The Zariski topology is a family of topologies in which algebraic equations, rather than metric inequalities, determine closed sets. It is the basic language for treating solution sets of polynomial equations as geometric spaces. The affine version lives on $\mathbb A^n_k$, the spectrum version lives on $\operatorname{Spec}R$, the projective version lives on projective space using homogeneous equations, and varieties or schemes inherit the topology from these local models. These settings are related, but they are not the same ambient definition. What they share is the governing principle that closed subsets are the loci where algebraic functions vanish. The guiding principle is that an algebraic condition should define a closed locus. For example, the equation $x^2+y^2-1=0$ cuts out a curve in the affine plane. Several equations cut out the set where all of them vanish at the same time. The Zariski topology turns these common zero sets into the closed subsets of [affine space](/page/Affine%20Space). This viewpoint links topology with ideals, because a system of polynomial equations is best organized by the ideal it generates. ## Definition In classical algebraic geometry, the unqualified phrase "the Zariski topology" most often means the topology on an affine algebraic set. Spectrum and projective versions use the same guiding principle, but they have their own ambient definitions: spectra use containment in prime ideals, and projective space uses homogeneous equations. We start with the affine definition because it is the canonical model from which the later variants are compared. [definition: Zariski Topology] Let $k$ be a field, let $X\subset\mathbb A^n_k$ be an affine algebraic set, and let $k[x_1,\ldots,x_n]$ be the coordinate [polynomial ring](/page/Polynomial%20Ring) of the ambient affine space. The Zariski topology on $X$ is the topology whose closed subsets are precisely the sets \begin{align*} X\cap V(I)=\{a\in X:f(a)=0\text{ for every }f\in I\}, \end{align*} where $I$ ranges over ideals of $k[x_1,\ldots,x_n]$. [/definition] When $X=\mathbb A^n_k$, this says that the closed subsets are exactly the common zero loci $V(I)$. The spectrum and projective versions later in the page keep the same equation-controlled idea, but they change the ambient points and the kind of equations being used. The ambient affine case is the test model for every later version. When $X=\mathbb A^n_k$, the definition specializes to this concrete rule: the closed subsets are the sets \begin{align*} V(I)=\{a\in\mathbb A^n_k:f(a)=0\text{ for every }f\in I\}, \end{align*} where $I$ ranges over ideals of $k[x_1,\ldots,x_n]$. This definition is compact, but it hides the two ingredients that drive the rest of the page: how zero loci behave under operations on ideals, and how much information is lost when equations are replaced by their solution sets. We now unpack those ingredients from the affine case outward. ## Affine Algebraic Sets We begin over a field $k$, usually algebraically closed when geometric intuition is needed. Affine $n$-space over $k$, written $\mathbb A^n_k$, is the set $k^n$ together with its polynomial coordinate functions. The [coordinate ring](/page/Coordinate%20Ring) is $k[x_1,\ldots,x_n]$. A point $a=(a_1,\ldots,a_n)$ evaluates a polynomial $f$ by substitution: \begin{align*} f(a)=f(a_1,\ldots,a_n). \end{align*} The first construction turns a set of equations into its common zero locus. It gives the closed sets before we have named the topology. [definition: Vanishing Set] The vanishing-set map is \begin{align*} V:\mathcal P(k[x_1,\ldots,x_n])\to \mathcal P(\mathbb A^n_k). \end{align*} For $S\subset k[x_1,\ldots,x_n]$, \begin{align*} S\mapsto V(S)=\{a\in \mathbb A^n_k: f(a)=0\text{ for every }f\in S\}. \end{align*} [/definition] When $S=\{f\}$, it is standard to write $V(f)$ for $V(\{f\})$. Since $V(\{f\})=V((f))$, the same notation also means the vanishing set of the principal ideal generated by $f$. [example: A First Vanishing Set] In $\mathbb A^2_k$, write a point as $(a,b)$ with $a,b\in k$. Evaluating the polynomial $y-x^2$ at this point means substituting $x=a$ and $y=b$, so \begin{align*} (y-x^2)(a,b)=b-a^2. \end{align*} By the definition of a vanishing set, $(a,b)\in