Commutative algebra · The Lumenary

MotivationH

Equations, ideals and points

A system of polynomial equations $f_1 = \cdots = f_r = 0$ in $n$ unknowns has the same solutions as every equation $\sum h_i f_i = 0$ obtained from it, so what really matters is the ideal $I = (f_1, \dots, f_r) \subseteq k[x_1, \dots, x_n]$ . Three questions arise at once.

Finiteness. Is every system of polynomial equations — possibly infinitely many — equivalent to a finite one? Hilbert's basis theorem (1890) says yes: every ideal of $k[x_1, \dots, x_n]$ is finitely generated.
Geometry. When does an ideal have solutions, and which polynomials vanish on all of them? Hilbert's Nullstellensatz (1893) answers both over an algebraically closed field: solutions exist unless $1 \in I$ , and $f$ vanishes on all solutions exactly when some power $f^m$ lies in $I$ . So radical ideals and algebraic sets determine each other.
Local behaviour. How do we study a curve near one point — say the crossing point of the nodal cubic $y^2 = x^2(x + 1)$ — ignoring the rest? By localising: allowing division by functions that do not vanish at the point.

Commutative algebra is the theory of commutative rings and their modules that answers such questions. It is the algebraic foundation of algebraic geometry and of algebraic number theory, where $\Z$ and the rings of integers of number fields play the role of rings of functions on "arithmetic curves". This lesson assumes modules and quotient rings; all rings are commutative with $1$ .

IntuitionH

Elements are functions; prime ideals are points

The guiding picture, due in this generality to Grothendieck, treats every commutative ring $R$ as a ring of functions on a space, $\operatorname{Spec} R$ , whose points are the prime ideals.

The value of $f \in R$ at the point $\mathfrak{p}$ is the image of $f$ in the field $\kappa(\mathfrak{p}) = \operatorname{Frac}(R/\mathfrak{p})$ . For $R = \Z$ : the integer $6$ has value $0$ at the points $(2)$ and $(3)$ , value $1$ at $(5)$ (in $\mathbb{F}_5$ ), and value $6 \in \Q$ at the "generic point" $(0)$ . So $f$ vanishes at $\mathfrak{p}$ exactly when $f \in \mathfrak{p}$ .
An ideal $I$ cuts out the closed set $V(I) = \{\mathfrak{p} : \mathfrak{p} \supseteq I\}$ of points where all of $I$ vanishes.
A nilpotent $f$ ( $f^m = 0$ ) vanishes at every point but need not be $0$ : it records infinitesimal information. In $k[x]/(x^2)$ the class of $x$ is a non-zero function with value $0$ at the only point — it remembers the direction of a tangent (see the widget below).
Localising at $\mathfrak{p}$ allows denominators that do not vanish at $\mathfrak{p}$ , like passing from polynomials to rational functions defined near a point. Everything that happens away from $\mathfrak{p}$ is forgotten.

For $k[x_1, \dots, x_n]$ with $k = \C$ , the Nullstellensatz will show that the maximal ideals are exactly the points of $\C^n$ , while the other primes correspond to irreducible subvarieties (curves, surfaces, …), each a "generic point" of its variety.

DefinitionsH

Definitions

Definition— prime and maximal ideals; the spectrum

An ideal $\mathfrak{p} \neq R$ is prime if $ab \in \mathfrak{p}$ implies $a \in \mathfrak{p}$ or $b \in \mathfrak{p}$ ; equivalently $R/\mathfrak{p}$ is an integral domain. An ideal $\mathfrak{m} \neq R$ is maximal if no ideal lies strictly between $\mathfrak{m}$ and $R$ ; equivalently $R/\mathfrak{m}$ is a field. The spectrum $\operatorname{Spec} R$ is the set of prime ideals, with the Zariski topology whose closed sets are the $V(I) = \{\mathfrak{p} : \mathfrak{p} \supseteq I\}$ . A ring homomorphism $\varphi : R \to S$ induces a continuous map $\operatorname{Spec} S \to \operatorname{Spec} R$ , $\mathfrak{q} \mapsto \varphi^{-1}(\mathfrak{q})$ .

Definition— radicals

The radical of $I$ is $\sqrt{I} = \{f : f^m \in I \text{ for some } m \geq 1\}$ , and $I$ is radical if $I = \sqrt I$ . The nilradical $\sqrt{0}$ is the set of nilpotent elements; it equals the intersection of all prime ideals. The Jacobson radical $J(R)$ is the intersection of all maximal ideals; $x \in J(R)$ if and only if $1 + xy$ is a unit for every $y \in R$ .

