# Discrete valuation rings, normal rings and Serre's criterion

*Written by GPT-6.1 Sol (OpenAI) in Codex, Ultra setting, October 2026. Self-checked by the writing AI. Public domain (CC0).*

A normal curve has a particularly simple local algebra: at a closed point, one parameter measures every order of vanishing. In higher dimension, these one-dimensional local rings still detect whether a rational function has a pole. Extending across the remaining points requires a second condition, expressed by depth. We prove the local characterization, the codimension-one intersection theorem and the two Serre criteria, including rings with several components.

Rings are commutative with identity. A domain is nonzero, and finite means finitely generated. A **normal domain** is integrally closed in its fraction field. A **normal ring** is a ring whose localization at every prime is a normal domain; this permits finite products and the zero ring. Noetherian hypotheses will be explicit. We use the [integrality lesson](integral-extensions-lying-over-going-up-and-going-down.md), Theorems 1.1 and 2.2, for the determinant trick and localization of normal domains; the [associated-prime lesson](associated-primes-and-primary-decomposition.md), Theorems 1.2, 2.2, 3.1 and 3.2, for existence, finiteness, localization and minimal support; the [depth lesson](regular-sequences-depth-and-cohen-macaulay-modules.md), Lemma 2.2 and Theorem 2.3, for depth zero and quotient depth; and the [regular-local lesson](regular-local-rings.md) for regularity and its localization.

## 1. A single parameter and a discrete valuation

A **valuation domain** is a domain \(A\), with fraction field \(K\), such that every \(u\in K^*\) satisfies \(u\in A\) or \(u^{-1}\in A\). Equivalently its principal ideals are comparable: compare the fraction of their nonzero generators. In fact all its ideals are comparable. If \(I\not\subset J\), choose \(a\in I\setminus J\); comparison with every \(b\in J\) forces \(b\in aA\subset I\). Every finitely generated ideal is therefore generated by one of its generators. Nonunits form an ideal: if two principal ideals are comparable, their nonunit generators have a sum divisible by one nonunit generator. Multiplication preserves nonunits. This proper ideal is the unique maximal ideal, so valuation domains are local.

A **discrete valuation ring**, or **DVR**, is a domain of the form

\[
A=\{0\}\cup\{u\in K^*:v(u)\geq0\},
\qquad v:K^*\twoheadrightarrow\mathbb Z,
\tag{1}
\]

where \(v(uw)=v(u)+v(w)\) and \(v(u+w)\geq\min\{v(u),v(w)\}\) when the sum is nonzero. Choose a **uniformizer** \(\pi\) with \(v(\pi)=1\). An element of value zero is a unit. Every nonzero element of \(A\) is uniquely a unit times \(\pi^n\), with \(n\geq0\). Consequently every nonzero ideal has the form

\[
I=(\pi^n),\qquad n=\min\{v(a):0\ne a\in I\}.
\tag{2}
\]

The zero ideal is separate from these finite powers. Thus \(A\) is Noetherian, its maximal ideal is \((\pi)\), and its only primes are \((0)\) and \((\pi)\). In particular it has dimension one and is regular local. It is a principal ideal domain, hence normal by Proposition 2.3 of the integrality lesson.

The following elementary argument explains how normality forces a parameter.

**Lemma 1.1 (a fraction acting on the maximal ideal).** Let \((A,\mathfrak m)\) be a Noetherian normal local domain with \(\mathfrak m\ne0\). If \(u\in\operatorname{Frac}(A)\setminus A\) and \(u\mathfrak m\subset A\), then \(\mathfrak m\) is principal and \(A\) is a DVR.

**Proof.** If \(u\mathfrak m\subset\mathfrak m\), multiplication by \(u\) preserves the finite module \(\mathfrak m\subset\operatorname{Frac}(A)\). The determinant trick gives a monic polynomial \(P\in A[T]\) annihilating this multiplication. As \(\mathfrak m\ne0\) and \(A\) is a domain, \(P(u)=0\). Normality would give \(u\in A\), a contradiction. Hence the ideal \(u\mathfrak m\subset A\) contains an element outside \(\mathfrak m\), so equals \(A\). It follows that \(\mathfrak m=u^{-1}A\). Its generator is nonzero and a nonunit. The principal ideal theorem, Theorem 3.1 of the dimension lesson, gives \(\dim A\leq1\); the strict chain \((0)\subsetneq\mathfrak m\) gives equality. Thus \(A\) is regular of dimension one. Solution 7.2 of the regular-local lesson proves directly, by Krull intersection and powers of a generator, that such a ring is a DVR. \(\square\)

**Theorem 1.2 (DVR characterizations).** For a local domain \(A\), the following are equivalent:

1. \(A\) is a DVR.
2. \(A\) is a Noetherian valuation domain and is not a field.
3. \(A\) is regular local of dimension one.
4. \(A\) is Noetherian, normal and of dimension one.
5. \(A\) is Noetherian of dimension one and its maximal ideal is principal.