V(y-x^2)$ exactly when $(y-x^2)(a,b)=0$. Therefore \begin{align*} (a,b)\in V(y-x^2)\quad\Longleftrightarrow\quad b-a^2=0\quad\Longleftrightarrow\quad b=a^2. \end{align*} Thus every point of $V(y-x^2)$ has the form $(a,a^2)$ with $a\in k$. Conversely, for every $t\in k$, \begin{align*} (y-x^2)(t,t^2)=t^2-t^2=0, \end{align*} so $(t,t^2)\in V(y-x^2)$. Hence \begin{align*} V(y-x^2)=\{(t,t^2):t\in k\}. \end{align*} For the polynomial $xy$, substitution gives \begin{align*} (xy)(a,b)=ab. \end{align*} Again by the definition of a vanishing set, \begin{align*} (a,b)\in V(xy)\quad\Longleftrightarrow\quad ab=0. \end{align*} Since $k$ is a field, it has no zero divisors, so $ab=0$ holds exactly when $a=0$ or $b=0$. If $a=0$, then $(a,b)=(0,b)$ lies on the $y$-axis; if $b=0$, then $(a,b)=(a,0)$ lies on the $x$-axis. Therefore \begin{align*} V(xy)=\{(0,b):b\in k\}\cup\{(a,0):a\in k\}. \end{align*} So $V(y-x^2)$ is the parabola parametrized by $t\mapsto(t,t^2)$, while $V(xy)$ is the union of the two coordinate axes. [/example] This first example shows how single equations and reducible equations appear as closed sets. It also previews two recurring themes: equations may define geometric pieces of different shapes, and factorization in the coordinate ring affects the topology. The same [closed set](/page/Closed%20Set) is obtained from the ideal generated by $S$. Indeed, if a point kills every polynomial in $S$, it kills every polynomial combination of elements of $S$. For this reason, algebraic geometry usually writes $V(I)$ with $I$ an ideal. To make these algebraic sets into the closed sets of a topology, two closure properties must be checked. First, unions of finitely many solution sets should again be a solution set; second, arbitrary intersections of solution sets should again be a solution set. The possible obstruction is that union and intersection are operations on sets of points, while the data defining those sets are ideals of polynomials. The key point is that products and sums of ideals translate these set-theoretic operations back into equations. [quotetheorem:9408] This result is the point where the algebraic definition becomes a topology. The empty set and the whole affine space arise from the extreme ideals $(1)$ and $(0)$, while products and sums of ideals encode finite unions and arbitrary intersections. Thus the closed-set axioms are not extra structure imposed on affine space; they are built into how polynomial equations combine. This topology behaves very differently from metric topologies. When $k$ is infinite, a nonzero polynomial in one variable has only finitely many roots. Consequently the Zariski closed subsets of $\mathbb A^1_k$ are exactly the finite subsets together with the whole line. The open subsets are therefore the complements of finite sets, plus the empty set. This cofinite behavior is a useful first warning: Zariski open sets are large, and Zariski closed sets are rigid. ## The Ideal-Geometry Dictionary The Zariski topology becomes powerful because it reverses algebraic containment. More equations mean fewer points. Fewer equations mean more possible solutions. This reversal is a recurring feature of algebraic geometry. ### Radical Ideals and Lost Equations The closed set attached to an ideal ignores nilpotent information in that ideal. To state this precisely, we need the radical operation. [definition: Radical Ideal] The radical operation is the map from ideals of $k[x_1,\ldots,x_n]$ to ideals of $k[x_1,\ldots,x_n]$ whose action is \begin{align*} I\mapsto \sqrt I. \end{align*} For an ideal $I\subset k[x_1,\ldots,x_n]$, its value is \begin{align*} \sqrt I=\{f\in k[x_1,\ldots,x_n]: f^m\in I\text{ for some integer }m\ge 1\}. \end{align*} The ideal $I$ is radical if $I=\sqrt I$. [/definition] The radical records which polynomials vanish on every solution forced by $I$. This matters because two different systems of equations can have the same solution set. To state the geometric side of this comparison, if $Y\subset\mathbb A^n_k$ is any subset, write \begin{align*} I(Y)=\{f\in k[x_1,\ldots,x_n]: f(y)=0\text{ for every }y\in Y\}. \end{align*} Thus $I(V(I))$ denotes the ideal of all polynomials vanishing on the common zero locus of $I$. The theorem