Definition— localisation

A subset $S \subseteq R$ is multiplicative if $1 \in S$ and $S$ is closed under products. The localisation $S^{-1}R$ consists of fractions $r/s$ ( $r \in R$ , $s \in S$ ) with $r/s = r'/s'$ if and only if $t(rs' - r's) = 0$ for some $t \in S$ , added and multiplied as fractions. The map $R \to S^{-1}R$ , $r \mapsto r/1$ , sends $S$ to units and is universal with this property: every ring homomorphism $R \to T$ sending $S$ to units factors uniquely through $S^{-1}R$ . For a module $M$ , $S^{-1}M$ is defined in the same way, and $S^{-1}M \cong S^{-1}R \otimes_R M$ . Important cases: $R_f = S^{-1}R$ for $S = \{1, f, f^2, \dots\}$ , and $R_\mathfrak{p} = S^{-1}R$ for $S = R \setminus \mathfrak{p}$ with $\mathfrak{p}$ prime.

Definition— local ring

A ring is local if it has exactly one maximal ideal $\mathfrak{m}$ ; then $R \setminus \mathfrak{m}$ is the set of units, and $k = R/\mathfrak{m}$ is its residue field.

Definition— Noetherian ring and module

A module is Noetherian if every ascending chain of submodules $N_1 \subseteq N_2 \subseteq \cdots$ is eventually constant, and a ring is Noetherian if it is Noetherian as a module over itself (ascending chains of ideals stabilise).

Definition— integral extension; Krull dimension

If $A \subseteq B$ are rings, $b \in B$ is integral over $A$ if it satisfies a monic polynomial $b^n + a_{n-1}b^{n-1} + \cdots + a_0 = 0$ with $a_i \in A$ ; $B$ is integral over $A$ if every element is. The Krull dimension $\dim R$ is the supremum of the lengths $n$ of chains of prime ideals $\mathfrak{p}_0 \subsetneq \mathfrak{p}_1 \subsetneq \cdots \subsetneq \mathfrak{p}_n$ .

Definition— algebraic sets

For a field $k$ and $T \subseteq k[x_1, \dots, x_n]$ , the algebraic set $V(T) = \{a \in k^n : f(a) = 0 \text{ for all } f \in T\}$ ; for $X \subseteq k^n$ , the ideal $I(X) = \{f : f(a) = 0 \text{ for all } a \in X\}$ . Always $I(X)$ is radical, $T \subseteq I(V(T))$ and $V(I(V(T))) = V(T)$ .

Worked examplesH

Examples and non-examples

Example— Spec ℤ and Spec k[x]

The primes of $\Z$ are $(0)$ and $(p)$ for primes $p$ ; the $(p)$ are maximal and $(0)$ is not, since $(0) \subsetneq (2) \subsetneq \Z$ . So $\dim \Z = 1$ . The closed sets are $\operatorname{Spec}\Z$ and the finite sets of non-zero primes; the point $(0)$ is dense — a generic point. Likewise for $k$ algebraically closed, $\operatorname{Spec} k[x] = \{(x - a) : a \in k\} \cup \{(0)\}$ : the affine line plus a generic point. In any PID, every non-zero prime is maximal.

Example— Spec k[x, y]

For $k$ algebraically closed the primes of $k[x, y]$ are: the maximal ideals $(x - a, y - b)$ (points of the plane, by the Nullstellensatz); the principal ideals $(f)$ with $f$ irreducible (irreducible curves, such as $(y - x^2)$ ); and $(0)$ . The chain $(0) \subsetneq (x) \subsetneq (x, y)$ shows that $(x)$ is prime but not maximal: $k[x,y]/(x) \cong k[y]$ is a domain, not a field. So $\dim k[x, y] \geq 2$ , with equality by the dimension theorem.

