Commutative algebra

Commutative algebra studies commutative rings, ideals, modules, and their spectra.

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

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 $r/s$.

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 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 $S^{-1}R$.

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.

For a multiplicative subset $S$ of a commutative ring $A$, localization preserves short exact sequences of $A$-modules. Surjectivity follows by lifting a numerator. If a fraction maps to zero, some element of $S$ annihilates its image, so multiplying its numerator by that element puts it into the original kernel. For an injective original map, the same criterion shows its localized kernel is zero. Thus $S^{-1}A$ is a flat module over $A$.

Let $R$ be Noetherian, $I\subseteq R$ an ideal, $S=1+I$, and $M$ a finitely generated $R$-module. ThenIf $(1+a)m=0$ with $a\in I$, then $m=(-a{)}^{j}m\in I^{j}M$ for every $j$. Conversely, apply the Artin-Rees lemma to $Rm\subseteq M$. For sufficiently large $c$ it givesso $m=am$ for some $a\in I$ and $(1-a)m=0$.

Letbe the semigroup algebra inside $k[t^{\mathbb{Q}}]$, and let $I$ be generated by the $t^q$ with $q>0$. This integral domain is not Noetherian, sinceMoreover $I^2=I$, because $t^q=(t^{q/2}{)}^2$. Thus ${\bigcap }_{j\ge 1}{I}^j=I\ne 0$, whereas the map $R\to (1+I{)}^{-1}R$ is injective because $R$ is a domain.

For a set $\pi$ of prime numbers, the integer localizationconsists of fractions whose denominator has prime factors only in $\pi$. As an additive group it is divisible by every prime in $\pi$.

If $A_{\neg p}=\mathbb{Z}[q^{-1}:q\ne p]$, its finite quotients are exactly the cyclic $p$-groups $\mathbb{Z}/p^n\mathbb{Z}$. Divisibility by every $q\ne p$ excludes other prime torsion, while a quotient of exponent $p^n$ factors through $A_{\neg p}/p^n{A}_{\neg p}\cong \mathbb{Z}/p^n\mathbb{Z}$.

Extension and contraction give inverse inclusion-preserving bijections between prime ideals of $S^{-1}R$ and prime ideals of $R$ disjoint from $S$:

For $S=1+I$, the prime ideal correspondence for localization identifies $\mathrm{Spec}{S}^{-1}R$ withIndeed, $\mathfrak{p}\cap (1+I)=\varnothing$ exactly when no element of $I$ is congruent to $-1$ modulo $\mathfrak{p}$, which is exactly $\mathfrak{p}+I\ne R$.

Every localization of a Noetherian ring is Noetherian. Each ideal $J\subseteq S^{-1}R$ is the extension of its contraction to $R$, whose finite generating set consequently generates $J$.

If $R$ is a Dedekind domain and $0\notin S$, then $S^{-1}R$ is an integrally closed Noetherian domain of dimension at most one. It is therefore either a Dedekind domain or, when its dimension is zero, a field.

Localization at a prime ideal $\mathfrak{p}$ means inverting $R\setminus \mathfrak{p}$. The resulting local ring $R_{\mathfrak{p}}$ has maximal ideal $\mathfrak{p}{R}_{\mathfrak{p}}$.

An element $a$ of an $R$-algebra is integral over $R$ when it satisfies a monic polynomial with coefficients in $R$.

An element $x$ of an $A$-algebra is an integral element precisely when $A[x]$ is a finite $A$-module. More generally, a finite $A$-submodule containing one and stable under multiplication by $x$ proves integrality by the determinant trick. Conversely a monic polynomial equation reduces all powers of $x$ to finitely many generators.

An extension $R\subseteq A$ is integral when every element of $A$ is integral over $R$.

If a field $L$ is integral over a subdomain $A$, every nonzero $a\in A$ has an inverse in $L$ satisfying a monic equation over $A$. Multiplying that equation by $a^{r-1}$ expresses $a^{-1}$ as a polynomial in $a$ with coefficients in $A$. Hence $a^{-1}\in A$, and $A$ is itself a field. The embedding and integrality assumptions are essential; a fraction field is not generally integral over its source domain.

An integral extension $A\subset B$ need not give an integral extension $A_{P\cap A}\subset B_P$. For example, take $\mathbb{Z}\subset \mathbb{Z}[i]$ and $P=(2+i)$, whose contraction is $(5)$. The element $1/(2-i)$ belongs to $\mathbb{Z}[i{]}_P$ but is not integral over ${\mathbb{Z}}_{(5)}$: multiplying a hypothetical monic equation by a power of $2-i$ and reducing modulo the other Gaussian prime ideal $(2-i)$ gives $1=0$ in ${\mathbb{F}}_5$. Localizing both rings by the same multiplicative subset does preserve integrality.

