---
title: A Glimpse of Commutative Algebra and Algebraic Geometry
module: "Capstone: Where Algebra Goes Next"
moduleNumber: 12
lessonNumber: 1
order: 1201
summary: >
  Commutative algebra reads geometry off the ring of polynomial functions. The
  dictionary runs through Noetherian rings and the ascending chain condition,
  Hilbert's Basis Theorem, affine algebraic sets and the two maps connecting
  ideals to zero sets, radicals, the Zariski topology, and Hilbert's
  Nullstellensatz, which over an algebraically closed field makes radical ideals
  and algebraic sets the same object.
topics: ["Capstone: Where Algebra Goes Next"]
sources:
  - book: Dummit & Foote
    ref: "Ch. 15 — Commutative Rings and Algebraic Geometry; §15.1 Noetherian Rings and Affine Algebraic Sets"
  - book: Dummit & Foote
    ref: "§15.2 Radicals and Affine Varieties; §15.3 Integral Extensions and Hilbert's Nullstellensatz"
draft: false
---

Solving a system of polynomial equations and studying an ideal in a polynomial
ring are the same activity viewed from two sides. The equations $f_1 = \cdots =
f_m = 0$ carve out a set of points; the polynomials $f_1, \dots, f_m$ generate an
ideal. Commutative algebra studies that correspondence; algebraic
geometry is its geometric side. The dictionary between them rests on the ring
theory of
[ideals and quotient rings](/abstract-algebra/ring-theory/ideals-quotients-and-homomorphisms)
and on the
[Gröbner-basis methods](/abstract-algebra/factorization-and-polynomials/irreducibility-criteria-and-groebner)
that make the computations effective. Throughout, $R$ denotes a commutative ring
with $1 \neq 0$, and $k$ denotes a field.

## Noetherian rings

The polynomial rings $k[x_1, \dots, x_n]$ are the ambient rings of algebraic
geometry, and their central finiteness property is a condition on ascending
chains of ideals.

> **Definition (Noetherian ring).** A commutative ring $R$ is **Noetherian**, or
> satisfies the **ascending chain condition** (A.C.C.) on ideals, if every
> increasing chain of ideals
> $$
> I_1 \subseteq I_2 \subseteq I_3 \subseteq \cdots
> $$
> is eventually constant: there is an $m$ with $I_k = I_m$ for all $k \geq m$.

This is the ring-theoretic special case of the module condition from the
[structure theorem over a PID](/abstract-algebra/modules-over-pids/structure-theorem-over-pids),
applied to $R$ as a module over itself, where the submodules are the
ideals themselves.[^df-noeth] Three conditions turn out to be equivalent, and each is used in
practice.

> **Theorem (Characterizations of Noetherian).** For a commutative ring $R$ the
> following are equivalent:
> - $R$ satisfies A.C.C. on ideals;
> - every nonempty set of ideals of $R$ has a maximal element under inclusion;
> - every ideal of $R$ is finitely generated.

$$
% caption: In a Noetherian ring every ascending chain of ideals is eventually
% constant; past some index the nested ideals stop growing.
\begin{tikzpicture}[font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[acc, fill=acc!8] (0,0) rectangle (5.0,3.2);
\draw[acc, fill=acc!11] (0.45,0.4) rectangle (4.2,2.75);
\draw[acc, fill=acc!14] (0.9,0.8) rectangle (3.4,2.35);
\draw[black, fill=acc!16] (1.35,1.25) rectangle (2.6,1.95);
\node[anchor=west] at (0.08,2.98) {$I_4 = I_5 = I_6$};
\node[anchor=west] at (0.5,2.55) {$I_3$};
\node[anchor=west] at (0.95,2.15) {$I_2$};
\node at (1.97,1.6) {$I_1$};
\node[anchor=west, black] at (5.3,1.6) {the chain stops at $I_4$};
\end{tikzpicture}
$$

The equivalence with finite generation is what connects A.C.C. to geometry: a
Noetherian ring has no ideal that needs infinitely many generators. The
class of Noetherian rings is broad and closed under the constructions that build
polynomial rings.