**Proof.** Formula (2) proves that a DVR satisfies all the listed properties. For (2), finite generation and comparability make the nonzero maximal ideal principal. The principal ideal theorem gives dimension one, proving (5). Conditions (3) and (5) are equivalent by the definition of embedding dimension, and the cited direct uniformizer proof gives (3) implies (1).

Suppose (4). Choose \(0\ne a\in\mathfrak m\). Every prime containing \(a\) is \(\mathfrak m\), so \(\sqrt{aA}=\mathfrak m\). Finite generation gives \(\mathfrak m^n\subset aA\) for some \(n\geq1\). Take the least such \(n\) and choose

\[
b\in\mathfrak m^{n-1}\setminus aA,
\qquad u=b/a.
\tag{3}
\]

Then \(u\notin A\) but \(u\mathfrak m\subset A\). Lemma 1.1 applies. This proves the remaining implication. Fields are normal of dimension zero and are excluded from every DVR characterization here. \(\square\)

See [Stacks, Tags 00PD and 00II]. For familiar examples, \(\mathbb Z_{(p)}\) has uniformizer \(p\), \(k[x]_{(x)}\) has uniformizer \(x\), and \(k[[t]]\) has uniformizer \(t\). In the last ring the valuation is the least exponent with nonzero coefficient; multiplication adds these exponents, and a series with nonzero constant term has an inverse obtained recursively coefficient by coefficient.

## 2. From ideal factorization to Dedekind domains

We use the ideal-theoretic definition: a **Dedekind domain** is a domain in which every nonzero ideal factors uniquely, up to order, as a finite product of nonzero prime ideals. The unit ideal is the empty product. No Noetherian assumption is included in this definition, and fields qualify. [Stacks, Tag 034W.]

Two short arguments will supply the missing finiteness.

**Lemma 2.1.** If every prime ideal of a ring is finitely generated, the ring is Noetherian.

**Proof.** Suppose some ideal is not finitely generated. Zorn's lemma gives a maximal such ideal \(I\): the union of a chain of these ideals is still not finitely generated, since finitely many proposed generators would belong to one chain member. The ideal \(I\) is proper. If it were not prime, choose \(a,b\notin I\) with \(ab\in I\). Both \(I+(a)\) and \((I:a)\) strictly contain \(I\), the latter because it contains \(b\), so both are finitely generated. Write

\[
I+(a)=(a,c_1,\ldots,c_r),\quad c_i\in I,
\qquad (I:a)=(d_1,\ldots,d_s).
\]

Such a first list is obtained by subtracting multiples of \(a\) from any finite generating list. For \(z\in I\), write \(z=ta+\sum t_i c_i\). Then \(t\in(I:a)\), so the \(c_i\) and \(ad_j\) generate \(I\), a contradiction. Thus \(I\) is a non-finitely-generated prime, contradicting the hypothesis. \(\square\)

**Lemma 2.2.** In a domain, if \(IJ=(a)\) with \(a\ne0\), both \(I\) and \(J\) are finitely generated.

**Proof.** Express \(a=\sum_{i=1}^r u_i v_i\) with \(u_i\in I\), \(v_i\in J\). For \(u\in I\), write \(uv_i=ac_i\). Then \(ua=a\sum u_i c_i\); cancellation gives \(u=\sum u_i c_i\). The \(u_i\) generate \(I\), and the symmetric argument handles \(J\). \(\square\)

These finiteness lemmas are [Stacks, Tags 05KG and 09ME].

**Theorem 2.3 (Dedekind characterizations).** For a ring \(R\), the following are equivalent:

1. \(R\) is a Dedekind domain in the ideal-theoretic sense above.
2. \(R\) is a Noetherian domain and \(R_{\mathfrak m}\) is a DVR for every nonzero maximal ideal \(\mathfrak m\).
3. \(R\) is a Noetherian normal domain with \(\dim R\leq1\).

**Proof.** Assume (1). For a nonzero prime \(\mathfrak p\), uniqueness gives \(\mathfrak p\ne\mathfrak p^2\): the factorizations with one and two copies would otherwise agree. Choose \(x\in\mathfrak p\setminus\mathfrak p^2\). For any \(y\in\mathfrak p\), factor the nonzero ideal \((x,y^2)\). At least one prime factor is contained in \(\mathfrak p\), because a product contained in a prime has a factor contained there. There is at most one such factor, counting multiplicity, since two would put \(x\) in \(\mathfrak p^2\). The other factors become unit ideals at \(\mathfrak p\). Thus \((x,y^2)R_{\mathfrak p}\) is prime. It contains \(y^2\), hence \(y\), and an expression

\[
y=cx+dy^2,
\qquad (1-dy)y=cx
\quad\text{in }R_{\mathfrak p}
\tag{4}
\]

shows \(y\in xR_{\mathfrak p}\), since \(1-dy\) is a unit. We have proved \(\mathfrak pR_{\mathfrak p}=xR_{\mathfrak p}\).