below is needed to decide when the topology has lost no information beyond radicals: it identifies the equations recoverable from a closed set. Over an [algebraically closed field](/page/Algebraically%20Closed%20Field), [Hilbert's Nullstellensatz](/theorems/2124) identifies this algebraic radical with the geometric vanishing condition and makes the ideal-closed-set dictionary exact. [quotetheorem:2124] This theorem says that affine algebraic sets correspond to radical ideals. The correspondence is order reversing: \begin{align*} I\subset J \quad \Longrightarrow \quad V(J)\subset V(I). \end{align*} The direction matches the intuition that adding equations can remove points. ### Zariski Closure This dictionary also explains closures. With the notation $I(Y)$ introduced above, the Zariski closure of $Y\subset\mathbb A^n_k$ is $V(I(Y))$. Thus a point lies in the closure of $Y$ exactly when every polynomial equation holding on $Y$ also holds at that point. This notion of closure is algebraic rather than analytic. For instance, an infinite subset of $\mathbb A^1_k$ is Zariski dense when $k$ is algebraically closed. No polynomial in one variable can vanish on infinitely many points without being the zero polynomial. In contrast, the same subset may be very sparse in the usual topology when $k=\mathbb C$. ## Basic Open Sets Open sets in the Zariski topology are complements of algebraic sets. The most important open sets come from a single inequality $f\ne 0$. They are the affine pieces on which one polynomial is allowed to be inverted. This construction is central because it connects topology with localization. The places where $f$ does not vanish are exactly the places where $1/f$ can be treated as a regular function. [definition: Principal Open Set] The principal-open map is \begin{align*} D:k[x_1,\ldots,x_n]\to \mathcal P(\mathbb A^n_k). \end{align*} For $f\in k[x_1,\ldots,x_n]$, \begin{align*} f\mapsto D(f)=\mathbb A^n_k\setminus V(f)=\{a\in \mathbb A^n_k:f(a)\ne 0\}. \end{align*} [/definition] Knowing that $D(f)$ is open is only the first step. Local arguments need a supply of open neighborhoods small enough to fit inside any given [open set](/page/Open%20Set), but still algebraic enough that rings can be computed on them. The next theorem is needed for that reason: it shows that principal opens are not merely examples of open sets, but the basic neighborhoods from which all affine Zariski opens are assembled. [quotetheorem:9609] Conceptually, the theorem says that affine Zariski neighborhoods can be chosen by imposing one nonvanishing condition at a time. Their stability under finite intersections means that several local nonvanishing requirements can still be treated as a single localization condition, so the topology is well matched to the algebraic operation of inverting functions. The same idea appears in coordinate rings. If $X=V(I)\subset\mathbb A^n_k$, its coordinate ring is $k[X]=k[x_1,\ldots,x_n]/I(X)$. For a regular function $f\in k[X]$, the principal open $D(f)\cap X$ has coordinate ring $k[X]_f$, the localization obtained by inverting powers of $f$. This is why principal opens are the standard local neighborhoods of algebraic geometry. ## Irreducibility and Generic Points The Zariski topology is not usually Hausdorff. This failure is a feature rather than a defect. Non-Hausdorff behavior allows a space to remember specialization, containment of subvarieties, and generic behavior. To make that memory precise, we need a way to ask whether a closed set is a single algebraic piece or a union of smaller closed pieces. Irreducibility is the topological test for that question. It is the notion that lets the topology detect the same indivisibility that integral domains detect in coordinate rings. [definition: Irreducible Topological Space] A nonempty [topological space](/page/Topological%20Space) $X$ is irreducible if it cannot be written as $X=Y\cup Z$, where $Y$ and $Z$ are proper closed subsets of $X$. [/definition] In affine algebraic geometry, irreducibility is an algebraic condition on the coordinate ring. This connection matters because it lets us recognize geometric components by checking whether products of functions can vanish without either factor vanishing. The