If $A\subseteq B$ is an integral extension, then $\dim A=\dim B$. The incomparability theorem for integral extensions contracts every strict prime chain in $B$ to a strict chain in $A$, giving $\dim B\le \dim A$. Conversely, the Lying-over theorem starts over the bottom of any prime chain in $A$, and repeated use of the Going-up theorem lifts the whole chain to $B$.

If $f\in k[Y][X]$ is monic of positive degree in $X$, monic division makes $k[Y][X]/(f)$ free of rank ${\mathrm{deg}}_{X}f$ over $k[Y]$. The inclusion of $k[Y]$ is injective and the quotient is an integral extension, so its Krull dimension is one. This argument applies even when the polynomial factors or the field has positive characteristic.

If $A\subseteq B$ is an integral extension, then every prime ideal $\mathfrak{p}$ of $A$ is the contraction of a prime ideal $\mathfrak{q}$ of $B$. Thus the contraction map $\mathrm{Spec}B\to \mathrm{Spec}A$ is surjective.

For an integral extension $A\subseteq B$, primes ${\mathfrak{p}}_1\subseteq {\mathfrak{p}}_2$ of $A$, and a prime ${\mathfrak{q}}_1$ of $B$ over ${\mathfrak{p}}_1$, there is a prime ${\mathfrak{q}}_2\supseteq {\mathfrak{q}}_1$ over ${\mathfrak{p}}_2$.

If $A\subseteq B$ is integral and ${\mathfrak{q}}_1\subseteq {\mathfrak{q}}_2$ are primes of $B$ with equal contractions to $A$, then ${\mathfrak{q}}_1={\mathfrak{q}}_2$. Consequently strict prime chains remain strict under contraction.

Let $A\subseteq B$ be integral and let $I\subseteq J$ be ideals of $B$. If $I$ is prime and $I\cap A=J\cap A$, then $I=J$. After quotienting by $I$, a nonzero element of $J/I$ has an integral equation of least degree whose nonzero constant term lies in $(J/I)\cap (A/(I\cap A))$, a contradiction.

If $A\subseteq B$ is integral and $\mathfrak{n}$ is a maximal ideal of $B$, then $\mathfrak{n}\cap A$ is maximal in $A$. Indeed, $B/\mathfrak{n}$ is an integral domain integral over $A/(\mathfrak{n}\cap A)$, and a subring over which a field is integral is itself a field.

For an integral extension $A\subseteq B$ and $\mathfrak{p}\in \mathrm{Spec}A$, localization gives a bijectionIt combines the prime ideal correspondence for localization with the contraction of a maximal ideal under an integral extension.

If $a_1,\dots ,a_n$ are integral over $R$, then $R[a_1,\dots ,a_n]$ is a finitely generated $R$-module. In particular, every element of this subalgebra is integral over $R$.

An integral domain is integrally closed when every element of its fraction field that is integral over the domain already belongs to the domain.

The integral closure of $R$ in an $R$-algebra $A$ is the subring of elements of $A$ integral over $R$.

For a normal Noetherian ring $A$ that is an integral domain, the integral closure in a finite separable field extension of its fraction field is finite as an $A$-module. An integral field basis by denominator clearing gives a finite lattice. Integral traces place the closure in its finite trace-dual lattice; Noetherianity makes that submodule finitely generated.

If $A$ is an integral domain with fraction field $K$ and $L/K$ is finite, a $K$-basis of $L$ can be rescaled to consist of integral elements over $A$. For a monic algebraic equation, choose $a\in A$ clearing the coefficient denominators; replacing $u$ by $au$ multiplies the coefficient of degree $r-j$ by $a^j$, putting it in $A$.

For a domain $R$ with fraction field $K$, its integral closure is the intersection of all valuation rings of $K$ containing $R$. Integral elements belong to every such ring because valuation rings are integrally closed. For a nonintegral $x$, the ideal $x^{-1}R[x^{-1}]$ is proper, since its containing $1$ would give a monic equation for $x$. Localize at a maximal ideal containing $x^{-1}$ and apply the valuation domination lemma. The resulting ring contains $R$ but not $x$, proving the reverse inclusion.

If a graded ring $A$ is a graded extension of a graded subring $R$, then the integral closure of $R$ in $A$ is graded: every homogeneous component of an integral element is integral. This follows by separating the extremal degrees in a monic integral equation and inducting on the number of nonzero homogeneous components.

Let $R\subseteq A$ be an integral extension of domains with $R$ integrally closed. Given primes ${\mathfrak{p}}_0\subseteq {\mathfrak{p}}_1$ of $R$ and a prime ${\mathfrak{q}}_1$ of $A$ over ${\mathfrak{p}}_1$, there is a prime ${\mathfrak{q}}_0\subseteq {\mathfrak{q}}_1$ over ${\mathfrak{p}}_0$.

