Commutative ring ($ab=ba$)

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 $a/s=0$.