Example— Localisations

$\Z_{(p)} = \{a/b : p \nmid b\}$ is local with maximal ideal $p\Z_{(p)}$ ; its primes are $(0)$ and $(p)$ — all other primes of $\Z$ became units.
$\Z[1/2] = R_f$ for $R = \Z$ , $f = 2$ , has the primes $(0)$ and $(p)$ for odd $p$ : the point $(2)$ has been removed from $\operatorname{Spec} \Z$ , an open subset.
$S^{-1}R = 0$ exactly when $0 \in S$ : inverting a nilpotent kills the ring, since $1/1 = f^m/f^m = 0$ .
For $n = p^e m$ with $p \nmid m$ , $(\Z/n)_{(p)} \cong \Z/p^e$ : by the Chinese remainder theorem $\Z/n \cong \Z/p^e \times \Z/m$ , the element $s = (1, 0)$ lies outside $(p)$ and kills the second factor, and every $a$ with $p \nmid a$ is already a unit modulo $p^e$ . The second experiment checks this from the definition for all $n \leq 60$ .

Example— Rings that are not Noetherian

$k[x_1, x_2, x_3, \dots]$ : the chain $(x_1) \subsetneq (x_1, x_2) \subsetneq \cdots$ never stabilises (formalised below), and the ideal $(x_1, x_2, \dots)$ is not finitely generated. It is a subring of the Noetherian field $k(x_1, x_2, \dots)$ , so subrings of Noetherian rings need not be Noetherian.
The ring of continuous functions $C([0, 1])$ : the ideals $I_n = \{f : f = 0 \text{ on } [0, 1/n]\}$ increase strictly.
The ring $\overline{\Z}$ of all algebraic integers: $(2) \subsetneq (2^{1/2}) \subsetneq (2^{1/4}) \subsetneq \cdots$ .

Example— Integral extensions

$\Z[i]$ is integral over $\Z$ ( $i^2 + 1 = 0$ , and sums and products of integral elements are integral). $\Q$ is not integral over $\Z$ : a rational root of a monic integer polynomial is an integer (rational root theorem). The subring $A = k[t^2, t^3] \subseteq k[t]$ , the coordinate ring of the cusp $y^2 = x^3$ (with $x = t^2$ , $y = t^3$ ), has $k[t]$ integral over it ( $t$ is a root of $T^2 - t^2$ ), and $t = t^3/t^2$ lies in the fraction field of $A$ but not in $A$ : the cusp is not normal, and $k[t]$ is its normalisation.

ProofH

The main theorems

Theorem— Proposition 1 (primes, maximals, radicals)

$\mathfrak{m}$ is maximal iff $R/\mathfrak{m}$ is a field; $\mathfrak{p}$ is prime iff $R/\mathfrak{p}$ is a domain. Hence every maximal ideal is prime.
Every proper ideal is contained in a maximal ideal (using Zorn's lemma).
The nilradical is the intersection of all prime ideals.

Proof

(1) Ideals of $R/\mathfrak{m}$ correspond to ideals of $R$ containing $\mathfrak{m}$ ; a non-zero ring is a field iff its only ideals are $0$ and itself. The statement for primes is the definition read in $R/\mathfrak{p}$ . A field is a domain.

(2) The proper ideals containing $I$ , ordered by inclusion, form a non-empty set in which every chain has an upper bound (its union, which is proper because it does not contain $1$ ); Zorn's lemma gives a maximal element.

(3) A nilpotent $f$ satisfies $f^m = 0 \in \mathfrak{p}$ , so $f \in \mathfrak{p}$ for every prime. Conversely, if $f$ is not nilpotent then $S = \{1, f, f^2, \dots\}$ does not contain $0$ , so $R_f \neq 0$ has a maximal ideal $\mathfrak{n}$ by (2), and its preimage in $R$ is a prime not containing $f$ .

∎

Theorem— Proposition 2 (primes of a localisation; local properties)

Let $S \subseteq R$ be multiplicative and $\iota : R \to S^{-1}R$ .

$\mathfrak{q} \mapsto \iota^{-1}(\mathfrak{q})$ is an inclusion-preserving bijection from the primes of $S^{-1}R$ to the primes of $R$ disjoint from $S$ , with inverse $\mathfrak{p} \mapsto S^{-1}\mathfrak{p}$ . In particular $R_\mathfrak{p}$ is local with maximal ideal $\mathfrak{p}R_\mathfrak{p}$ , and its primes correspond to the primes of $R$ contained in $\mathfrak{p}$ .
Localisation is exact: if $M' \to M \to M''$ is exact, so is $S^{-1}M' \to S^{-1}M \to S^{-1}M''$ .
For a module $M$ , $M = 0$ if and only if $M_\mathfrak{m} = 0$ for every maximal ideal $\mathfrak{m}$ .

