwise, so this generates the actual ideal without first assuming it is closed. Iterating gives the result for finitely many variables.

The formal power series ring $R={\mathbb{Z}}_p[[T]]$ is complete for two useful topologies. In the $p$-adic topology, convergence means that all coefficient errors are uniformly divisible by arbitrarily large powers of $p$. In the $(p,T)$-adic topology, only finitely many coefficients are controlled at each precision. The latter topology is compact, since $R/(p,T{)}^n$ is finite and $R$ is its inverse limit. The former is finer and is not compact: the series $T^n$ converge to zero in the latter but remain pairwise separated modulo $p$ in the former. Coleman norm contraction uses the former; compact interpolation arguments use the latter.

A formal power series with zero constant coefficient has a compositional inverse over $R$ exactly when its linear coefficient is a unit. Starting with $b_1=a_1^{-1}$, the degree-$k$ coefficient of $f(b_{1}T+\cdots +b_k{T}^k)$ is $a_1{b}_k$ plus an expression already determined by lower coefficients, so recursion gives a unique inverse. For a homomorphism of formal group laws, cancellation in the homomorphism identity proves that this inverse is also a formal-group homomorphism.

Let $R$ be a commutative unital ring and $f(T)=a_{1}T+a_2{T}^2+\cdots$ with $a_1$ a unit. Set $g(T)={\sum }_{n\ge 1}{b}_n{T}^n$. The coefficient of $T$ in $f(g(T))$ gives $b_1=a_1^{-1}$. At every degree $n>1$, the coefficient is $a_1{b}_n$ plus an expression involving only $b_1,\dots ,b_{n-1}$; choose $b_n$ to cancel that expression. This gives a unique right compositional inverse of a formal power series.

Similarly, the coefficient of $T^n$ in $h(f(T))$ is $a_1^n{h}_n$ plus previously fixed terms, so the same recursion gives a left inverse $h$. Composition of zero-constant-term formal power series is associative, since every coefficient involves finitely many terms. Therefore $h=h\circ (f\circ g)=(h\circ f)\circ g=g$, proving $f\circ g=g\circ f=T$.

If $(A,\mathfrak{m})$ is a local ring, then $B=A[[x]]$ is local with maximal ideal $I=\mathfrak{m}B+xB$ and residue field $A/\mathfrak{m}$. A series $f=\sum a_n{x}^n$ is a unit if and only if $a_0$ is a unit. Necessity follows by taking constant coefficients of an inverse; sufficiency follows from the recursion $b_0=a_0^{-1}$ and $b_n=-a_0^{-1}{\sum }_{i=1}^n{a}_i{b}_{n-i}$ for its inverse. Thus the nonunits are precisely series with $a_0\in \mathfrak{m}$, which is exactly $I$, since $f=a_0+xg$. The constant-coefficient map followed by reduction modulo $\mathfrak{m}$ has kernel $I$ and image $A/\mathfrak{m}$.

For a local ring $(A,\mathfrak{m})$ with residue field $k$, put $B=A[[x]]$ and $I=\mathfrak{m}B+xB$. There is a natural $k$-vector space isomorphism $I/I^2\cong (\mathfrak{m}/{\mathfrak{m}}^2)\oplus k$. The inverse sends $(\bar{a},\bar{b})$ to the class of $a+bx$. It is well-defined because changing $a$ by ${\mathfrak{m}}^2$ or $b$ by $\mathfrak{m}$ changes the result by $I^2$. It is surjective since every $f\in I$ is congruent modulo $x^{2}B\subseteq I^2$ to its constant and linear terms. As $I^2={\mathfrak{m}}^{2}B+x\mathfrak{m}B+x^{2}B$, those coefficients of an element of $I^2$ lie in ${\mathfrak{m}}^2$ and $\mathfrak{m}$, respectively. This proves injectivity. Thus adjoining one formal variable increases finite embedding dimension by one.

A formal power series ring is a reduced ring if and only if its coefficient ring is reduced. A nonzero constant nilpotent proves one direction. In the other direction, a nonzero series with first nonzero coefficient $a_r$ has first coefficient $a_r^m$ in its $m$th power; this is nonzero over a reduced ring.