next theorem is one of the cleanest examples of the topology-ring dictionary. [quotetheorem:9418] Irreducible closed subsets are the geometric counterparts of prime ideals. For an algebraic set $X$, closed irreducible subsets of $X$ correspond to prime ideals in $k[X]$. This is one motivation for replacing classical affine varieties by prime spectra. ## The Spectrum Viewpoint The modern form of the Zariski topology is defined on the prime spectrum of a ring. This version works for arbitrary commutative rings and is the topological foundation of schemes. It also explains why prime ideals, rather than maximal ideals alone, are natural points. ### Closed Sets in Prime Spectra The classical affine picture has a limitation: if we keep only closed points, irreducible subvarieties appear as subsets but not as individual points. A line inside a plane is visible as a closed set, yet there is no point whose closure is exactly that line. Prime ideals repair this by adding points for irreducible algebraic pieces, including generic points that remember whole components at once. This is why the definition begins with the set of prime ideals rather than the set of maximal ideals. Closed conditions still come from ideals, but now the phrase "vanishes at a point" means "is contained in a prime ideal." The resulting topology keeps the old equation-closed-set dictionary while making specialization and irreducible structure part of the space itself. [definition: Zariski Topology on the Prime Spectrum] Let $R$ be a commutative ring with identity. The prime spectrum $\operatorname{Spec}R$ is the set of prime ideals of $R$. The closed-set assignment is the map from ideals of $R$ to subsets of $\operatorname{Spec}R$ given by \begin{align*} I\mapsto V(I)=\{\mathfrak p\in\operatorname{Spec}R:I\subset\mathfrak p\}. \end{align*} The sets $V(I)$ are the closed subsets of the Zariski topology on $\operatorname{Spec}R$. [/definition] When $R=k[x_1,\ldots,x_n]/I$ with $k$ algebraically closed, [Hilbert's Nullstellensatz](/theorems/2940) identifies the classical points of the affine algebraic set $V(I)$ with the maximal ideals of $R$. Under that identification, the closed subsets seen by maximal ideals match the affine Zariski closed sets. The full spectrum is larger: it also includes nonmaximal prime ideals, which serve as generic points for irreducible closed sets. For example, the zero ideal is a point of $\operatorname{Spec} k[x]$. Its closure is the whole affine line, so it acts as the generic point of the line. ### Localization on Principal Opens The corresponding basic opens are again principal. The principal-open assignment is the map from $R$ to subsets of $\operatorname{Spec}R$ given by \begin{align*} f\mapsto D(f)=\{\mathfrak p\in\operatorname{Spec}R:f\notin\mathfrak p\}. \end{align*} These sets form a base, and $D(f)\cap D(g)=D(fg)$. The ring of functions on $D(f)$ is the localization $R_f$. Thus the topology is designed so that algebraic localization and topological restriction match. ## Separation, Density, and Dimension Zariski spaces are often $T_0$ but not $T_1$. In $\operatorname{Spec}R$, the closure of a point $\mathfrak p$ is \begin{align*} \overline{\{\mathfrak p\}}=V(\mathfrak p). \end{align*} This contains the prime ideals that contain $\mathfrak p$. The containment relation on prime ideals becomes the specialization relation in the topology. This specialization order gives geometric meaning to algebraic inclusion. A smaller prime ideal represents a more generic point. A larger prime ideal represents a more special point. For the commutative rings with identity considered here, a point $\mathfrak p\in\operatorname{Spec}R$ is closed exactly when $\mathfrak p$ is a maximal ideal: if a larger prime ideal existed, it would lie in the closure $V(\mathfrak p)$, while if $\mathfrak p$ were not maximal, some maximal ideal containing it would give such a larger prime. Generic points of irreducible components are minimal prime ideals; the component attached to a minimal prime $\mathfrak p$ is $V(\mathfrak p)$, whose generic point is $\mathfrak p$. Dimension is also visible in this topology. In noetherian affine settings, chains of irreducible closed subsets correspond to chains of prime