Proof

(1) The preimage of a prime is a prime, and it misses $S$ because $\iota(S)$ consists of units. Conversely, let $\mathfrak{p}$ be prime with $\mathfrak{p} \cap S = \varnothing$ . Then $S^{-1}\mathfrak{p} = \{a/s : a \in \mathfrak{p}\}$ is a proper ideal: if $a/s = 1$ then $t(a - s) = 0$ for some $t \in S$ , so $ts = ta \in \mathfrak{p}$ , impossible as $ts \in S$ . It is prime: if $(a/s)(b/s') = c/s''$ with $c \in \mathfrak{p}$ then $t(abs'' - css') = 0$ , so $tabs'' \in \mathfrak{p}$ and, since $t, s'' \notin \mathfrak{p}$ , $a \in \mathfrak{p}$ or $b \in \mathfrak{p}$ . The same computation shows $\iota^{-1}(S^{-1}\mathfrak{p}) = \mathfrak{p}$ , and every ideal $J$ of $S^{-1}R$ satisfies $J = S^{-1}\iota^{-1}(J)$ (if $a/s \in J$ then $a/1 \in J$ ). For $S = R \setminus \mathfrak{p}$ , the primes of $R_\mathfrak{p}$ correspond to primes contained in $\mathfrak{p}$ , and the largest, $\mathfrak{p}R_\mathfrak{p}$ , is the unique maximal ideal.

(2) If $g \circ f = 0$ then the localised composite is $0$ . Let $m/s \in S^{-1}M$ map to $0$ : then $t\,g(m) = g(tm) = 0$ for some $t \in S$ , so $tm = f(m')$ , and $m/s = f(m')/(ts)$ .

(3) Suppose $M \neq 0$ and pick $m \neq 0$ . Its annihilator $\{r : rm = 0\}$ is a proper ideal, contained in some maximal $\mathfrak{m}$ . Then $m/1 \neq 0$ in $M_\mathfrak{m}$ , because $tm = 0$ would put $t \in \operatorname{Ann}(m) \subseteq \mathfrak{m}$ .

∎

Part (3), with exactness, is the local–global principle: injectivity, surjectivity and exactness of maps of modules can all be checked after localising at every maximal ideal. This is how problems are reduced to local rings, where Nakayama's lemma is available.

Theorem— Proposition 3 (Noetherian rings)

For a ring $R$ the following are equivalent: (a) $R$ is Noetherian; (b) every ideal of $R$ is finitely generated; (c) every non-empty set of ideals has a maximal element. Quotients and localisations of Noetherian rings are Noetherian, and finitely generated modules over a Noetherian ring are Noetherian.

Proof

(a ⇒ c) If a non-empty set $\Sigma$ had no maximal element, we could choose $I_1 \subsetneq I_2 \subsetneq \cdots$ in $\Sigma$ (dependent choice). (c ⇒ b) Among the finitely generated ideals contained in $I$ choose a maximal one, $J$ ; if $a \in I \setminus J$ then $J + (a)$ is larger and finitely generated, so $J = I$ . (b ⇒ a) Given a chain $I_1 \subseteq I_2 \subseteq \cdots$ , the union $I$ is an ideal, generated by finitely many elements, which all lie in some $I_n$ ; then $I_n = I_{n+1} = \cdots = I$ . Ideals of $R/I$ are ideals of $R$ containing $I$ , and ideals of $S^{-1}R$ are $S^{-1}J$ by Proposition 2, so (b) passes to both. For modules: $R^n$ is Noetherian by induction using $0 \to R^{n-1} \to R^n \to R \to 0$ (a chain in $R^n$ stabilises if its intersections with $R^{n-1}$ and its images in $R$ do), and quotients of Noetherian modules are Noetherian.

∎

Theorem— Theorem 4 (Hilbert's basis theorem)

If $R$ is Noetherian, then so is $R[x]$ . Hence $R[x_1, \dots, x_n]$ — in particular $k[x_1, \dots, x_n]$ and $\Z[x_1, \dots, x_n]$ — and every finitely generated $R$ -algebra are Noetherian.