The Krull dimension of a commutative ring is the supremum of the lengths of strict chains of prime ideals.

A strictly increasing finite sequence of prime ideals. Its length is the number of strict inclusions, so the displayed chain has length $s$. The Krull dimension of a ring is the supremum of such lengths; the height of a prime ideal restricts to chains ending at that prime.

If $A$ is a finitely generated algebra over a field $k$, then there are algebraically independent elements $y_1,\dots ,y_d\in A$ such that $A$ is finite, hence integral, over $k[y_1,\dots ,y_d]$.

To normalize $k[a_1,\dots ,a_n]$ over an arbitrary field, choose a nonzero polynomial relation and a base $M$ larger than all its exponents. The substitutions $b_i=a_i-a_n^{M^i}$ give distinct top weights to its monomials, so its highest power of $a_n$ has a nonzero scalar coefficient. It becomes a monic polynomial equation, making the algebra finite over $k[b_1,\dots ,b_{n-1}]$. Induction proves the Noether normalization lemma, including over finite fields.

For a nonconstant polynomial over any field, use triangular substitutions $x_i=y_i+y_n^{B^i}$, $x_n=y_n$, where $B$ exceeds every exponent in its support. Base-$B$ weights distinguish all original monomials, so the transformed polynomial has a unique highest $y_n$ power with coefficient in $k^{\times }$. Rescale to make it monic. The monic polynomial quotient is finite free over $k[y_1,\dots ,y_{n-1}]$, yielding a finite flat projection. The nonlinear substitution works over finite fields as well as infinite ones; a generic linear-direction argument alone would not do so.

Letover a field in which $3$ is invertible. Put $u=x-z$ and $v=y-z$. ThensoDivision by this monic quadratic makes $T$ a free $k[u,v]$-module with basis $1,z$. Thus $k[u,v]$ is a polynomial algebra and $T$ is integral over it.

The height of an ideal $I$ is the infimum of the heights of prime ideals containing it:The height of a prime is the supremum of the lengths of strict prime chains ending at that prime.

The height of a prime ideal is the supremum of lengths of strict prime chains ending at it, equivalently the Krull dimension of its localization at a prime ideal. The Krull height theorem bounds the height of a minimal prime over an $n$-generated ideal by $n$.

For a prime ideal $P$ of a polynomial ring over a field, the quotient's Krull dimension is the transcendence degree of its fraction field, and its height is $n$ minus that degree. This dimension theorem identifies algebraic codimension with the length of prime chains. It must not be applied indiscriminately to every Noetherian ring without its polynomial-ring hypotheses.

If an ideal $I$ is contained in a finite union of prime ideals ${\mathfrak{p}}_1\cup \cdots \cup {\mathfrak{p}}_r$, then $I$ is contained in one of the ${\mathfrak{p}}_i$. Equivalently, if none of the primes contains $I$, one can choose an element of $I$ outside all of them.

In a Noetherian ring, every prime ideal minimal over a proper principal ideal has height at most one.

For a proof, suppose instead that ${\mathfrak{p}}_0⊊\mathfrak{q}⊊\mathfrak{p}$ and that $\mathfrak{p}$ is minimal over $(a)$. Quotient by ${\mathfrak{p}}_0$ and localize at $\mathfrak{p}$; one may assume that $R$ is a Noetherian local domain, $0⊊\mathfrak{q}⊊\mathfrak{p}$ with $\mathfrak{p}$ maximal, and $\mathfrak{p}$ is the only prime containing $(a)$. For the symbolic powerseach ${\mathfrak{q}}^{(n)}$ is $\mathfrak{q}$-primary. The quotient $R/(a)$ is a zero-dimensional Noetherian ring and hence Artinian, so the descending chain $({\mathfrak{q}}^{(n)}+(a))/(a)$ stabilizes. For large $n$, writing an element of ${\mathfrak{q}}^{(n)}$ modulo ${\mathfrak{q}}^{(n+1)}$ and using $a\notin \mathfrak{q}$ givesThe Nakayama lemma applied to ${\mathfrak{q}}^{(n)}/{\mathfrak{q}}^{(n+1)}$ gives ${\mathfrak{q}}^{(n)}={\mathfrak{q}}^{(n+1)}$. Localizing at $\mathfrak{q}$ would therefore make the powers of the nonzero maximal ideal $\mathfrak{q}{R}_{\mathfrak{q}}$ stabilize at a nonzero ideal, contradicting the Krull intersection theorem for the Noetherian local domain $R_{\mathfrak{q}}$.