ideals. For affine algebraic varieties, this recovers Krull dimension. For example, $\mathbb A^n_k$ has dimension $n$, matching the longest possible chain of irreducible closed sets. ## Concrete Zariski Spaces ### The Affine Line [example: The Affine Line Has Cofinite Zariski Topology] Let $k$ be algebraically closed. Since $k[x]$ is a [Euclidean domain](/page/Euclidean%20Domain), every ideal $I\subset k[x]$ is principal, so $I=(g)$ for some $g\in k[x]$. If $g=0$, then every $t\in k$ satisfies $0(t)=0$, hence \begin{align*} V(I)=V(0)=\mathbb A^1_k. \end{align*} Now suppose $g\ne 0$. Algebraic closedness gives a factorization \begin{align*} g=c(x-a_1)^{m_1}\cdots(x-a_r)^{m_r} \end{align*} with $c\in k^\times$, distinct $a_i\in k$, and $m_i\ge 1$. For $t\in k$, evaluation gives \begin{align*} g(t)=c(t-a_1)^{m_1}\cdots(t-a_r)^{m_r}. \end{align*} Because $c\ne 0$ and $k$ is a field, this product is $0$ exactly when one factor $(t-a_i)^{m_i}$ is $0$. Since $m_i\ge 1$, the equality $(t-a_i)^{m_i}=0$ holds exactly when $t-a_i=0$, equivalently $t=a_i$. Therefore \begin{align*} V(g)=\{a_1,\ldots,a_r\}. \end{align*} Thus every proper closed subset of $\mathbb A^1_k$ is finite. Conversely, let $F=\{a_1,\ldots,a_r\}\subset k$ be finite and put \begin{align*} h=(x-a_1)\cdots(x-a_r). \end{align*} For $t\in k$, \begin{align*} h(t)=(t-a_1)\cdots(t-a_r). \end{align*} Since $k$ is a field, $h(t)=0$ exactly when $t-a_i=0$ for some $i$, which is exactly when $t\in F$. Hence \begin{align*} F=V(h). \end{align*} Also $V(1)=\varnothing$, because $1(t)=1\ne 0$ for every $t\in k$. Therefore the closed subsets of $\mathbb A^1_k$ are exactly the finite subsets and the whole line. It follows that every nonempty open subset has the form $\mathbb A^1_k\setminus F$ with $F$ finite. The field $k$ is infinite: if $k=\{b_1,\ldots,b_q\}$ were finite, then the polynomial \begin{align*} p(x)=1+\prod_{i=1}^q(x-b_i) \end{align*} would satisfy $p(b_j)=1+0=1$ for every $j$, so $p$ would have no root in $k$, contradicting algebraic closedness. Hence $\mathbb A^1_k\setminus F$ is infinite. Since the only proper closed subsets are finite, this infinite open set is not contained in any proper closed subset, so its closure is all of $\mathbb A^1_k$. In particular, if two polynomials $f,g\in k[x]$ agree on an infinite subset $S\subset k$, then $f-g$ vanishes at every point of $S$. A nonzero polynomial in $k[x]$ has only finitely many roots by the factorization argument above, so $f-g=0$. Thus $f=g$ as polynomials, and the equality holds at every point of $\mathbb A^1_k$. [/example] ### Curves in the Affine Plane The affine line is the cleanest place to see how coarse the topology can be. The affine plane shows the next phenomenon: proper closed sets need not be finite, because a single equation can cut out an entire curve. [example: A Parabola and a Reducible Pair of Axes] Over an algebraically closed field $k$, evaluate $y-x^2$ at a point $(a,b)\in\mathbb A^2_k$ by substituting $x=a$ and $y=b$: \begin{align*} (y-x^2)(a,b)=b-a^2. \end{align*} Hence, by the definition of a vanishing set, \begin{align*} (a,b)\in V(y-x^2)\Longleftrightarrow b-a^2=0. \end{align*} The equation $b-a^2=0$ is equivalent to $b=a^2$, so every point of $V(y-x^2)$ has the form $(a,a^2)$. Conversely, for every $a\in k$, \begin{align*} (y-x^2)(a,a^2)=a^2-a^2=0. \end{align*} Therefore \begin{align*} V(y-x^2)=\{(a,a^2):a\in k\}. \end{align*} To check irreducibility algebraically, define a $k$-algebra homomorphism $\theta:k[x,y]\to k[x]$ by $\theta(x)=x$ and $\theta(y)=x^2$. Then \begin{align*} \theta(y-x^2)=\theta(y)-\theta(x)^2=x^2-x^2=0, \end{align*} so $(y-x^2)\subset\ker\theta$. For any $p\in k[x,y]$, regard $p$ as a polynomial in $y$ with coefficients in $k[x]$. Since $y-x^2$ is monic in $y$, polynomial division in $k[x][y]$ gives \begin{align*} p=q(y-x^2)+r(x) \end{align*} with $q\in k[x,y]$ and $r(x)\in k[x]$. Applying $\theta$ gives \begin{align*} \theta(p)=\theta(q)\theta(y-x^2)+\theta(r(x)). \end{align*} Using $\theta(y-x^2)=0$ and $\theta(r(x))=r(x)$, this becomes \begin{align*} \theta(p)=\theta(q)\cdot 0+r(x)=r(x). \end{align*} Thus if $p\in\ker\theta$, then $r(x)=0$, and the division expression reduces to $p=q(y-x^2)$, so $p\in(y-x^2)$. Hence $\ker\theta=(y-x^2)$. The [first isomorphism theorem](/theorems/791) gives \begin{align*} k[x,y]/(y-x^2)\cong k[x]. \end{align*} Since $k[x]$ is an [integral domain](/page/Integral%20Domain), the ideal $(y-x^2)$ is prime and therefore radical. By Hilbert's Nullstellensatz and the closure dictionary, $I(V(y-x^2))=\sqrt{(y-x^2)}=(y-x^2)$. Therefore the reduced coordinate ring of the parabola is \begin{align*} k[V(y-x^2)]\cong k[x,y]/(y-x^2)\cong k[x]. \end{align*} Since $k[x]$ is an integral domain, the [irreducibility criterion](/theorems/2426) for coordinate rings shows that $V(y-x^2)$ is irreducible. For $V(xy)$, evaluation at $(a,b)\in\mathbb A^2_k$ gives \begin{align*} (xy)(a,b)=ab. \end{align*} Therefore \begin{align*} (a,b)\in V(xy)\Longleftrightarrow ab=0. \end{align*} Because $k$ is a field, it has no zero divisors, so $ab=0$ holds exactly when $a=0$ or $b=0$. If $a=0$, then $(a,b)=(0,b)\in V(x)$; if $b=0$, then $(a,b)=(a,0)\in V(y)$. Conversely, every point with $a=0$ or $b=0$ satisfies $ab=0$. Hence \begin{align*} V(xy)=\{(0,b):b\in k\}\cup\{(a,0):a\in k\}=V(x)\cup V(y). \end{align*} Inside $V(xy)$, the subsets $V(x)$ and $V(y)$ are closed because they are intersections of $V(xy)$ with affine Zariski closed sets. They are proper subsets of $V(xy)$: the point $(1,0)$ satisfies $xy=0$ and $y=0$ but not $x=0$, while $(0,1)$ satisfies $xy=0$ and $x=0$ but not $y=0$. Thus $V(xy)$ is the union of two proper closed subsets, so it is reducible. The parabola is one irreducible algebraic piece, while $V(xy)$ splits into the two coordinate axes. [/example] ### Principal Opens [example: The Punctured Affine Line] In $\mathbb A^1_k$, the coordinate function $x$ sends a point $a\in k$ to $x(a)=a$. Therefore the vanishing set of $x$ is \begin{align*} V(x)=\{a\in k:x(a)=0\}=\{a\in k:a=0\}=\{0\}. \end{align*} By the definition of a principal open set, the open subset determined by $x$ is \begin{align*} D(x)=\mathbb A^1_k\setminus V(x)=k\setminus\{0\}=\{a\in k:a\ne 0\}. \end{align*} Thus $D(x)$ is the affine line with the origin removed. On $D(x)$, the function $x$ never evaluates to $0$. Algebraically, this is reflected by localizing $k[x]$ at the multiplicative set $\{1,x,x^2,\ldots\}$: \begin{align*} k[x]_x=\left\{\frac{p(x)}{x^m}:p(x)\in k[x]\text{ and }m\ge 0\right\}. \end{align*} For every fraction in this localization, \begin{align*} \frac{p(x)}{x^m}=p(x)\cdot x^{-m}. \end{align*} Since $p(x)$ is a finite sum $p(x)=c_0+c_1x+\cdots+c_rx^r$, multiplying by $x^{-m}$ gives \begin{align*} p(x)x^{-m}=c_0x^{-m}+c_1x^{1-m}+\cdots+c_rx^{r-m}, \end{align*} which is a Laurent polynomial. Conversely, if $c_sx^s$ is a Laurent monomial, then for $s\ge 0$ it already lies in $k[x]$, and for $s<0$ we have \begin{align*} c_sx^s=\frac{c_s}{x^{-s}}, \end{align*} with $-s\ge 1$. Hence every finite Laurent polynomial lies in $k[x]_x$, so \begin{align*} k[x]_x=k[x,x^{-1}]. \end{align*} In this localized ring, \begin{align*} x\cdot x^{-1}=1, \end{align*} so the algebraic operation of passing from $k[x]$ to $k[x]_x$ exactly records the geometric restriction to the locus where $x$ is nonzero. [/example] ### Arithmetic Spectra [example: The Spectrum of the Integers] In $\operatorname{Spec}\mathbb Z$, the points are the prime ideals of $\mathbb Z$. Every ideal of $\mathbb Z$ has the form $(n)$ with $n\ge 0$. The ideal $(0)$ is prime because if $ab\in(0)$, then $ab=0$, and since $\mathbb Z$ is an integral domain, either $a=0$ or $b=0$. Equivalently, either $a\in(0)$ or $b\in(0)$. For $n\ge 2$, the ideal $(n)$ is prime exactly when $n$ is a prime number. If $n=p$ is prime and $ab\in(p)$, then $ab=pm$ for some $m\in\mathbb Z$, so $p\mid ab$. By Euclid's lemma, $p\mid a$ or $p\mid b$. Thus $a\in(p)$ or $b\in(p)$, so $(p)$ is prime. Conversely, suppose $n$ is not prime. Since $n\ge 2$, there are integers $r,s$ with $1<r<n$, $1<s<n$, and $n=rs$. Then \begin{align*} rs=n=1\cdot n\in(n). \end{align*} But $r\notin(n)$ because $r=mn$ would force either $m=0$, giving $r=0$, or $|m|\ge 1$, giving $|r|\ge n$, both impossible for $1<r<n$. Similarly $s\notin(n)$. Hence $(n)$ is not prime. The ideal $(1)=\mathbb Z$ is not a point of the spectrum because prime ideals are proper. Therefore \begin{align*} \operatorname{Spec}\mathbb