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In an affine variety with coordinate ring $A$, every dense open subset $U$ contains a dense principal open subset. Write its closed complement as $V(I)$. Density says that $I$ is contained in none of the finitely many minimal prime ideals of the reduced ring $A$. Prime avoidance supplies $s\in I$ outside all these primes. Thus $D(s)\subseteq U$, and $s$ is a non-zero-divisor. Noetherian Zariski topology also makes $U$ a finite union of principal opens; one can add $D(s)$ to this cover. An arbitrary existing cover need not contain a dense member when the variety is reducible.

If an ideal of a Noetherian ring is generated by $n$ elements, every prime ideal minimal over it has height at most $n$. The proof inducts on $n$, using the Krull principal ideal theorem in a quotient and prime avoidance to choose a prime chain compatible with the induction.

Let $X$ be an affine variety over an algebraically closed field $k$, with reduced coordinate ring $A=k[X]$. On an open subset $U$, define ${\mathcal{O}}_X(U)$ to be the functions $h:U\to k$ which, near every point, have an expression $a/b$ with $a,b\in A$ and $b$ nowhere zero on that neighbourhood. Restriction is ordinary restriction of functions. Compatible local functions glue uniquely and remain locally quotients, so this defines the structure sheaf. On the principal open subset $D(f)$ it gives $A_f$, and at $P$ it gives the local ring $A_{{\mathfrak{m}}_P}$. The principal-open identification can also be seen by identifying $D(f)$ with the affine algebraic set obtained by adjoining a coordinate $t$ and the equation $tf=1$; its coordinate ring is $A[t]/(tf-1)\cong A_f$. This construction permits several irreducible components; no division by a function vanishing identically on a component is assumed.

An abstract algebraic variety is a locally ringed space locally isomorphic to such affine models, with a finite affine cover; we use the usual separated convention. Reducible varieties are allowed here. The separation condition says that the diagonal morphism is a closed immersion. Two varieties are related by a birational map with an inverse when some dense open subset of one is isomorphic as a locally ringed space to a dense open subset of the other.

Define the ring of rational functions on a reduced variety by pairs $(U,h)$, where $U\subseteq X$ is dense open and $h\in {\mathcal{O}}_X(U)$, identifying two pairs when their restrictions agree on a common dense open subset. Finite intersections of dense open subsets are dense, so addition and multiplication on these intersections are well defined. For an irreducible variety this gives the usual function field; for a reducible variety it need not be a field. Restriction to any dense open subset $V$ induces an isomorphismIndeed, intersecting domains with $V$ defines the restriction, and a dense open subset of $V$ is also dense open in $X$, which defines its inverse. An isomorphism between dense opens therefore induces an isomorphism of their rational rings. Thus birational varieties have isomorphic rings of rational functions, also in the reducible case.

For the finite-cover assertion, write the closed complement of $U$ as $V(I)$, where $I=(f_1,\dots ,f_r)$ is an ideal of the Noetherian ring $A$. ThenLet ${\mathfrak{p}}_1,\dots ,{\mathfrak{p}}_t$ be the minimal prime ideals of $A$, corresponding to its irreducible components. Since $U$ is dense, it meets every component, so $I$ is contained in none of these primes. Prime avoidance supplies $s\in I$ outside their union. Consequently $D(s)\subseteq U$ meets a dense open subset of every component and is dense in $X$. Add $D(s)$ to the finite cover if necessary. This proves that a finite principal-open cover can be chosen with a dense member. It does not say that every principal-open cover already has such a member: for $X=V(xy)$, the dense open $X\setminus \{(0,0)\}=D(x)\cup D(y)$ has neither displayed member dense, although adding $D(x+y)$ supplies a dense member.