Proof

Let $I \subseteq R[x]$ be an ideal; we show it is finitely generated. For $d \geq 0$ let $L_d \subseteq R$ be the set of leading coefficients of polynomials in $I$ of degree $d$ , together with $0$ . Each $L_d$ is an ideal (sums and multiples of leading coefficients of degree- $d$ elements are leading coefficients of degree- $d$ elements, or $0$ ), and $L_d \subseteq L_{d+1}$ (multiply by $x$ ). Since $R$ is Noetherian, the chain stabilises at some $L_N$ , and each of $L_0, \dots, L_N$ is finitely generated. For each $d \leq N$ and each generator of $L_d$ , choose a polynomial in $I$ of degree $d$ with that leading coefficient; let $J \subseteq I$ be the ideal generated by these finitely many polynomials.

We show by induction on degree that every $f \in I$ lies in $J$ . If $f = 0$ there is nothing to do. Let $\deg f = d$ with leading coefficient $a \in L_d$ . If $d \leq N$ , write $a = \sum r_i a_i$ with $a_i$ the chosen generators of $L_d$ and $g_i \in J$ the chosen polynomials of degree $d$ ; then $f - \sum r_i g_i \in I$ has degree less than $d$ . If $d > N$ , then $a \in L_d = L_N$ , so $a = \sum r_i a_i$ with $a_i$ the generators of $L_N$ , and $f - \sum r_i x^{d - N} g_i \in I$ has degree less than $d$ . By induction this difference lies in $J$ , so $f \in J$ . The statements about several variables follow by induction, since $R[x_1, \dots, x_n] = R[x_1, \dots, x_{n-1}][x_n]$ , and finitely generated algebras are quotients of polynomial rings.

∎

The proof is non-constructive — it does not say how to find generators — which famously troubled Paul Gordan. Gröbner bases (see the computation lens below) make it effective for polynomial rings over a field.

Theorem— Theorem 5 (Nakayama's lemma)

Let $M$ be a finitely generated $R$ -module and $I$ an ideal contained in the Jacobson radical $J(R)$ . If $IM = M$ then $M = 0$ . In particular, over a local ring $(R, \mathfrak{m})$ : if $\mathfrak{m}M = M$ then $M = 0$ ; and elements $m_1, \dots, m_n \in M$ generate $M$ if and only if their images span the $R/\mathfrak{m}$ -vector space $M/\mathfrak{m}M$ .

Proof

The determinant trick. Let $m_1, \dots, m_n$ generate $M$ . Since $M = IM$ , each $m_i = \sum_j a_{ij} m_j$ with $a_{ij} \in I$ . So the matrix $A = (a_{ij})$ satisfies $(\mathrm{Id} - A)\,\mathbf{m} = 0$ , where $\mathbf{m} = (m_1, \dots, m_n)^{\mathsf T}$ . Multiplying by the adjugate of $\mathrm{Id} - A$ gives $\det(\mathrm{Id} - A)\, m_i = 0$ for every $i$ . Expanding the determinant, $r = \det(\mathrm{Id} - A) = 1 + x$ with $x \in I$ . So $r M = 0$ with $r - 1 \in I$ .

Conclusion. Since $x = r - 1 \in J(R)$ , the element $r = 1 + x$ is a unit, and $M = r^{-1} r M = 0$ .

Generators over a local ring. If the images of $m_1, \dots, m_n$ span $M/\mathfrak{m}M$ , let $N = Rm_1 + \cdots + Rm_n$ . Then $M = N + \mathfrak{m}M$ , so the finitely generated module $M/N$ satisfies $\mathfrak{m}(M/N) = M/N$ , hence $M/N = 0$ . The converse is clear.

∎

So over a local ring, all minimal generating sets of a finitely generated module have the same size, $\dim_k M/\mathfrak{m}M$ — a basis-free substitute for dimension. Finite generation is essential: $\Q$ is a $\Z_{(p)}$ -module with $p\Q = \Q \neq 0$ .

Theorem— Theorem 6 (Hilbert's Nullstellensatz)

Let $k$ be an algebraically closed field.

(Weak form) The maximal ideals of $k[x_1, \dots, x_n]$ are exactly the ideals $\mathfrak{m}_a = (x_1 - a_1, \dots, x_n - a_n)$ , $a \in k^n$ . Consequently, if $I$ is a proper ideal then $V(I) \neq \varnothing$ .
(Strong form) For every ideal $I$ , $\;I(V(I)) = \sqrt{I}$ .

Hence $V$ and $I$ are mutually inverse, inclusion-reversing bijections between radical ideals of $k[x_1, \dots, x_n]$ and algebraic subsets of $k^n$ ; prime ideals correspond to irreducible algebraic sets, and maximal ideals to points.