Now factor \((x)\). Its unique factor \(\mathfrak q\subset\mathfrak p\) has \(\mathfrak qR_{\mathfrak p}=\mathfrak pR_{\mathfrak p}\). Contraction gives \(\mathfrak q=\mathfrak p\), by prime correspondence. Lemma 2.2 makes this factor of the nonzero principal ideal finitely generated. The zero prime is already finitely generated. Lemma 2.1 proves \(R\) Noetherian. Every nonzero maximal ideal localizes to a nonzero principal maximal ideal, so the principal ideal theorem and Theorem 1.2 make its localization a DVR. This proves (2) without assuming finiteness prematurely.

Assume (2). Every nonzero prime is contained in a nonzero maximal ideal, whose localization has only one nonzero prime. Prime correspondence therefore makes every nonzero prime maximal. Thus \(\dim R\leq1\). The maximal localizations are normal, including the field case, so the integrality lesson, Theorem 2.2, gives normality. Conversely (3), localized at a nonzero maximal ideal, has dimension one and satisfies Theorem 1.2. Hence (2) and (3) are equivalent.

Finally assume these conditions and let \(I\) be a nonzero proper ideal. Choose \(0\ne a\in I\). There are only finitely many maximal ideals containing \(a\): they are the minimal primes over \((a)\), since every nonzero prime is maximal. Let \(\mathfrak m_1,\ldots,\mathfrak m_r\) be those containing \(I\). At each such maximal ideal write

\[
I R_{\mathfrak m_i}=(\pi_i^{n_i}),\quad n_i>0,
\qquad J=\prod_{i=1}^r\mathfrak m_i^{n_i}.
\tag{5}
\]

Distinct maximal ideals become unit ideals in each other's localizations, so \(I\) and \(J\) agree at every maximal ideal. Local module detection, applied to \((I+J)/I\) and \((I+J)/J\), gives \(I=J\). Each exponent is forced by (2), and every nonzero prime factor must be one of the maximal ideals containing \(I\). This proves uniqueness. The unit ideal and fields use the empty product. \(\square\)

This supplies both finiteness and uniqueness in [Stacks, Tag 034X]. In dimension one, ideal factorization assembles the separate orders of vanishing at the closed points.

## 3. Normal domains and codimension-one detection

**Lemma 3.1 (associated primes detect a module element).** Over a Noetherian ring, if \(0\ne z\in M\), there is \(\mathfrak p\in\operatorname{Ass}(M)\) such that \(z/1\ne0\) in \(M_{\mathfrak p}\).

**Proof.** The nonzero cyclic submodule \(Rz\) has an associated prime \(\mathfrak p\). Choose \(0\ne w\in Rz\) with annihilator \(\mathfrak p\). It survives localization at \(\mathfrak p\), since no element outside \(\mathfrak p\) kills it. If \(z\) vanished there, so would \(w\). The inclusion of the submodule puts \(\mathfrak p\) in \(\operatorname{Ass}(M)\), by Proposition 2.1 of the associated-prime lesson. \(\square\)

The detection lemma is [Stacks, Tag 0311].

**Theorem 3.2.** Let \(R\) be a Noetherian normal domain and \(a\ne0\). All associated primes of \(R/aR\) have height one, and \(R/aR\) has no embedded primes.

**Proof.** A unit \(a\) gives the zero module, so suppose an associated prime \(\mathfrak p\) exists. Localize: \(A=R_{\mathfrak p}\) is a Noetherian normal domain, and its maximal ideal \(\mathfrak m\) is associated to \(A/aA\). Choose a nonzero class \(b+aA\) whose annihilator is \(\mathfrak m\). Then

\[
u=b/a\notin A,\qquad u\mathfrak m\subset A.
\tag{6}
\]

The maximal ideal contains the nonzero element \(a\), so Lemma 1.1 applies. It gives \(\dim R_{\mathfrak p}=1\), that is, \(\operatorname{ht}\mathfrak p=1\). An associated prime is minimal in the support precisely when it is minimal among associated primes. A strict containment between two of these height-one primes would force height at least two above the chain starting at \((0)\). Thus all are minimal and none is embedded. \(\square\)

In particular these associated primes are exactly the minimal primes over \((a)\), by the minimal-support theorem. This height-one mechanism also occurs in the proof of [Stacks, Tag 031S].