For clarity, the zero divisors in this reduced Noetherian ring are exactly ${\bigcup }_i{\mathfrak{p}}_i$. If $ab=0$ and $a$ avoids every ${\mathfrak{p}}_i$, reduction in the domain $A/{\mathfrak{p}}_i$ forces $b\in {\mathfrak{p}}_i$ for all $i$, hence $b=0$. Conversely, if $a\in {\mathfrak{p}}_i$, choose for every $j\ne i$ an element $b_j\in {\mathfrak{p}}_j\setminus {\mathfrak{p}}_i$, possible because distinct minimal primes are incomparable. Their product $b$ is not in ${\mathfrak{p}}_i$ but belongs to all the other primes. Then $b\ne 0$ and $ab\in {\bigcap }_j{\mathfrak{p}}_j=0$. For a single minimal prime, reducedness makes that prime zero and this argument uses $b=1$. It follows that

The total quotient ring $S^{-1}A$, where $S$ consists of these non-zero-divisors, maps to $\mathrm{Rat}(X)$ byThe equivalence relation for localization makes this a well-defined $k$-algebra homomorphism. For surjectivity, a rational class represented on $U$ restricts to the dense principal open $D(s)\subseteq U$ constructed above. Its section there belongs to $A_s$, so it is $a/s^r$ and lies in the image. For injectivity, if such a fraction is zero on a dense open subset, its numerator vanishes on that dense subset. The numerator's zero set is closed, so it vanishes everywhere on $X$ and is zero in the reduced coordinate ring. ThusIn particular $A\to S^{-1}A$ is canonically injective: $a/1=0$ implies $sa=0$ for some non-zero-divisor $s$, hence $a=0$.

Finally let $Y$ be the reduced closed union of exactly those irreducible components of $X$ which contain $P$. Removing all the other components gives an open neighbourhood of $P$ on which $X$ and $Y$ coincide, including their structure sheaves. Hence ${\mathcal{O}}_{X,P}\cong {\mathcal{O}}_{Y,P}$. Write $B=k[Y]$ and let ${\mathfrak{n}}_P$ be its maximal ideal at $P$. Every minimal prime of $B$ is contained in ${\mathfrak{n}}_P$, because every component of $Y$ contains $P$. If $b\notin {\mathfrak{n}}_P$, it avoids each such prime and is a non-zero-divisor. Therefore the map of localizationsis defined and injective: if its value is zero, a non-zero-divisor annihilates $a$, so $a=0$. We obtain the desired local ring embeds in rational functions on incident components:At an intersection point one must retain all incident components. For example, the local ring of $V(xy)$ at the origin has two nonzero germs $x,y$ with $xy=0$; it cannot embed into the function field of just one of its axes.

A prime ideal $P$ is a proper ideal such that $ab\in P$ implies $a\in P$ or $b\in P$. Equivalently $R/P$ is an integral domain.

For each $i$, and each $j\ne i$, choose $a_{ji}\in P_j\setminus P_i$, using incomparability. PutThen $b_i$ belongs to the other two prime ideals and does not belong to $P_i$, by primality. In particular each $b_i$ lies in the union. But $b_1+b_2+b_3\notin P_i$ for every $i$: modulo $P_i$, the sum equals $b_i\ne 0$. The union is not closed under addition and is not an ideal. This is a concrete instance of prime avoidance.

The union of three incomparable nonprime ideals example can retain pairwise incomparability. In the ringtakeThese distinct two-element ideals are pairwise incomparable, yetwhich is an ideal. None is prime: its quotient ring still contains a nonzero element of square zero. The three one-dimensional vector subspaces cover the two-dimensional vector space over ${\mathbb{F}}_2$.

We prove prime avoidance by contradiction. Remove any prime ideal contained in another one on the list; this leaves the same union. If the ideal $I$ is contained in none of the remaining prime ideals, choose $a_i\in I\setminus P_i$. For every $j\ne i$, incomparability lets us choose $b_{ij}\in P_j\setminus P_i$. SetFor a one-element list, use the empty product $b_1=1$. By the defining property of a prime ideal, $b_i\notin P_i$, and also $a_i{b}_i\notin P_i$. But $b_j\in P_i$ for every $j\ne i$. Thus all summands except $a_i{b}_i$ lie in $P_i$, and consequently $c\notin P_i$ for every $i$. On the other hand, $c\in I$ because each $a_i\in I$. This contradicts the assumed containment in the union. Therefore

An integral domain $A$ is an integrally closed domain if every element of its fraction field which is an integral element over $A$ belongs to $A$. Explicitly, if $z\in \mathrm{Frac}A$ satisfiesthen $z\in A$. Thus $\boxed{A\text{ is integrally closed exactly when its integral closure in its fraction field is }A}$.