Z=\{(0)\}\cup\{(p):p\text{ is a prime number}\}. \end{align*} We next compute the closures of these points. For any prime ideal $\mathfrak p\in\operatorname{Spec}\mathbb Z$, the closed sets containing $\mathfrak p$ are exactly the sets $V(I)$ with $I\subset\mathfrak p$. If $V(I)$ contains $\mathfrak p$, then $I\subset\mathfrak p$ by the definition of $V(I)$. Conversely, if $I\subset\mathfrak p$, then $\mathfrak p\in V(I)$. The smallest closed set containing $\mathfrak p$ is therefore \begin{align*} V(\mathfrak p)=\{\mathfrak q\in\operatorname{Spec}\mathbb Z:\mathfrak p\subset\mathfrak q\}. \end{align*} For $\mathfrak p=(0)$, every prime ideal contains $(0)$, so \begin{align*} V((0))=\{\mathfrak q\in\operatorname{Spec}\mathbb Z:(0)\subset\mathfrak q\}=\operatorname{Spec}\mathbb Z. \end{align*} Thus $(0)$ is a generic point whose closure is the whole space. Now let $p$ be a prime number. Then \begin{align*} V((p))=\{\mathfrak q\in\operatorname{Spec}\mathbb Z:(p)\subset\mathfrak q\}. \end{align*} The prime ideals of $\mathbb Z$ are $(0)$ and $(q)$ for prime numbers $q$. We have $(p)\nsubseteq(0)$ because $p\in(p)$ but $p\notin(0)$. For a prime number $q$, the containment $(p)\subset(q)$ holds exactly when $p\in(q)$, which means $q\mid p$. Since $p$ and $q$ are prime numbers, $q\mid p$ holds exactly when $q=p$. Hence \begin{align*} V((p))=\{(p)\}. \end{align*} So the closed points are exactly the ideals $(p)$ for prime numbers $p$, while $(0)$ is a non-closed point whose closure records the whole arithmetic spectrum. [/example] ### How to Read the Topology - A closed condition means that a collection of polynomial equations is being imposed. - An open condition often means that some polynomial is being required to be nonzero. - Passing from $D(f)$ to its coordinate ring means inverting $f$. - Taking closure means retaining every polynomial equation that already holds on the subset. - Irreducible closed subsets behave like single algebraic pieces. - Reducible closed subsets remember algebraic factorization through their components. - Generic points represent whole irreducible closed subsets at once. - Specialization records containment of prime ideals. - Maximal ideals are the closed points of $\operatorname{Spec}R$. - In $\operatorname{Spec}R$, minimal prime ideals are the generic points of irreducible components. - Dense open sets are common because Zariski closed sets are rigid. - The topology is coarser than analytic topology but better aligned with polynomial algebra. - Local calculations use principal opens because they match localization. - The spectrum construction extends these ideas from polynomial rings to arbitrary commutative rings. ## Common Pitfalls The Zariski topology is not meant to approximate Euclidean closeness. Two complex points may be very near in the analytic sense, but this has no direct bearing on Zariski closure. Zariski closure asks which polynomial equations vanish on the set. Another common mistake is to confuse closed sets with finite sets in all dimensions. In $\mathbb A^1_k$, proper closed sets are finite when $k$ is algebraically closed. In $\mathbb A^2_k$, a curve such as $V(y-x^2)$ is an infinite proper closed set. Higher-dimensional affine space has closed subsets of many dimensions. It is also important not to discard non-closed points in spectra. Generic points are part of the topology, not a bookkeeping device. They make irreducible closed subsets visible as closures of single points. This feature is essential in schemes and in the local study of algebraic varieties. [example: Maximal Ideals Miss a Generic Point] Assume $k$ is algebraically closed. For each $a\in k$, define $\operatorname{ev}_a:k[x]\to k$ by $\operatorname{ev}_a(f)=f(a)$. If $f\in(x-a)$, then $f=(x-a)q$ for some $q\in k[x]$, and hence \begin{align*} f(a)=(a-a)q(a)=0. \end{align*} Conversely, if $f(a)=0$, the [factor theorem](/theorems/3235) gives $f=(x-a)q$ for some $q\in k[x]$, so $f\in(x-a)$. Thus \begin{align*} \ker(\operatorname{ev}_a)=(x-a). \end{align*} The map $\operatorname{ev}_a$ is surjective because every $c\in k$ is the value of the constant polynomial $c$. Therefore the first isomorphism theorem gives \begin{align*} k[x]/(x-a)\cong k. \end{align*} Since