**Theorem 3.3 (algebraic Hartogs).** A Noetherian normal domain \(R\), with fraction field \(K\), satisfies

\[
\boxed{R=\bigcap_{\operatorname{ht}\mathfrak p=1}R_{\mathfrak p}\ \subset K.}
\tag{7}
\]

Every ring in this intersection is a DVR.

**Proof.** Each height-one localization is normal of dimension one, hence a DVR by Theorem 1.2. The inclusion from left to right is immediate. For the reverse inclusion write \(u=b/a\), \(a\ne0\). If \(u\notin R\), the class of \(b\) in \(R/aR\) is nonzero. Lemma 3.1 and Theorem 3.2 provide a height-one prime where that class survives, so \(u\notin R_{\mathfrak p}\). This proves (7). If there are no height-one primes, the intersection is interpreted inside \(K\) as all of \(K\); the same argument shows \(R=K\). \(\square\)

Corresponding results are [Stacks, Tags 031S and 031T]. The mechanism is explicit: a failed denominator is detected by an associated prime of its quotient, and normality forces that detecting prime into codimension one.

There is also a depth consequence we will need.

**Proposition 3.4.** A Noetherian normal local domain \(A\) has depth at least \(\min\{2,\dim A\}\).

**Proof.** In dimension zero it is a field; in dimension one it is a DVR and its uniformizer is regular. In dimension at least two, choose \(0\ne a\in\mathfrak m\). Its multiplication is injective. The finite associated-prime set of \(A/aA\) consists of height-one primes by Theorem 3.2, so none is \(\mathfrak m\). Prime avoidance, Solution 8.5 of the associated-prime lesson, provides \(b\in\mathfrak m\) outside their union. It is a nonzerodivisor on \(A/aA\), by the zero-divisor theorem. The quotient by \((a,b)\) is nonzero since this ideal is proper. Thus \(a,b\) is a regular sequence. \(\square\)

## 4. Serre conditions and the normality criterion

For a Noetherian ring \(R\) and an integer \(k\geq0\), define

\[
\begin{aligned}
(R_k)&:\ R_{\mathfrak p}\text{ is regular whenever }
                 \operatorname{ht}\mathfrak p\leq k,\\
(S_k)&:\ \operatorname{depth}R_{\mathfrak p}
          \geq\min\{k,\operatorname{ht}\mathfrak p\}
          \quad\text{for every }\mathfrak p.
\end{aligned}
\tag{8}
\]

Here \(\operatorname{ht}\mathfrak p=\dim R_{\mathfrak p}\). For a finite module \(M\), the condition \((S_k)\) uses \(\dim\operatorname{Supp}(M_{\mathfrak p})\) in place of the ring dimension, at primes in its support; outside the support it holds vacuously, with \(\operatorname{depth}0=\infty\). Thus \((R_1)\) includes height zero, and \((S_2)\) asks for a regular pair only when the local dimension is at least two. [Stacks, Tag 031P.]

**Proposition 4.1.** A finite module \(M\) over a Noetherian ring satisfies \((S_1)\) if and only if it has no embedded associated primes.

**Proof.** By associated-prime localization and the depth-zero test,

\[
\operatorname{depth}M_{\mathfrak p}=0
\quad\Longleftrightarrow\quad
\mathfrak p\in\operatorname{Ass}(M)
\qquad(\mathfrak p\in\operatorname{Supp}M).
\tag{9}
\]

Indeed the maximal ideal of \(R_{\mathfrak p}\) can be the extension of an associated prime only from \(\mathfrak p\) itself. The local support has dimension zero exactly when \(\mathfrak p\) is minimal in the global support: its primes correspond to support primes contained in \(\mathfrak p\). Such primes are the minimal associated primes. Consequently \((S_1)\) forbids exactly the depth-zero points which are not minimal. \(\square\)

See [Stacks, Tag 031Q].

**Theorem 4.2 (reducedness criterion).** A Noetherian ring is reduced if and only if it satisfies \((R_0)\) and \((S_1)\).

**Proof.** In a reduced ring, a minimal-prime localization is a reduced zero-dimensional local ring, hence a field: its only prime is its maximal ideal, which is also its nilradical and so is zero. This gives \((R_0)\). Solution 8.4 of the associated-prime lesson proves that a reduced Noetherian ring has only its minimal primes as associated primes. Proposition 4.1 gives \((S_1)\).

Conversely these conditions say the minimal localizations are fields and there are no embedded associated primes. If the nilradical \(N\) were nonzero, it would have an associated prime, which as a submodule of \(R\) is one of the minimal primes of \(R\). Its annihilator representative survives there. But \(N\) localizes to zero at a minimal prime, because that localized ring is a field. This contradiction proves reducedness. The zero ring has empty spectrum and satisfies both conditions. \(\square\)

This is [Stacks, Tag 031R]. To pass from reducedness to normality we must also account for components.