- **Every PID is Noetherian**, since every ideal is generated by one element. In
  particular $\mathbb{Z}$, $k[x]$, and the Gaussian integers $\mathbb{Z}[i]$ are
  Noetherian; this recovers the
  [Euclidean-domain hierarchy](/abstract-algebra/factorization-and-polynomials/euclidean-domains-pids-ufds)
  from the finiteness side.
- **Quotients stay Noetherian.** Any homomorphic image $R/I$ of a Noetherian ring
  is Noetherian, because a chain in $R/I$ pulls back to a chain in $R$.
- **Polynomial rings stay Noetherian**, by the next theorem.

> **Theorem (Hilbert's Basis Theorem).** If $R$ is a Noetherian ring, then the
> polynomial ring $R[x]$ is Noetherian.

Applied $n$ times, the Basis Theorem shows that $k[x_1, \dots, x_n]$ is
Noetherian, so **every ideal of a polynomial ring over a field is finitely
generated**.[^df-basis] Not every ring is Noetherian. The polynomial ring
$\mathbb{Z}[x_1, x_2, \dots]$
in infinitely many variables fails A.C.C.: the ideal $(x_1, x_2, \dots)$ needs
infinitely many generators. The ring of continuous real functions on $[0,1]$ also
fails it. A Noetherian ring may still have infinite _descending_ chains — in
$\mathbb{Z}$,
$$
(2) \supsetneq (4) \supsetneq (8) \supsetneq \cdots,
$$
so A.C.C. and the descending chain condition are genuinely different; rings with
the descending condition are called Artinian and form a separate, smaller class.

> **Worked example (finitely many equations suffice).** Consider the infinite set
> $$
> S = \{\, y - x,\; y - x^2,\; y - x^3,\; \dots \,\} \subseteq k[x, y].
> $$
> A point lies in $Z(S)$ iff $y = x^n$ for every $n \geq 1$; comparing $n = 1$ with
> $n = 2$ forces $x = x^2$, so $x \in \{0, 1\}$ and $Z(S) = \{(0,0),\, (1,1)\}$. The
> Basis Theorem guarantees the ideal $(S)$ is finitely generated, and two generators
> already suffice. Subtracting the first two elements gives $(y - x) - (y - x^2) =
> x^2 - x$, and every generator lies in $(y - x,\; x^2 - x)$ because
> $$
> y - x^n = (y - x) - (x^n - x), \qquad x^n - x = (x^2 - x)(x^{n-2} + \cdots + 1).
> $$
> So $(S) = (y - x,\; x^2 - x)$: the infinite system collapses to the two equations
> $y = x$ and $x^2 = x$.[^df-basis]

### Finitely generated algebras

A ring $R$ containing a field $k$ in its center is a **$k$-algebra**. It is
**finitely generated** as a $k$-algebra if $R$ is generated as a ring by $k$
together with finitely many elements $r_1, \dots, r_n$. This is equivalent to
having a surjective $k$-algebra homomorphism
$$
k[x_1, \dots, x_n] \twoheadrightarrow R, \qquad x_i \mapsto r_i,
$$
so every finitely generated $k$-algebra is a quotient of a polynomial ring, hence
Noetherian.[^df-basis] These are the coordinate rings that the geometry attaches
to its spaces.

## Affine algebraic sets

Fix a field $k$ and let $\mathbb{A}^n$ denote **affine $n$-space**, the set of
$n$-tuples of elements of $k$. Each polynomial $f \in k[x_1, \dots, x_n]$ is a
$k$-valued function on $\mathbb{A}^n$ by evaluation, and a set $S$ of polynomials
picks out the points where all of them vanish.