Proof

We prove it for $k = \C$ (the same argument works for every uncountable algebraically closed field); the general case needs Zariski's lemma — a field that is finitely generated as a $k$ -algebra is a finite extension of $k$ — proved in Atiyah–Macdonald, Chapters 5 and 7, or Eisenbud, Theorem 4.19.

Weak form. Each $\mathfrak{m}_a$ is maximal, being the kernel of the surjection $f \mapsto f(a)$ onto the field $\C$ . Conversely let $\mathfrak{m}$ be maximal and $K = \C[x_1, \dots, x_n]/\mathfrak{m}$ , a field containing $\C$ . As a $\C$ -vector space, $K$ is spanned by the images of the countably many monomials. Suppose some $t \in K$ is not in $\C$ . If $t$ were transcendental over $\C$ , the uncountably many elements $1/(t - c)$ , $c \in \C$ , would be linearly independent: a relation $\sum_{i=1}^{r} \lambda_i/(t - c_i) = 0$ with distinct $c_i$ gives, after clearing denominators, $\sum_i \lambda_i \prod_{j \neq i}(t - c_j) = 0$ , a polynomial identity in the transcendental $t$ , and substituting $c_i$ for $t$ in that polynomial gives $\lambda_i = 0$ . This contradicts countable dimension. So $t$ is algebraic: $p(t) = 0$ for some non-zero $p \in \C[T]$ , which factors as $p = c\prod (T - c_j)$ ; since $K$ is a field, $t = c_j$ for some $j$ , a contradiction. Hence $K = \C$ , and if $a_i \in \C$ is the image of $x_i$ , then $x_i - a_i \in \mathfrak{m}$ for all $i$ , so $\mathfrak{m}_a \subseteq \mathfrak{m}$ , and equality holds because $\mathfrak{m}_a$ is maximal. Finally, a proper ideal $I$ lies in some maximal $\mathfrak{m}_a$ (Proposition 1), and then every element of $I$ vanishes at $a$ .

Strong form (Rabinowitsch's trick). Clearly $\sqrt{I} \subseteq I(V(I))$ . Conversely, let $f$ vanish on $V(I)$ . By Theorem 4, $I = (g_1, \dots, g_r)$ . In $\C[x_1, \dots, x_n, t]$ consider $J = (g_1, \dots, g_r, 1 - tf)$ . It has no zeros: at a common zero $(a, b)$ of the $g_i$ we have $f(a) = 0$ , so $1 - bf(a) = 1$ . By the weak form $J$ is not proper: $1 = \sum_i h_i\, g_i + h\,(1 - tf)$ for some polynomials $h_i, h$ . Substitute $t = 1/f$ in the field $\C(x_1, \dots, x_n)$ (if $f = 0$ there is nothing to prove): $1 = \sum_i h_i(x, 1/f)\, g_i(x)$ . Multiplying by a high power $f^N$ clears all denominators and gives $f^N = \sum_i \tilde h_i g_i \in I$ .

The bijection follows: $V(I(X)) = X$ for algebraic sets and $I(V(I)) = I$ for radical ideals. $X$ is irreducible iff $I(X)$ is prime, by the usual argument with $X \subseteq V(f) \cup V(g) = V(fg)$ .

∎

The hypothesis "algebraically closed" is essential: over $\R$ , the ideal $(x^2 + 1)$ of $\R[x]$ is maximal (with $\R[x]/(x^2 + 1) \cong \C$ ) but not of the form $\mathfrak{m}_a$ , and $V(x^2 + 1) = \varnothing$ although $1 \notin (x^2 + 1)$ .

Theorem— Theorem 7 (dimension; stated)

Let $k$ be a field. Then $\dim k[x_1, \dots, x_n] = n$ . More generally, if $A$ is a finitely generated $k$ -algebra and a domain, then $\dim A$ equals the transcendence degree of $\operatorname{Frac}(A)$ over $k$ , and every maximal chain of primes in $A$ has this length.

