Commutative algebra

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.

If $B$ is a commutative $A$-algebra which is a finite free module over $A$, its norm sends $b$ to the determinant of multiplication $m_b:B\to B$. The determinant is independent of the chosen basis and is multiplicative because $m_{bc}=m_b{m}_c$. It commutes with scalar extension. On a separable field extension it becomes the product over embeddings, recovering the field norm. An invertible $b$ has invertible norm.

A unital commutative ring generated by finitely many elements under addition and multiplication, starting from the image of the integers. Equivalently it is a quotient of a finite-variable polynomial ring over $\mathbb{Z}$. localization of a ring at one element remains finite type, since $R[1/r]\cong R[T]/(rT-1)$. Generation here is as an algebra, not merely as a field extension when the ring happens to be a field.

A field that is a finite-type integer algebra has prime characteristic or zero. Prime characteristic and Zariski lemma make it a finite algebraic extension of a finite prime field. In characteristic zero, the same lemma makes it a number field; clearing finitely many coefficient denominators makes it integral over $\mathbb{Z}[1/N]$. The integral field extension forces the base domain to be a field, but a prime not dividing $N$ is not invertible there. This contradiction excludes characteristic zero.

For a commutative algebra $R$ over $k$, put $D_k^{-1}(R)=0$ and define $D_k^j(R)$ to be the $k$-linear endomorphisms $\theta$ with $[\theta ,m_a]\in D_k^{j-1}(R)$ for every multiplication map $m_a$. The union is a ring under composition, with $D_k^0(R)=R$ and $D_k^r{D}_k^s\subseteq D_k^{r+s}$. This commutator definition applies to singular rings and fields of arbitrary characteristic, unlike a definition restricted to polynomial coefficients and partial derivatives.

The least index in the commutator filtration of the algebraic differential operator ring containing a nonzero operator. The zero operator has order $-\infty$ by convention. Equivalently every $(j+1)$-fold iterated commutator with multiplication maps vanishes for an operator of order at most $j$.

If $\theta$ has order at most one, set $c=\theta (1)$ and $\delta =\theta -m_c$. Then $\delta (1)=0$ and $[\delta ,m_a]=m_{\delta (a)}$, which is the Leibniz rule. Thus $\delta$ is a derivation of an algebra. Conversely any derivation satisfies this commutator identity. The sum is direct because a derivation that is multiplication must vanish at $1$.

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.

An algebra over a commutative ring $A$ is finitely generated if it is a quotient ring of a polynomial ring $A[t_1,\dots ,t_n]$ in finitely many variables. Equivalently, finitely many elements generate it under addition, multiplication, and the action of $A$. The induced map of affine schemes $\mathrm{Spec}B\to \mathrm{Spec}A$ is a morphism of finite type exactly when this holds. Finite generation as an algebra does not require finite generation as an $A$-module.

An algebra over a commutative ring that is also a coalgebra, whose comultiplication and counit preserve the multiplication and unit. Equivalently it is a bimonoid in $k$-modules.

A bialgebra equipped with a coquasitriangular structure, namely a convolution-invertible scalar form $r:H\otimes H\to k$ satisfying $r(ab,c)=\sum r(a,c_{(1)})r(b,c_{(2)})$, $r(a,bc)=\sum r(a_{(1)},c)r(a_{(2)},b)$ and $\sum r(a_{(1)},b_{(1)})a_{(2)}{b}_{(2)}=\sum b_{(1)}{a}_{(1)}r(a_{(2)},b_{(2)})$, with unit normalization. The induced right-comodule braiding is $v\otimes w\mapsto \sum w_{(0)}\otimes v_{(0)}r(v_{(1)},w_{(1)})$.

A normalized scalar map $\gamma :H\otimes H\to k$ satisfying $\sum \gamma (a_{(1)},b_{(1)})\gamma (a_{(2)}{b}_{(2)},c)=\sum \gamma (b_{(1)},c_{(1)})\gamma (a,b_{(2)}{c}_{(2)})$, and $\gamma (1,a)=\epsilon (a)=\gamma (a,1)$. Some conventions require convolution invertibility; that extra hypothesis distinguishes strong from lax tensor transformations.

With the convention $m\ast \gamma =\gamma \ast n$ and invertible bialgebra scalar cocycle $\gamma$ for $n$, the same coalgebra has $n=\bar{\gamma }\ast m\ast \gamma$, or equivalently $m=\gamma \ast n\ast \bar{\gamma }$. Thus $n(a,b)=\sum \bar{\gamma }(a_{(1)},b_{(1)})m(a_{(2)},b_{(2)})\gamma (a_{(3)},b_{(3)})$. Specifying the equation prevents opposite twist conventions from being conflated.

The tensor of left modules over a bialgebra has action $h(u\otimes v)=\sum (h_{(1)}u)\otimes (h_{(2)}v)$. The unit module $k$ has action through the counit.

A ring extension $A\subseteq B$ is an inclusion of rings that makes $B$ an $A$-algebra.

A ring extension $A\subseteq B$ is module-finite, or finite, when $B$ is a finitely generated $A$-module. Every module-finite extension is integral.

For a ring homomorphism $A\to B$ and a prime $\mathfrak{p}\subset A$, the fiber ring at $\mathfrak{p}$ isIts prime ideals correspond to the primes of $B$ whose contraction is $\mathfrak{p}$.

The fiber over $\mathfrak{p}\in \mathrm{Spec}A$ of $\mathrm{Spec}B\to \mathrm{Spec}A$ is the set of primes $\mathfrak{q}\subset B$ satisfying $\mathfrak{q}\cap A=\mathfrak{p}$. It is naturally the spectrum of the fiber ring $B{\otimes }_A\kappa (\mathfrak{p})$.