> **Definition (Affine algebraic set).** For $S \subseteq k[x_1, \dots, x_n]$, its
> **zero locus** is
> $$
> Z(S) = \{\, (a_1, \dots, a_n) \in \mathbb{A}^n \mid f(a_1, \dots, a_n) = 0
> \text{ for all } f \in S \,\}.
> $$
> A subset $V \subseteq \mathbb{A}^n$ is an **affine algebraic set** if $V = Z(S)$
> for some $S$.

Since $Z(S) = Z(I)$ where $I = (S)$ is the generated ideal, only ideals matter, and
because $k[x_1, \dots, x_n]$ is Noetherian every ideal is generated by finitely
many $f_1, \dots, f_q$. Consequently
$$
V = Z(I) = Z(f_1) \cap Z(f_2) \cap \cdots \cap Z(f_q),
$$
so **every affine algebraic set is the intersection of finitely many
hypersurfaces**. A purely geometric statement — a solution set needs only finitely
many equations — falls out of an algebraic one, the Basis Theorem.[^df-zi]

Examples fix the picture:

- **Points and finite sets.** The single point $(a_1, \dots, a_n)$ is
  $Z(x_1 - a_1, \dots, x_n - a_n)$, and any finite set is algebraic.
- **Linear sets.** Lines, planes, and coordinate subspaces are loci of degree-one
  polynomials; the $x$-axis in $\mathbb{A}^3$ is $Z(y, z)$.
- **Hypersurfaces.** The parabola $Z(y - x^2)$, the circle $Z(x^2 + y^2 - 1)$, and
  the hyperbola $Z(xy - 1)$ are algebraic sets in the plane.

$$
% caption: An affine variety is a common zero set: the two points where a circle
% and a line meet are cut out simultaneously by the two defining polynomials.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=0.95]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black, ->] (-2.7,0) -- (2.9,0) node[right]{$x$};
\draw[black, ->] (0,-2.7) -- (0,2.9) node[above]{$y$};
\draw[acc, thick] (0,0) circle (2);
\draw[black, thick] (-2.4,-2.4) -- (2.4,2.4);
\fill[black] (1.414,1.414) circle (2.3pt);
\fill[black] (-1.414,-1.414) circle (2.3pt);
\node[acc, anchor=west] at (2.02,0.7) {$Z(f)$};
\node[black, anchor=west] at (1.0,2.35) {$Z(g)$};
\end{tikzpicture}
$$

### The two maps

The correspondence runs in both directions. Going one way, an ideal determines its
zero locus. Going the other, a set of points determines the ideal of everything
that vanishes on it.

> **Definition (The ideal of a set).** For $A \subseteq \mathbb{A}^n$,
> $$
> I(A) = \{\, f \in k[x_1, \dots, x_n] \mid f(a) = 0 \text{ for all } a \in A \,\}.
> $$
> This is the unique largest ideal of functions vanishing identically on $A$.

The maps $Z$ and $I$ are both inclusion-reversing (contravariant), and they satisfy
$$
A \subseteq Z(I(A)), \qquad I \subseteq I(Z(I)), \qquad Z(I(Z(I))) = Z(I).
$$
They are mutually inverse once restricted to the algebraic sets $V = Z(I)$ on one
side and the ideals of the form $I(V)$ on the other.[^df-zi] Determining exactly
which ideals arise as $I(V)$ is the content of the Nullstellensatz below.

### The coordinate ring

The functions on $\mathbb{A}^n$ that restrict to functions on a subset $V$ are the
polynomials modulo those that vanish on $V$.

> **Definition (Coordinate ring).** For an affine algebraic set $V$, the
> **coordinate ring** is the quotient $k[V] = k[x_1, \dots, x_n] / I(V)$.