**Lemma 4.3 (components and the total quotient ring).** Let \(R\) be Noetherian and reduced, and let \(S\) be its set of nonzerodivisors. If \(\mathfrak p_1,\ldots,\mathfrak p_r\) are its minimal primes, then

\[
Q(R)=S^{-1}R\simeq
\prod_{i=1}^r\operatorname{Frac}(R/\mathfrak p_i).
\tag{10}
\]

Moreover the following are equivalent: \(R\) is normal; \(R\) is integrally closed in \(Q(R)\); \(R\) is a finite product of normal domains.

**Proof.** The reduced-ring associated-prime result and zero-divisor theorem give \(S=R\setminus\bigcup_i\mathfrak p_i\). A prime disjoint from \(S\) is contained in this finite union, hence in one \(\mathfrak p_i\) by prime avoidance, and thus equals it. The primes of \(Q(R)\) are exactly these minimal primes extended. This ring is Noetherian, reduced and zero-dimensional, hence Artinian. Its nilradical is the intersection of its finitely many maximal ideals and is zero. Distinct maximal ideals are comaximal, so the Chinese remainder theorem identifies it with the product of its field quotients. The factor belonging to \(\mathfrak p_i\) is also its localization there, namely \(R_{\mathfrak p_i}\). This field contains \(R/\mathfrak p_i\), and every element of it is a fraction from that domain; it is its fraction field. This proves (10).

Suppose \(R\) is integrally closed in (10). Each coordinate idempotent \(e_i\) satisfies \(T^2-T=0\), so belongs to \(R\). Thus \(R=\prod_i e_iR\), with each factor a domain inside its corresponding field. The fraction field of that factor is the field in (10), because projection of \(R\) is \(R/\mathfrak p_i\). If an element of this field is integral over \(e_iR\), place it in the \(i\)-th coordinate and put zero in the others. Its monic equation extends to a monic equation over \(R\), using \(T^n\) in the other coordinates. Integral closedness puts it in \(R\), and therefore in \(e_iR\). Each factor is normal.

A finite product of normal domains is normal: a prime selects one factor, and its localization is a localization of that normal domain, by the integrality lesson. Conversely suppose every prime localization of \(R\) is a normal domain. More generally this already forces \(R\) reduced, since a nilpotent vanishes in every localization and local detection makes it zero. For an integral fraction \(b/a\in Q(R)\), \(a\) a nonzerodivisor, its image is integral in the fraction field of every \(R_{\mathfrak p}\), hence belongs to that normal domain. Thus \(b+aR\) vanishes at every prime in \(R/aR\). Local detection gives \(b\in aR\), proving integral closedness. Empty products cover the zero ring. \(\square\)

This verifies the convention in [Stacks, Tags 030C and 030B]. In particular components of a normal Noetherian ring are disjoint, rather than merely generically normal.

**Theorem 4.4 (Serre's normality criterion).** A Noetherian ring is normal if and only if it satisfies \((R_1)\) and \((S_2)\).

**Proof.** If \(R\) is normal, every local ring is a normal domain. In height zero it is a field; in height one Theorem 1.2 makes it a DVR and hence regular. Proposition 3.4, applied at every prime, gives \((S_2)\).

Conversely suppose \((R_1)\) and \((S_2)\). They imply \((R_0)\) and \((S_1)\), so Theorem 4.2 makes \(R\) reduced. Take an integral fraction \(b/a\in Q(R)\), where \(a\) is a nonzerodivisor. For any \(\mathfrak p\in\operatorname{Ass}(R/aR)\), localization and (9) give depth zero for \(R_{\mathfrak p}/aR_{\mathfrak p}\). The regular-element depth formula of the depth lesson gives

\[
\operatorname{depth}R_{\mathfrak p}=1.
\tag{11}
\]

Condition \((S_2)\) therefore forces \(\operatorname{ht}\mathfrak p\leq1\). This prime contains \(a\), and \(a\) belongs to no minimal prime, because those are associated primes of the reduced ring. Hence its height is exactly one. By \((R_1)\), \(R_{\mathfrak p}\) is regular of dimension one, so is a normal DVR. The integral fraction belongs to it. Thus the class of \(b\) in \(R/aR\) vanishes at every associated prime. Lemma 3.1 makes this class zero. We have proved integral closedness in \(Q(R)\), and Lemma 4.3 gives normality, including the separation of components. \(\square\)

This proves the full ring version of [Stacks, Tag 031S]. There is no irreducibility assumption.

**Corollary 4.5.** Every Noetherian regular ring is normal. A Noetherian Cohen–Macaulay ring is normal if and only if it satisfies \((R_1)\).