Proof sketch. The chain $0 \subsetneq (x_1) \subsetneq (x_1, x_2) \subsetneq \cdots \subsetneq (x_1, \dots, x_n)$ consists of primes (each quotient is a polynomial ring, a domain), so $\dim \geq n$ . For $\leq n$ one uses Noether normalisation — $A$ is integral over a polynomial subring $k[y_1, \dots, y_d]$ with $d = \operatorname{trdeg}$ — and the theorems of lying over, going up and incomparability (Cohen–Seidenberg) for integral extensions, which show that $A$ and $k[y_1, \dots, y_d]$ have the same dimension; then induct on $d$ . See Atiyah–Macdonald, Chapters 5 and 11, or Eisenbud, Chapters 8–13. Mathlib proves $\dim R[x_1, \dots, x_n] = \dim R + n$ for every Noetherian $R$ (see the formal section).

Theorem— Theorem 8 (primary decomposition; stated)

In a Noetherian ring every ideal is a finite intersection of primary ideals ( $\mathfrak{q}$ is primary if $ab \in \mathfrak{q}$ and $a \notin \mathfrak{q}$ imply $b^m \in \mathfrak{q}$ for some $m$ ). In a minimal decomposition $I = \mathfrak{q}_1 \cap \cdots \cap \mathfrak{q}_s$ the primes $\sqrt{\mathfrak{q}_i}$ are distinct and uniquely determined by $I$ (Lasker 1905 for polynomial rings, Noether 1921).

For example, in $k[x, y]$ , $(x^2, xy) = (x) \cap (x^2, y)$ . The zero set is the line $x = 0$ , but the decomposition also records an embedded primary component at the origin: $x$ vanishes on the line, $x \notin (x^2, xy)$ , yet $x^2 \in (x^2, xy)$ , and $y$ times the nilpotent $x$ is $0$ — the ideal remembers a fuzzy point stuck to the line. The first experiment detects exactly this with Gröbner bases.

Explore

Explore: tangency is a nilpotent

The parabola $y = x^2$ meets the horizontal line $y = c$ where $x^2 = c$ . Algebraically, the intersection is described by the ring

k[x, y]/(y - x^2,\; y - c) \;\cong\; k[x]/(x^2 - c).

Move the slider. For $c > 0$ there are two intersection points $(\pm\sqrt{c}, c)$ , and $k[x]/(x^2 - c) \cong k \times k$ (Chinese remainder theorem): two reduced points. As $c \to 0$ they merge, and at $c = 0$ the ring becomes $k[x]/(x^2)$ , which is not reduced: the class of $x$ is a non-zero nilpotent, remembering that the two points collided along the $x$ -direction — the line is tangent. For $c < 0$ there are no real points, but over $\C$ there are still two, at $x = \pm i\sqrt{-c}$ : this is why the Nullstellensatz needs an algebraically closed field.

Two points merging into a double point

At c = 0 the intersection ring k[x]/(x²) has a nilpotent: the line is tangent.

y = x²
y = c

cc = 1

Trace along xx = 1

y = x²: y = 1
y = c: y = 1

The first experiment confirms this: $x \notin (y - x^2, y)$ but $x^2 \in (y - x^2, y)$ , whereas for the secant $y = 1$ the ideal is already radical.

Formal mathematicsF

Commutative algebra in Lean and Mathlib

Mathlib's commutative algebra library is large. We start with the difference between prime and maximal ideals, proved directly, and the library fact that maximal ideals are prime:

zero_prime_not_maximalLean 4Proved here from basic factsFormally verified

/-- **A prime ideal need not be maximal.** In `ℤ` the zero ideal is prime (because `ℤ` is an
integral domain: `ab = 0` forces `a = 0` or `b = 0`), but it is not maximal, since
`(0) ⊊ (2) ⊊ ℤ`. In a PID every *non-zero* prime ideal is maximal, so `(0)` is the only
exception in `ℤ`. -/
theorem zero_prime_not_maximal : (⊥ : Ideal ℤ).IsPrime ∧ ¬ (⊥ : Ideal ℤ).IsMaximal := by
  refine ⟨Ideal.isPrime_bot, fun hmax => ?_⟩
  -- `(2)` lies strictly between `(0)` and `ℤ`
  have h2 : (⊥ : Ideal ℤ) < Ideal.span {2} := by
    rw [bot_lt_iff_ne_bot, Ne, Ideal.span_singleton_eq_bot]
    norm_num
  have htop : Ideal.span {(2 : ℤ)} = ⊤ := hmax.1.2 _ h2
  rw [Ideal.span_singleton_eq_top] at htop
  exact absurd (Int.isUnit_iff.mp htop) (by norm_num)