Two polynomials define the same function on $V$ exactly when their difference lies
in $I(V)$, so $k[V]$ is genuinely the ring of polynomial functions on $V$. It is a
finitely generated $k$-algebra, generated by the restricted coordinate functions.
Maps between algebraic sets translate perfectly into maps between coordinate rings:
a **morphism** $\varphi: V \to W$ (given by polynomials in each coordinate) induces
a $k$-algebra homomorphism $\varphi^{*}: k[W] \to k[V]$ by $f \mapsto f \circ
\varphi$, and every $k$-algebra homomorphism arises this way from a unique
morphism.[^df-zi] The correspondence is contravariant, and it is an isomorphism of
algebraic sets exactly when the induced map of coordinate rings is an isomorphism.
Gröbner bases make this computational: the reduced remainder after division
by a Gröbner basis for $I(V)$ gives a canonical representative for each coset in
$k[V]$, which is how one computes in coordinate rings, tests membership in an
ideal, and finds kernels and images of $k$-algebra maps in practice.[^df-zi]

## Radicals

An algebraic set does not determine its defining ideal uniquely: the zeros of $f$
are the zeros of $f^2$, so $Z(f) = Z(f^k)$ for all $k \geq 1$, and $(x), (x^2),
(x^3), \dots$ all cut out the same $y$-axis. The redundancy is captured by the
radical.

> **Definition (Radical).** For an ideal $I$ of a commutative ring $R$, the
> **radical** is
> $$
> \operatorname{rad} I = \{\, a \in R \mid a^k \in I \text{ for some } k \geq 1 \,\}.
> $$
> The radical of the zero ideal is the **nilradical**, the set of nilpotent
> elements. An ideal is a **radical ideal** if $I = \operatorname{rad} I$.

The radical is again an ideal containing $I$, and $(\operatorname{rad} I)/I$ is the
nilradical of $R/I$; so $R/I$ has no nonzero nilpotents exactly when $I$ is
radical.[^df-rad]

- **Radical as an intersection of primes.** For a proper ideal $I$,
  $\operatorname{rad} I$ equals the intersection of all prime ideals containing
  $I$. In particular the nilradical is the intersection of all primes of $R$, and
  prime (hence maximal) ideals are radical.
- **In a UFD, radicals kill exponents.** If $a = p_1^{e_1} \cdots p_r^{e_r}$ is the
  factorization into distinct primes, then $\operatorname{rad}(a) = (p_1 \cdots
  p_r)$ is the squarefree part. For instance $\operatorname{rad}(180) = (30)$ in
  $\mathbb{Z}$.

$$
% caption: Different ideals can share a zero set. Both $(x)$ and $(x^2)$ cut out
% the $y$-axis, and taking radicals collapses them to the same ideal $(x)$.
\begin{tikzpicture}[>=stealth, font=\footnotesize,
  b/.style={draw, minimum width=16mm, minimum height=8mm}]
\definecolor{acc}{HTML}{4A6FA5}
\node[b] (i1) at (0,1.1) {$(x)$};
\node[b] (i2) at (0,-1.1) {$(x^2)$};
\begin{scope}[shift={(5,0)}]
  \draw[black,->] (-1.4,0)--(1.4,0) node[right]{$x$};
  \draw[black,->] (0,-1.6)--(0,1.6) node[above]{$y$};
  \draw[acc, very thick] (0,-1.5)--(0,1.5);
\end{scope}
\draw[->, acc, thick] (i1) to[out=0,in=150] (3.9,0.45);
\draw[->, acc, thick] (i2) to[out=0,in=210] (3.9,-0.45);
\node[acc] at (2.5,1.05) {$Z$};
\node[acc] at (2.5,-1.05) {$Z$};
\end{tikzpicture}
$$

Because $k$ has no nilpotents, the coordinate ring $k[V] = k[\mathbb{A}^n]/I(V)$ has
none either, so **$I(V)$ is always a radical ideal**.[^df-rad] Over an arbitrary
field the converse fails: $(x^2 + 1)$ is maximal, hence radical, in
$\mathbb{R}[x]$, but $x^2 + 1$ has no real zeros, so it is not $I(V)$ for any $V$.
The obstruction comes down to polynomials with no roots, and it
disappears over an algebraically closed field.

## The Zariski topology and varieties

The algebraic sets are closed under finite union and arbitrary intersection, and
they include $\emptyset = Z(1)$ and $\mathbb{A}^n = Z(0)$. Those are precisely the
axioms for the closed sets of a topology.