**Proof.** Regularity of all prime localizations supplies \((R_1)\); their Cohen–Macaulayness, Theorem 1.1 of the regular-local lesson, gives \((S_2)\). For a Cohen–Macaulay ring, the equality of depth and dimension at every prime gives \((S_2)\) directly. Apply Theorem 4.4 in both cases. \(\square\)

The regular-ring implication is [Stacks, Tag 0567]. In particular a normal Noetherian local domain of dimension at most two is Cohen–Macaulay: Proposition 3.4 reaches its full dimension.

## 5. Curves and two-dimensional examples

Let \(\theta=(1+\sqrt{-3})/2\). It satisfies \(\theta^2-\theta+1=0\), but does not belong to \(\mathbb Z[\sqrt{-3}]\). Thus the latter is not normal. Solution 7.1 of the integrality lesson computes the full quadratic normalization as \(\mathbb Z[\theta]\). It is finite over \(\mathbb Z\), so Noetherian, and integral dimension invariance makes its dimension one. It is normal, hence Dedekind by Theorem 2.3. The failure of unique element factorization in a Dedekind domain does not interfere with unique ideal factorization.

The cusp has coordinate ring \(k[t^2,t^3]\). At its origin the dimension is one but the two cotangent classes \(t^2,t^3\) are independent, so it fails \((R_1)\). The fraction \(t\) supplies an explicit integral element outside the ring. Solution 6.1 below calculates its normalization in every characteristic.

For the cone and the meeting planes, write

\[
C=k[x,y,z]/(xy-z^2),\qquad
B=k[x,y,z,w]/(xz,xw,yz,yw).
\tag{12}
\]

The cone is a Cohen–Macaulay domain of dimension two. Away from its origin, the charts obtained by inverting \(x\) or \(y\) are regular; the origin has height two. Hence the cone satisfies \((R_1)\) and \((S_2)\), and is normal. Its local ring at the origin has embedding dimension three and is not regular. The regular-local lesson, Section 6, also proves it is not a UFD. All these conclusions hold in characteristic two as well.

The two components of \(B\) are planes, meeting only at the origin. Elsewhere every point has a regular chart, so \((R_1)\) holds. At the origin the local dimension is two and the depth is one. This is the obstruction to \((S_2)\) and normality. Solutions 6.4 and 6.5 give the complete chart and depth calculations, so neither assertion relies on a picture of the singularity.

## 6. Exercises

**Exercise 6.1 (easy: the cusp).** Over an arbitrary field, show that \(A=k[x,y]/(y^2-x^3)\) is not normal and compute its normalization. Verify the failure of \((R_1)\) at the origin.

**Exercise 6.2 (medium: a hypersurface test).** Let \((R,\mathfrak m)\) be Noetherian regular local and \(0\ne f\in\mathfrak m\). Prove that \(R/fR\) is normal if and only if it satisfies \((R_1)\).

**Exercise 6.3 (medium: punctured-plane Hartogs).** Prove directly that a regular function on \(\operatorname{Spec}k[x,y]\setminus\{(x,y)\}\) is a polynomial. Interpret regular functions as compatible functions locally represented by fractions whose denominators are invertible. Use the cover \(D(x)\cup D(y)\).

**Exercise 6.4 (medium: the cone).** Prove that the affine ring \(C\) in (12) is normal over every field. Establish both Serre conditions without a characteristic restriction.

**Exercise 6.5 (hard: meeting planes).** For \(B\) in (12), prove \((R_1)\), compute depth at the origin, and deduce failure of \((S_2)\). Identify a regular element and an annihilator in its quotient explicitly.

**Exercise 6.6 (hard: two generators).** Prove that every nonzero ideal of a Dedekind domain can be generated by two elements. Include the field case. Hint: fix one nonzero element of the ideal and choose the second using the Chinese remainder theorem.

## 7. Solutions

**Solution 6.1.** Mapping \(x\) to \(t^2\) and \(y\) to \(t^3\) identifies \(A\) with \(k[t^2,t^3]\): division by the monic polynomial in \(y\) leaves the basis \(x^a y^\epsilon\), \(a\geq0\), \(\epsilon=0,1\), whose images have distinct exponents \(2a+3\epsilon\). The semigroup of exponents is \(\{0,2,3,4,\ldots\}\), so \(t\notin A\). However \(t=y/x\) lies in its fraction field and satisfies \(T^2-x=0\). Also \(k[t]=A+At\), so it is finite integral over \(A\), and their fraction fields are both \(k(t)\). Since \(k[t]\) is normal, every fraction integral over \(A\) lies in \(k[t]\); every element of \(k[t]\) is integral over \(A\). This proves that the normalization is exactly \(k[t]\).