For a multiplicatively closed set $S$, the localization $S^{-1}R$ formally makes every element of $S$ invertible. Elements are represented by fractions $r/s$.

The localization of a ring $A$ at the powers of $f$ is generated over $A$ by $1/f$. The presentation $A_f\cong A[t]/(ft-1)$ follows from the universal property of localization: both sides universally receive a ring homomorphism from $A$ making $f$ invertible. This proof works with zero divisors and includes the zero ring when $f$ is nilpotent. In particular, a principal open subscheme of an affine scheme has an inclusion which is a morphism of finite type.

For a prime ideal $P$, suppose the kernel of a ring homomorphism $K=\ker (R\to R_P)$ is a finitely generated ideal, say $(a_1,\dots ,a_t)$. Choose $s_i\notin P$ killing $a_i$ and let $f={\prod }_i{s}_i$. Then $f\notin P$ and $fK=0$. If a fraction $a/f^n$ maps to zero in $R_P$, then $a\in K$, so $fa=0$ and the fraction was already zero in $R_f$. Thus the canonical ring homomorphism $R_f\to R_P$ is injective. The empty product $f=1$ handles $K=0$. Every Noetherian ring satisfies the finite generation hypothesis.

Here $S$ consists of the non-zero-divisors of a commutative ring $A$. The localization map $A\to Q(A)$ is injective: a fraction $a/1$ is zero only if some $s\in S$ annihilates $a$. For a reduced ring with finitely many minimal prime ideals, its zero divisors form the union of these primes. Indeed, if $ab=0$ and $a$ avoids every minimal prime, reduction to each domain quotient forces $b$ into their intersection, which is zero. Conversely, for $a$ in a minimal prime ${\mathfrak{p}}_i$, choose $b$ in every other minimal prime but outside ${\mathfrak{p}}_i$, using their incomparability and multiplying suitable elements. Then $b\ne 0$ and $ab=0$.

For a multiplicatively closed set $T$ in a commutative ring $A$, the map sends $(m\otimes n)/t$ to $(m/t)\otimes (n/1)$. An inverse sends $(m/s)\otimes (n/t)$ to $(m\otimes n)/(st)$. The fraction relations and the tensor product of modules balancing relation make these maps well-defined and mutually inverse.

An element $a$ of a commutative ring $R$ is zero if and only if its image in every localization at a prime ideal is zero. If $a\ne 0$, its proper annihilator lies in a maximal ideal $\mathfrak{m}$. No element outside $\mathfrak{m}$ can annihilate $a$, so $a/1\ne 0$ in $R_{\mathfrak{m}}$. This also applies to elements of an $R$-module using localization of a module.

A multiplicative subset $S$ of a ring contains $1$ and is closed under multiplication. It may contain zero, in which case its localization is the zero ring.

A multiplicative subset is saturated if every factor of a product in the subset also lies in it. We include $1$ in a multiplicative subset. A saturated subset containing $0$ is the whole ring. The units form a saturated subset in a commutative ring.

In $\mathbb{Z}$, the saturated multiplicative subset $\{-1,1\}$ has complement equal to the union of $(p)$ over all positive primes. No finite subcollection of prime ideals covers this complement: a prime outside the finite list is a nonunit not in any listed ideal.

The complement of a multiplicative subset is a union of prime ideals exactly when the subset is saturated. If $a\notin S$, saturation ensures $(a)\cap S=\varnothing$; extend this to a prime ideal avoiding a multiplicative subset. Conversely, a product in $S$ cannot have a factor in a prime ideal contained in the complement.

The localization map $\iota :R\to S^{-1}R$ sends every element of $S$ to a unit. If a ring homomorphism $f:R\to T$ also sends every element of $S$ to a unit, there is a unique homomorphism $\tilde{f}:S^{-1}R\to T$ with $\tilde{f}(r/s)=f(r)f(s{)}^{-1}$ and $\tilde{f}\iota =f$.

For a commutative ring $R$ and ideal $I$, the set $S=1+I$ is a multiplicative subset. For $i\in I$ and $s,t\in S$, the numerator $st-ri$ of $1-(r/t)(i/s)$ again lies in $S$. Thus this difference is a unit in the localization of a ring. The unit criterion for the Jacobson radical proves $S^{-1}I\subseteq J(S^{-1}R)$. This requires no Noetherian hypothesis.

For a multiplicative subset $S\subseteq R$ and an $R$-module $M$, the localization $S^{-1}M$ consists of fractions $m/s$, with $m/s=m^{\prime }/s^{\prime }$ when some $u\in S$ satisfies $u(s^{\prime }m-s{m}^{\prime })=0$. It is an $S^{-1}R$-module, and the canonical map is $m\mapsto m/1$. If $M$ is finitely generated over a Noetherian ring, then $S^{-1}M$ is a Noetherian module over $S^{-1}R$.

For a finitely generated module $M=(m_1,\dots ,m_t)$ and a prime ideal $P$, if $M_P=0$, choose $s_i\notin P$ annihilating $m_i$. Their product $f\notin P$ annihilates all of $M$. Conversely such an $f$ becomes a unit in $R_P$, forcing $M_P=0$. ThusThis identifies the support of a module $M$ with the Zariski-closed set $V({\mathrm{Ann}}_R(M))$. For $M=0$, use $f=1$. No Noetherian assumption on $R$ is required.

The fraction relation defining localization of a module gives this criterion. For localization at powers of a single element $f$, a fraction vanishes exactly when some power of $f$ annihilates its numerator. Finitely many such vanishing relations can be cleared with one common exponent. This is distinct from detecting global zero by localizing at every prime ideal.