> **Definition (Zariski topology).** The **Zariski topology** on $\mathbb{A}^n$ is
> the topology whose closed sets are the affine algebraic sets.

The topology is coarse. On $\mathbb{A}^1$ the only closed sets are the finite sets,
the empty set, and the whole line, so any two nonempty open sets over an infinite
field meet. Points are closed, but distinct points cannot be separated by disjoint
opens, so the Zariski topology is not Hausdorff. Over $\mathbb{R}$ it is strictly
coarser than the Euclidean topology.[^df-zariski] Within it, the natural building
blocks are the sets that do not split.

> **Definition (Irreducible set, variety).** A nonempty algebraic set $V$ is
> **irreducible** if it is not the union of two proper algebraic subsets. An
> irreducible affine algebraic set is called an **affine variety**.

Irreducibility is a ring-theoretic condition on the ideal: $V = Z(I)$ is
irreducible if and only if $I$ is a prime ideal, equivalently the coordinate ring
$k[V]$ is an integral domain.[^df-zariski] Since $k[x_1, \dots, x_n]$ is
Noetherian, every algebraic set decomposes into finitely many irreducible pieces.

> **Theorem (Irreducible decomposition).** Every affine algebraic set $V$ is a
> finite union $V = V_1 \cup \cdots \cup V_m$ of irreducible components, unique
> once redundant pieces (those contained in another) are dropped.

$$
% caption: A radical ideal splits its variety into irreducible components; the
% zero set here is a parabola together with a line, each an irreducible piece.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=0.95]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black,->] (-2.6,0)--(2.8,0) node[right]{$x$};
\draw[black,->] (0,-0.7)--(0,3.0) node[above]{$y$};
\draw[acc, thick] (-1.6,2.4) .. controls (-0.6,-0.5) and (0.6,-0.5) .. (1.6,2.4);
\draw[black, thick] (-2.3,0.35)--(2.3,2.6);
\node[acc, anchor=west] at (1.35,1.0) {$V_1$};
\node[black, anchor=south] at (1.55,2.75) {$V_2$};
\end{tikzpicture}
$$

The decomposition mirrors the algebra exactly: over an algebraically closed field a
radical ideal is a finite intersection of prime ideals, and the union of the
corresponding varieties is the decomposition of the algebraic set into irreducible
components.[^df-zariski]

> **Worked example (a reducible set and its ideal).** In $\mathbb{A}^2$ take $V =
> Z(xy)$, the union of the two coordinate axes. A product vanishes iff a factor
> does, so $Z(xy) = Z(x) \cup Z(y)$ splits $V$ into the $y$-axis $Z(x)$ and the
> $x$-axis $Z(y)$. Since $xy$ is squarefree, $(xy)$ is a radical ideal and $I(V) =
> (xy)$. It is not prime — $xy \in (xy)$ while $x \notin (xy)$ and $y \notin (xy)$ —
> matching the reducibility of $V$. The coordinate ring
> $$
> k[V] = k[x, y] / (xy)
> $$
> accordingly has zero divisors: the classes of $x$ and $y$ are nonzero, yet their
> product is $0$. The two components correspond to the primes $(x)$ and $(y)$, both
> containing $(xy)$, and
> $$
> (xy) = (x) \cap (y)
> $$
> realizes $I(V)$ as the intersection of primes promised by the
> decomposition.[^df-zariski]

## Hilbert's Nullstellensatz

Over an algebraically closed field the correspondence closes into a clean
bijection. The theorem comes in a weak form, about points and maximal ideals, and a
strong form, about arbitrary ideals and their radicals.

> **Theorem (Nullstellensatz, weak form).** Let $k$ be algebraically closed. The
> maximal ideals of $k[x_1, \dots, x_n]$ are the ideals
> $(x_1 - a_1, \dots, x_n - a_n)$ for $(a_1, \dots, a_n) \in \mathbb{A}^n$, and no others. Hence
> $Z$ and $I$ give a bijection between points of $\mathbb{A}^n$ and maximal ideals,
> and every proper ideal has $Z(I) \neq \emptyset$.