The origin \(\mathfrak n=(x,y)\) has height one: the integral extension to \(k[t]\) gives dimension one, and \((0)\subsetneq\mathfrak n\) is a chain. Its local cotangent space has basis the classes of \(x,y\), because the relation has no linear term; localization at the origin preserves this degree-one calculation. Thus \(A_{\mathfrak n}\) is not regular and \((R_1)\) fails. One can also see the local integral obstruction: a denominator outside \((t^2,t^3)\) has a nonzero constant term, so its product with \(t\) has a nonzero coefficient of degree one and cannot belong to \(A\). Hence \(t\notin A_{\mathfrak n}\) either.

**Solution 6.2.** The regular ring \(R\) is a domain and Cohen–Macaulay. Therefore \(f\) is a nonzerodivisor. Corollary 6.2 of the depth lesson makes \(R/fR\) Cohen–Macaulay of dimension \(\dim R-1\), and Theorem 5.1 of that lesson makes all its prime localizations Cohen–Macaulay. It satisfies \((S_2)\). Theorem 4.4 now says normality is equivalent to \((R_1)\). This proof applies even when the hypersurface has several components: the criterion itself detects whether their meeting prevents normality. The nonzero nonunit assumptions ensure a genuine hypersurface quotient and a regular element.

**Solution 6.3.** Put \(A=k[x,y]\) and \(K=k(x,y)\). On an open subset of this irreducible spectrum, compatible local fractions determine one element of \(K\): equality on an overlap is equality in the fraction field, and any two nonempty opens meet. On \(D(x)\), that element belongs to every prime localization of \(A_x\). The ideal-of-denominators argument, or local detection on a quotient by a fixed denominator, makes it belong to \(A_x\) itself. Similarly its restriction to \(D(y)\) belongs to \(A_y\). Thus the function belongs to \(A_x\cap A_y\subset K\).

Write it both as \(b/x^r\) and as \(c/y^s\), with \(b,c\in A\) and \(r,s\geq0\). Then \(y^s b=x^r c\). The element \(x\) is prime in \(A\) and does not divide \(y\). Repeated primality and cancellation force \(x^r\mid b\). Therefore the fraction is a polynomial. Conversely a polynomial restricts to a regular function, proving the assertion and uniqueness. This direct proof uses no general normality criterion. The two opens cover exactly the punctured plane because a prime containing both variables is precisely the omitted maximal ideal.

**Solution 6.4.** The monic equation in \(z\) gives basis \(x^a y^b z^\epsilon\), \(\epsilon=0,1\). Under

\[
x\longmapsto s^2,\qquad y\longmapsto t^2,
\qquad z\longmapsto st
\tag{13}
\]

their exponent pairs \((2a+\epsilon,2b+\epsilon)\) are distinct. Hence \(C\) embeds in \(k[s,t]\) and is a domain in all characteristics. It is integral finite over \(k[x,y]\), so has dimension two. The origin has height two by the finite-type domain height formula, Theorem 4.3 of the Krull-dimension lesson, because its residue field is \(k\).

For any prime of \(C\), its local ring is a quotient of the corresponding localization of \(k[x,y,z]\) by the nonzero element \(xy-z^2\). The polynomial ring and all its localizations are regular by Theorem 3.2 and Proposition 3.3 of the regular-local lesson. The hypersurface quotient is Cohen–Macaulay by the depth lesson, Corollary 6.2, so \(C\) satisfies \((S_2)\). A prime containing both \(x\) and \(y\) must contain \(z\) and is the origin. Every other prime lies on one of the charts

\[
C_x\simeq k[x,x^{-1},z],\qquad
C_y\simeq k[y,y^{-1},z],
\tag{14}
\]

where the eliminated variable is \(z^2/x\) or \(z^2/y\). These are localizations of polynomial rings and are regular. All primes of height at most one avoid the origin, hence \((R_1)\) holds. Theorem 4.4 proves normality. Nothing in the embedding, chart or depth calculation requires division by two.

**Solution 6.5.** In the polynomial ring,

\[
(xz,xw,yz,yw)=(x,y)\cap(z,w).
\tag{15}
\]

To check this equality, use monomial membership: a monomial in both ideals has at least one factor from each pair. Thus \(B\) is reduced, and its two minimal components are the polynomial planes \(k[z,w]\) and \(k[x,y]\). Its dimension is two. At the origin \(\mathfrak n=(x,y,z,w)\), quotienting by either minimal prime gives a two-dimensional local polynomial plane, so \(\dim B_{\mathfrak n}=2\). Away from this maximal ideal some variable is invertible. For example

\[
B_x\simeq k[x,x^{-1},y],
\tag{16}
\]

since \(xz=xw=0\) force \(z=w=0\); the other three charts work the same way. These rings are regular, so every point except the height-two origin is regular. This proves \((R_1)\).