Suppose $A$ is a unique factorization domain and $z=a/b$ is integral, with $a,b\in A$, $b\ne 0$, and with no common irreducible element as a factor. Multiplying its monic equation by $b^d$ givesHence $b$ divides $a^d$. Every irreducible element dividing $b$ is a prime element, so it would divide $a$, contrary to the choice of the fraction. Therefore $b$ is a unit, and $z\in A$. We have proved

The going-down theorem is the following. Let $A\subseteq B$ be an integral extension of integral domains with $A$ an integrally closed domain. IfthenWe first prove this for a module-finite ring extension and then remove that extra hypothesis. We use the Lying-over theorem, the going-up theorem, and the incomparability theorem for integral extensions: primes exist over every base prime, prime chains can be extended upwards, and distinct primes over the same base prime cannot contain one another. The going-up theorem follows by applying the Lying-over theorem to the quotient by the lower prime. These results concern arbitrary integral extensions; none assumes the going-down theorem.

For the finite case, let $K=\mathrm{Frac}A$ and $L=\mathrm{Frac}B$. Since $B$ is module-finite, $L/K$ is a finite field extension. Take a finite normal field extension $E/K$ containing $L$, obtained as a splitting field of the minimal polynomials of finitely many field generators. The extension $E/K$ need not be a separable field extension. Let $C$ be the integral closure of $A$ in $E$. Then $B\subseteq C$, and $C$ is integral over both $A$ and $B$.

We need one auxiliary fact: the finite group $G={\mathrm{Aut}}_K(E)$ acts transitively on the prime ideals of $C$ over any fixed prime ideal $\mathfrak{p}$ of $A$. First observe that an element $w\in E$ fixed by all of $G$ is a purely inseparable algebraic element over $K$. Indeed, all $K$-embeddings of $K(w)$ into an algebraic closure extend to $K$-automorphisms of the normal field extension $E$. Hence the minimal polynomial of an algebraic element $w$ has only one distinct root. In characteristic zero this says $w\in K$; in characteristic $p>0$ it says $w^{p^e}\in K$ for some $e\ge 0$.

Now let $Q,Q^{\prime }$ be primes of $C$ over $\mathfrak{p}$, and suppose $Q$ is different from every $\sigma (Q^{\prime })$, $\sigma \in G$. By the incomparability theorem for integral extensions, $Q$ is contained in none of these conjugate primes. The prime avoidance lemma suppliesThe productis fixed by $G$, is integral over $A$, and lies in $Q$ because one factor is $x$. Choose an exponent $e_0$ equal to $1$ in characteristic zero, or a power of $p$ in characteristic $p$, such that $h^{e_0}\in K$. Since $A$ is an integrally closed domain, $h^{e_0}\in A$. It also lies in $Q\cap A=\mathfrak{p}$, hence in $Q^{\prime }$. Because $Q^{\prime }$ is a prime ideal, some factor $\sigma (x)$ lies in $Q^{\prime }$. This means $x\in {\sigma }^{-1}(Q^{\prime })$, contradicting our choice. Thus $Q=\sigma (Q^{\prime })$ for some $\sigma$, proving transitivity. The power $e_0$ accounts for the possible purely inseparable field extension and makes the argument valid in every characteristic.

Lift the prescribed ${\mathfrak{q}}_2$ to a prime $Q_2$ of $C$ by the Lying-over theorem for $B\subseteq C$. Separately, choose a prime $Q_1^{\prime }$ over ${\mathfrak{p}}_1$ and use the going-up theorem to find $Q_2^{\prime }\supseteq Q_1^{\prime }$ over ${\mathfrak{p}}_2$. Transitivity gives $\sigma \in G$ with $\sigma (Q_2^{\prime })=Q_2$. ThenContracting to $B$ gives ${\mathfrak{q}}_1=Q_1\cap B\subseteq {\mathfrak{q}}_2$ with the required contraction. This proves the going-down theorem for the module-finite case.