The last clause is the origin of the name — _Nullstellensatz_, "zero-locus
theorem": any proper system of polynomial equations over an algebraically closed
field has a common solution. The proof runs through **Noether's Normalization
Lemma**, which presents any finitely generated $k$-algebra as a finite (integral)
extension of a polynomial subalgebra.[^df-null] An element $s$ of a ring extension
$S \supseteq R$ is **integral over $R$** if it satisfies a monic polynomial with
coefficients in $R$; the integral elements form a subring, and integrality is
transitive. This is the ring-theoretic sibling of an
[algebraic field extension](/abstract-algebra/field-theory/field-extensions-and-algebraic-elements),
and it supplies the finiteness that drives the maximal-ideal statement.

> **Theorem (Hilbert's Nullstellensatz).** Let $k$ be algebraically closed. Then
> $$
> I(Z(I)) = \operatorname{rad} I
> $$
> for every ideal $I$ of $k[x_1, \dots, x_n]$, and $Z$, $I$ restrict to mutually
> inverse bijections
> $$
> \{\text{affine algebraic sets}\} \;\longleftrightarrow\;
> \{\text{radical ideals}\}.
> $$

Passing to the zero set and back to the ideal recovers the radical of $I$ rather
than $I$ itself, the largest ideal with the same zero set. Over $\mathbb{C}$, then,
the "geometrically defined" ideals $I(V)$ are precisely the radical ideals, a
condition that is purely algebraic.[^df-null] A computational corollary makes
radical membership decidable: for a proper ideal $I = (f_1, \dots, f_s)$ in
$k[x_1, \dots, x_n]$,
$$
f \in \operatorname{rad} I
\iff
(f_1, \dots, f_s,\, 1 - yf) = k[x_1, \dots, x_n, y],
$$
the **Rabinowitsch trick**, which reduces a radical-membership question to whether
an ideal in one extra variable is the whole ring — a Gröbner-basis
computation.[^df-null]

> **Worked example (the Nullstellensatz recovers the radical).** Over $\mathbb{C}$
> take $I = (x^2, y) \subseteq \mathbb{C}[x, y]$. A point of $Z(I)$ satisfies $x^2 =
> 0$ and $y = 0$, hence $x = y = 0$, so $Z(I) = \{(0,0)\}$ is a single point. Its
> ideal is the maximal ideal at the origin, $I(Z(I)) = (x, y)$. Computing the radical
> directly: $x^2 \in I$ gives $x \in \operatorname{rad} I$, and $y \in I \subseteq
> \operatorname{rad} I$, so $(x, y) \subseteq \operatorname{rad} I$; since $(x, y)$ is
> maximal and contains $I$, equality holds. Thus
> $$
> I(Z(I)) = (x, y) = \operatorname{rad} I \;\supsetneq\; (x^2, y) = I,
> $$
> exactly as the Nullstellensatz predicts: passing to the zero set and back enlarges
> $I$ to its radical. The ideal $I$ is not radical ($x \notin I$ but $x^2 \in I$), so
> it is not $I(V)$ for any algebraic set $V$ — the "fat point" $(x^2, y)$ carries a
> multiplicity that the reduced point $\{(0,0)\}$ has dropped.[^df-null]

## The dictionary

Everything above collapses into a single translation table. Over an algebraically
closed field, geometry on the left becomes commutative algebra on the right.