For a vector space $M$ over a field $k$, its formal power series module is $M[[t]]={\underleftarrow{\lim }}_{n}M{\otimes }_{k}k[t]/(t^n)$. It is complete for the filtration by $t^{n}M[[t]]$. The ordinary tensor product $M{\otimes }_{k}k[[t]]$ embeds in it and consists of series whose coefficients span a finite-dimensional vector subspace; it equals $M[[t]]$ if $M$ is finite-dimensional. The inclusion is proper for infinite-dimensional $M$, as shown by a series with linearly independent coefficients.

A formal power series ring $R[[x]]$ is Noetherian if and only if $R$ is Noetherian. In the forward direction, use the quotient ring $R[[x]]/(x)\cong R$. In the reverse direction, for an ideal $I$ consider the ascending sequence of ideals of $R$ formed by the coefficient of $x^n$ in $I\cap x^{n}R[[x]]$. Stabilization and finite generation give a finite list of series that generate $I$ by successive coefficient cancellation.

For an ideal $H\subseteq R[[X]]$, let $A_n$ consist of coefficients of $X^n$ in members of $H$ with all lower coefficients zero. These are ideals of $R$ and $A_n\subseteq A_{n+1}$, by multiplication by $X$. If $R$ is Noetherian, this chain stabilizes. Choose series whose leading coefficients generate the finitely many distinct coefficient ideals, then cancel coefficients successively. The accumulated multipliers are formal power series, giving an ordinary finite ideal generating set for $H$, rather than merely a dense subideal.

This is the power-series specialization of a generating function.

The ordinary generating function of a sequence $(a_n{)}_{n\ge 0}$ is the formal power series $A(z)={\sum }_{n\ge 0}{a}_n{z}^n$. Algebraic identities for $A$ encode recurrences and counting constructions for the coefficients.

A reduced ring is a commutative ring with no nonzero nilpotent element. Equivalently, its nilradical is zero.

The subset of the spectrum of a ring where the localization at a prime ideal is reduced. Since the nilradical commutes with localization, its complement is the support of a module $N=\mathrm{Nil}(R)$. When $N$ is a finitely generated module, vanishing of a finitely generated module near a prime givesso the locus is a Zariski-open subset. In particular this holds for every Noetherian ring.

If a nonzero nilpotent element $a$ existed in $A$, its proper annihilator would lie in a maximal ideal $\mathfrak{m}$. No element outside $\mathfrak{m}$ annihilates $a$, so $a/1$ is nonzero in the localization at a prime ideal $A_{\mathfrak{m}}$. Conversely, if $(a/s{)}^n=0$ in a localization, some $u$ in the multiplicative subset satisfies $u{a}^n=0$. Then $(ua{)}^n=0$, so reducedness of $A$ gives $ua=0$ and hence $a/s=0$.

A graded ring is a direct sum $R={⨁}_{n\ge 0}{R}_n$ with $R_m{R}_n\subseteq R_{m+n}$.

A graded module over a graded ring $R={⨁}_i{R}_i$ is an $R$-module with a direct sum decomposition $M={⨁}_d{M}_d$ satisfying $R_i{M}_d\subseteq M_{i+d}$. A graded homomorphism preserves these degrees. Over a polynomial ring with positive variable degrees, a finitely generated module of this kind is bounded below and has finite-dimensional graded components over the coefficient field.

A super vector space is a vector space graded by the two-element group. Degree-preserving maps and the usual graded tensor product form a monoidal category. Over a field of characteristic different from two, this category has two different symmetries: ordinary interchange of tensor factors and interchange multiplied by $(-1{)}^{pq}$ on homogeneous factors of degrees $p,q$. The latter is the Koszul sign rule.

If $f$ is homogeneous of degree one, multiplication by powers of the invertible element $f$ identifies every homogeneous component of $M_f$ with $(M_f{)}_0$. A homogeneous element $m$ of degree $a$ is written $f^a(f^{-a}m)$. Thus the displayed map is an isomorphism of graded modules. In particular, the degree-zero part of a tensor product of localized graded modules is the tensor product of their degree-zero parts over $(S_f{)}_0$. This proves tensor compatibility of graded sheafification on degree-one-generated Proj.