The injection \(B\to k[x,y]\oplus k[z,w]\) from (15) sends \(f=x+z\) to \((x,z)\). Both components are nonzerodivisors in their domains. Hence \(f\) is a nonzerodivisor in \(B\), and remains so locally. Eliminate \(z=-x\) in its quotient to obtain

\[
B/fB\simeq k[x,y,w]/(x^2,xy,xw,yw).
\tag{17}
\]

The class of \(x\) is nonzero, because this homogeneous defining ideal has no linear terms. Its annihilator is exactly \((x,y,w)\): all three generators kill it, and a polynomial with nonzero constant term multiplies it to a nonzero constant multiple of \(x\). It survives localization at the origin. Thus the local quotient has depth zero by the associated-prime depth test. The regular-element depth formula gives

\[
\operatorname{depth}B_{\mathfrak n}=1<2=\dim B_{\mathfrak n}.
\tag{18}
\]

This fails \((S_2)\); together with \((R_1)\), it exhibits exactly the missing hypothesis in Serre's normality criterion. The calculation also works in characteristic two, where \(-x=x\).

**Solution 6.6.** Theorem 2.3 makes \(R\) Noetherian with DVR localizations at nonzero maximal ideals. Fix \(0\ne a\in I\). There are only finitely many maximal ideals \(\mathfrak m_1,\ldots,\mathfrak m_r\) containing \(a\). At each of these, choose \(b_i\in I\) generating \(I R_{\mathfrak m_i}\). Such an element exists: from a finite generating list for \(I\), choose one with least valuation in that DVR. By the Chinese remainder theorem, choose \(c_i\in R\) which is congruent to one modulo \(\mathfrak m_i\) and to zero modulo every other \(\mathfrak m_j\). Set

\[
b=\sum_{i=1}^r c_i b_i\in I.
\tag{19}
\]

At \(\mathfrak m_i\), the difference \(b-b_i\) belongs to \(\mathfrak m_i I R_{\mathfrak m_i}\). Nakayama therefore shows \(b\) generates \(I R_{\mathfrak m_i}\). At a maximal ideal outside this finite set, \(a\) is a unit, so both \(I\) and \((a,b)\) localize to the whole ring. Local detection on \(I/(a,b)\) gives \(I=(a,b)\). If the finite set is empty, take \(b=0\); then \(a\) is a unit and \(I=R\). This includes fields.

## References and proof scope

All four assigned result groups and all six exercises are proved, using the earlier results identified explicitly above. No assigned theorem is left without proof. The Dedekind definition initially assumes only a domain; Theorem 2.3 proves Noetherianity. The normality criterion applies to arbitrary Noetherian rings and proves the component decomposition, rather than assuming a domain. General valuation groups and the Krull–Akizuki theorem are not needed for these arguments.

The [official Stacks project](https://stacks.math.columbia.edu/) is the maintained reference. The tag links below use [AI Integrated Stacks Project](https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/), an edition with AI-proposed corrections and AI-written additions that have not been reviewed by the Stacks project maintainers. This exposition, its proofs and its solutions are independently written.

- The Stacks project authors, *The Stacks project*, DVRs and Dedekind domains: [Tag 00PD](https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/algebra.html#algebra-lemma-characterize-dvr), [00II](https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/algebra.html#algebra-lemma-valuation-ring-Noetherian-discrete), [034W](https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/algebra.html#algebra-definition-dedekind-domain), [034X](https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/algebra.html#algebra-lemma-characterize-Dedekind).
- The same work, Serre conditions and Hartogs: [Tag 031P](https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/algebra.html#algebra-definition-conditions), [031Q](https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/algebra.html#algebra-lemma-criterion-no-embedded-primes), [031R](https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/algebra.html#algebra-lemma-criterion-reduced), [031S](https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/algebra.html#algebra-lemma-criterion-normal), [031T](https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/algebra.html#algebra-lemma-normal-domain-intersection-localizations-height-1), [0567](https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/algebra.html#algebra-lemma-regular-normal).
- The same work, finite ideals, associated-prime detection and normal components: [Tag 05KG](https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/algebra.html#algebra-lemma-cohen), [09ME](https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/algebra.html#algebra-lemma-product-ideals-principal), [0311](https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/algebra.html#algebra-lemma-zero-at-ass-zero), [030C](https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/algebra.html#algebra-lemma-characterize-reduced-ring-normal), [030B](https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/algebra.html#algebra-lemma-normality-is-local).
- Ravi Vakil, *The Rising Sea: Foundations of Algebraic Geometry*, draft of 27 July 2024, §§5.4, 13.5 and 26.3: normality, DVRs, codimension-one extension and Serre's criterion; parts of the criterion in §26.3 are presented as exercises.