For a general integral extension $A\subseteq B$, consider $B_{{\mathfrak{q}}_2}$ and the idealWe claim that $J$ is disjoint from $A\setminus {\mathfrak{p}}_1$. Otherwise, clearing the finitely many denominators in an expression for an element $a\in J\cap (A\setminus {\mathfrak{p}}_1)$ givesThe subalgebra $B_0=A[s,b_1,\dots ,b_h]$ is a module-finite ring extension of $A$, since all its generators are integral elements. Apply the finite case to ${\mathfrak{q}}_2\cap B_0$: there is a prime $Q_1\subseteq {\mathfrak{q}}_2\cap B_0$ contracting to ${\mathfrak{p}}_1$. The displayed identity puts $sa$ in $Q_1$, but $a\notin Q_1$. Thus $s\in Q_1\subseteq {\mathfrak{q}}_2$, a contradiction.

Localize $B_{{\mathfrak{q}}_2}$ further at the multiplicative subset $A\setminus {\mathfrak{p}}_1$. The extension of $J$ is a proper ideal, by the disjointness just proved. Choose a maximal ideal containing it and contract back to a prime ideal of $B_{{\mathfrak{q}}_2}$. This prime contains $J$ and avoids $A\setminus {\mathfrak{p}}_1$. Its contraction ${\mathfrak{q}}_1$ to $B$ lies inside ${\mathfrak{q}}_2$ by the prime ideal correspondence for localization, and its contraction to $A$ is exactly ${\mathfrak{p}}_1$. This completes the proof of the going-down theorem without any finite generation assumption on $B$.

For a prime ideal $\mathfrak{p}$, its height is the supremum of the lengths of strict chains of prime ideals ending at $\mathfrak{p}$. For a proper ideal $I$,This is the height of an ideal.

The Krull height theorem states that if $R$ is Noetherian, $I=(a_1,\dots ,a_m)$, and $\mathfrak{p}$ is minimal over $I$, then

We prove it by induction on $m$. The case $m=1$ is the Krull principal ideal theorem. For the induction step, let ${\mathfrak{q}}_1,\dots ,{\mathfrak{q}}_s$ be the finitely many minimal primes over $(a_2,\dots ,a_m)$ that lie below $\mathfrak{p}$. By induction each has height at most $m-1$; if one equals $\mathfrak{p}$, we are done.

Otherwise, suppose $\mathfrak{p}$ has finite height $d$ and choose a chainwhose first nonminimal term ${\mathfrak{p}}_1$ is contained in none of the ${\mathfrak{q}}_i$. Such a chain is obtained by prime avoidance and the principal ideal theorem: a three-term segment can be replaced by a prime minimal over ${\mathfrak{p}}_0+(b)$ for an element $b$ avoiding the finitely many unwanted primes. Choose

The prime $\mathfrak{p}$ is minimal over $(b,a_2,\dots ,a_m)$. Otherwise a prime strictly between some ${\mathfrak{q}}_i$ and $\mathfrak{p}$ would show that $\mathfrak{p}/(a_2,\dots ,a_m)$ has height at least two, although it is minimal over the principal ideal generated by $a_1$; this contradicts the principal ideal theorem. In $R/(b)$, the prime $\mathfrak{p}/(b)$ is therefore minimal over an ideal generated by $m-1$ elements, so induction bounds its height by $m-1$. The $d-1$ strict inclusions from ${\mathfrak{p}}_1/(b)$ to $\mathfrak{p}/(b)$ giveand hence $d\le m$. If the height were infinite, the same argument applied to arbitrarily long finite chains would give the same fixed bound, which is impossible. This completes the proof.

If $R$ is Noetherian and ${\mathfrak{p}}_0⊊{\mathfrak{p}}_1⊊{\mathfrak{p}}_2$ are prime ideals, there are infinitely many primes strictly between ${\mathfrak{p}}_0$ and ${\mathfrak{p}}_2$. After quotienting by ${\mathfrak{p}}_0$ and localizing at ${\mathfrak{p}}_2$, this reduces to a local Noetherian domain of dimension at least two. If it had only finitely many height-one primes, prime avoidance would give a nonzero element of the maximal ideal outside all of them, contradicting the Krull principal ideal theorem.

In ${\mathbb{F}}_2[u,v]/(u,v{)}^2$, the ideals $(u)$, $(v)$ and $(u+v)$ are pairwise incomparable and nonprime. Their union is the ideal $(u,v)$: the three lines cover its two-dimensional vector space over ${\mathbb{F}}_2$. This shows why a primality hypothesis matters in prime avoidance.