the quotient is a field, $(x-a)$ is maximal; every maximal ideal is prime, so $(x-a)$ is a point of $\operatorname{Spec}k[x]$. We now compute its closure. A closed set in $\operatorname{Spec}k[x]$ has the form \begin{align*} V(I)=\{\mathfrak p\in\operatorname{Spec}k[x]:I\subset\mathfrak p\}. \end{align*} For $I=(x-a)$, \begin{align*} V(x-a)=\{\mathfrak p\in\operatorname{Spec}k[x]:(x-a)\subset\mathfrak p\}. \end{align*} If $\mathfrak p$ is a prime ideal containing $(x-a)$, then $\mathfrak p/(x-a)$ is a prime ideal of $k[x]/(x-a)$. Under the isomorphism $k[x]/(x-a)\cong k$, this gives a prime ideal of the field $k$. The only ideals of $k$ are $(0)$ and $k$, and a prime ideal must be proper, so the only prime ideal of $k$ is $(0)$. Hence \begin{align*} \mathfrak p/(x-a)=(0). \end{align*} This means every element of $\mathfrak p$ lies in $(x-a)$, so $\mathfrak p\subset(x-a)$. Since we already have $(x-a)\subset\mathfrak p$, it follows that $\mathfrak p=(x-a)$. Therefore \begin{align*} V(x-a)=\{(x-a)\}. \end{align*} Thus each point $(x-a)$ is closed in $\operatorname{Spec}k[x]$. The ideal $(0)$ is also prime. Indeed, if $fg\in(0)$, then $fg=0$ in $k[x]$; since $k[x]$ is an integral domain, either $f=0$ or $g=0$, equivalently $f\in(0)$ or $g\in(0)$. Hence $(0)$ is another point of $\operatorname{Spec}k[x]$. A closed set $V(I)$ contains $(0)$ exactly when \begin{align*} I\subset(0). \end{align*} Since $(0)$ contains only the zero polynomial, this forces $I=(0)$. Therefore the only closed set containing $(0)$ is \begin{align*} V(0)=\{\mathfrak p\in\operatorname{Spec}k[x]:(0)\subset\mathfrak p\}=\operatorname{Spec}k[x]. \end{align*} So the closure of $\{(0)\}$ is all of $\operatorname{Spec}k[x]$. If we kept only the maximal ideals $(x-a)$, every remaining point would be closed, and no point would have closure equal to the whole affine line; the generic point $(0)$ is exactly the point that records the irreducible line as a single closure. [/example] ## Beyond and Connected Topics The Zariski topology is closely related to the coordinate ring, the Nullstellensatz, localization, and prime ideals. It is the topological structure behind affine varieties and affine schemes. It also provides the ambient language for sheaves of regular functions, morphisms of varieties, and scheme-theoretic gluing. The projective analogue lives on [Projective Space](/page/Projective%20Space): homogeneous polynomials replace ordinary affine equations, and closed sets must be compatible with scaling in projective coordinates. Compared with [Cambridge II Algebraic Geometry](/page/Cambridge%20II%20Algebraic%20Geometry), this page isolates the topology itself. Compared with [Cambridge IB Analysis and Topology](/page/Cambridge%20IB%20Analysis%20and%20Topology), it emphasizes that closed sets can come from algebraic equations rather than limit points. The algebraic-topology pages [Cambridge II Algebraic Topology](/page/Cambridge%20II%20Algebraic%20Topology) and [Cambridge III Algebraic Topology](/page/Cambridge%20III%20Algebraic%20Topology) use topology for homotopy and homology, while the Zariski topology uses topology to encode polynomial equations. The central message is that algebra controls geometry through closed sets. Ideals define loci, radical ideals recover loci over algebraically closed fields, prime ideals describe irreducible behavior, and principal opens implement localization. The Zariski topology is therefore not an auxiliary decoration on algebraic geometry. It is the topology built to make algebraic equations, local algebra, and geometric structure speak the same language. ## References Androma, [Cambridge II Algebraic Geometry](/page/Cambridge%20II%20Algebraic%20Geometry). Androma, [Cambridge IB Analysis and Topology](/page/Cambridge%20IB%20Analysis%20and%20Topology). Androma, [Cambridge II Algebraic Topology](/page/Cambridge%20II%20Algebraic%20Topology). Androma, [Cambridge III Algebraic Topology](/page/Cambridge%20III%20Algebraic%20Topology). Projective analogue: [Projective Space](/page/Projective%20Space). Atiyah and Macdonald, *Introduction to Commutative Algebra* (1969). Hartshorne, *Algebraic Geometry* (1977). Vakil, *The Rising Sea: Foundations of Algebraic Geometry* (2025).