$$
% caption: Over an algebraically closed field the maps Z and I are mutually
% inverse, matching each geometric object with an algebraic one.
\begin{tikzpicture}[>=stealth, font=\footnotesize,
  g/.style={draw, minimum width=34mm, minimum height=8mm, align=center},
  a/.style={draw, minimum width=44mm, minimum height=8mm, align=center}]
\definecolor{acc}{HTML}{4A6FA5}
\node[font=\small] at (0,3.1) {Geometry};
\node[font=\small] at (6.0,3.1) {Algebra};
\node[g] (v1) at (0,2.2) {variety $V$};
\node[g] (v2) at (0,1.0) {points of $V$};
\node[g] (v3) at (0,-0.2) {subvarieties of $V$};
\node[g] (v4) at (0,-1.4) {morphism $V$ to $W$};
\node[a] (a1) at (6.0,2.2) {radical (prime) ideal $I(V)$};
\node[a] (a2) at (6.0,1.0) {maximal ideals of $k[V]$};
\node[a] (a3) at (6.0,-0.2) {prime ideals of $k[V]$};
\node[a] (a4) at (6.0,-1.4) {$k$-algebra map $k[W]$ to $k[V]$};
\draw[<->, acc, thick] (v1) -- (a1) node[midway, above, font=\scriptsize]{$I$ and $Z$};
\draw[->, acc, thick] (v2) -- (a2);
\draw[->, acc, thick] (v3) -- (a3);
\draw[->, acc, thick] (v4) -- (a4);
\end{tikzpicture}
$$

| Geometry | Algebra |
| --- | --- |
| affine algebraic set $V$ | radical ideal $I(V)$ |
| affine variety (irreducible) | prime ideal |
| single point | maximal ideal |
| $V$ contained in $W$ | $I(W)$ contained in $I(V)$ |
| union $V_1 \cup V_2$ | intersection $I(V_1) \cap I(V_2)$ |
| intersection $V_1 \cap V_2$ | radical of $I(V_1) + I(V_2)$ |
| coordinate ring $k[V]$ | $k[\mathbb{A}^n]/I(V)$ |
| morphism $\varphi: V \to W$ | $k$-algebra map $\varphi^{*}: k[W] \to k[V]$ |

The table is the reason commutative algebra and algebraic geometry are studied as
one subject. A question about the geometry of solution sets becomes a question
about ideals and quotient rings, and the Nullstellensatz guarantees that nothing is
lost in translation over an algebraically closed field. Dropping "algebraically
closed" leads to arithmetic geometry and, through the integral extensions above, to
the ring of integers of a number field and algebraic number theory. Allowing the
base to vary leads to schemes, where the prime spectrum $\operatorname{Spec} R$
replaces $\mathbb{A}^n$ and every commutative ring becomes a geometric space. The
[Chinese Remainder Theorem](/abstract-algebra/ring-theory/fractions-and-the-chinese-remainder-theorem)
reappears as the statement that a disjoint union of points corresponds to a product
of coordinate rings, and Gröbner bases make all of it computable. The
[representation and character theory](/abstract-algebra/capstone/representation-and-character-theory)
capstone applies the same modules-over-a-ring viewpoint to the group ring in place
of the polynomial ring.

[^df-noeth]: Dummit & Foote, §15.1 — Noetherian rings, the ascending chain condition, and the equivalence of A.C.C. with the maximal-element condition and with finite generation of every ideal.
[^df-basis]: Dummit & Foote, §15.1 — Hilbert's Basis Theorem (restated from §9.6) and its corollary that $k[x_1, \dots, x_n]$ and every finitely generated $k$-algebra are Noetherian.
[^df-zi]: Dummit & Foote, §15.1 — affine algebraic sets, the maps $Z$ and $I$, coordinate rings, and the correspondence between morphisms of algebraic sets and $k$-algebra homomorphisms, computed via Gröbner bases.
[^df-rad]: Dummit & Foote, §15.2 — the radical of an ideal, the nilradical, radical as the intersection of the primes containing $I$, and the fact that $I(V)$ is always radical.
[^df-zariski]: Dummit & Foote, §15.2 — the Zariski topology, irreducible sets and varieties, the correspondence between irreducibility and prime ideals, and the unique decomposition into irreducible components.
[^df-null]: Dummit & Foote, §15.3 — integral extensions, Noether's Normalization Lemma, and both the weak and strong forms of Hilbert's Nullstellensatz, with the radical-membership criterion.