Integral extensions: lying over, going up and going down

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

A polynomial equation can express two very different kinds of control. The equation \(ax=b\) allows division only where \(a\) is invertible. A monic equation for \(x\), by contrast, reduces every high power of \(x\) to a fixed finite list of lower powers. This finite control explains why integral extensions lift prime ideals, preserve chains of specializations, and give closed maps of spectra. Normality will provide the extra control needed to lift a specialization downward from a prescribed point upstairs.

All rings are commutative with identity, and maps preserve identity. No Noetherian hypothesis is imposed unless stated. We use the fraction construction, prime correspondence and residue-field fibres from Localization, local properties and support, and the Artinian structure theorem from Noetherian and Artinian rings. A finite algebra is finite as a module; an algebra of finite type has finitely many algebra generators. These conditions differ.

1. A monic equation is finite control

Let \(R\to S\) be a ring map. An element \(x\in S\) is integral over \(R\) if, for some \(n\geq1\),

where coefficients act through the ring map. The map is integral if every element of \(S\) is integral. Write \(R[x]\) for the image of the evaluation map \(R[X]\to S\); this notation does not require the map \(R\to S\) to be injective.

Theorem 1.1 (four tests for integrality). The following conditions on \(x\in S\) are equivalent:

1. \(x\) is integral over \(R\).
2. \(R[x]\) is a finite \(R\)-module.
3. Some \(R\)-subalgebra \(C\subset S\), containing \(x\) and the identity of \(S\), is a finite \(R\)-module.
4. There is a faithful \(R[x]\)-module \(M\) which is finite as an \(R\)-module.

Proof. A monic equation of degree \(n\) expresses \(x^n\) in the span of \(1,x,\ldots,x^{n-1}\). Multiplying this relation repeatedly by \(x\) reduces every higher power, so that span is \(R[x]\). Thus (1) implies (2), and (2) implies (3) by taking \(C=R[x]\). If (3) holds, multiplication makes \(C\) a faithful \(R[x]\)-module: an element acting as zero kills \(1\), hence is zero. This gives (4).

For (4), choose generators \(m_1,\ldots,m_r\) over \(R\), adding a zero generator if necessary so \(r\geq1\). Write

The matrix \(xI-(a_{ij})\) kills the column of these generators. Multiplication by its adjugate shows that the monic polynomial

satisfies \(P(x)m_i=0\) for every \(i\). It kills all of \(M\), and faithfulness over \(R[x]\) gives \(P(x)=0\) in \(R[x]\), hence in \(S\). This proves (1). \(\square\)

The faithfulness in the fourth test is over \(R[x]\). The determinant first proves an equation for an operator on \(M\); faithfulness is what turns it into an equation for the element of the algebra. The third test also requires a subalgebra containing \(1\), so that the action detects elements.

Theorem 1.2 (closure, transitivity and finiteness). The elements of \(S\) integral over \(R\) form an \(R\)-subalgebra, called the integral closure of \(R\) in \(S\). Integrality is transitive: if \(R\to S\to T\) are integral, then \(R\to T\) is integral. An \(R\)-algebra is finite if and only if it is integral and of finite type.

Proof. If \(x_1,\ldots,x_r\) are integral, choose a monic equation of degree \(n_i\) for each. Reducing powers shows that the monomials

span \(R[x_1,\ldots,x_r]\) as an \(R\)-module. Every element of this finite subalgebra is integral by Theorem 1.1. In particular, sums, negatives and products of integral elements are integral, and elements from \(R\) satisfy degree-one monic equations. This proves the subalgebra assertion.

If \(t\in T\) is integral over \(S\), only finitely many coefficients \(s_1,\ldots,s_r\) occur in a monic equation for it. The algebra \(C=R[s_1,\ldots,s_r]\subset S\) is finite over \(R\). The image algebra \(C[t]\subset T\) is finite over \(C\). Products of a finite list of \(R\)-module generators of \(C\) with a finite list of \(C\)-module generators of \(C[t]\) generate \(C[t]\) over \(R\). Theorem 1.1 now proves that \(t\) is integral over \(R\).

An integral algebra with finitely many algebra generators is finite by the bounded-monomial argument. Conversely, if \(S\) is finite over \(R\), Theorem 1.1 applied to the subalgebra \(S\) proves every element integral. Module generators also generate \(S\) as an algebra, since every element is already their \(R\)-linear combination. \(\square\)

For example, \(\mathbb Q\) is generated by fractions, but is not integral over \(\mathbb Z\): a monic equation for \(1/2\), after multiplication by the appropriate power of \(2\), would say that an odd integer is zero. In the other direction, the algebraic closure of \(\mathbb Q\) is integral over \(\mathbb Q\) but is not finite. If its dimension were \(N\), the irreducible polynomials \(X^m-2\), for primes \(m>N\), would give a subfield of degree \(m\), a contradiction. Their irreducibility follows from Eisenstein at \(2\): reducing a hypothetical factorization of primitive integer polynomials modulo \(2\) forces both constant terms divisible by \(2\), contradicting the constant term \(-2\). The primitive-factor reduction used here is also proved below.

Proposition 1.3 (quotients and base localization). If \(R\to S\) is integral, then \(R\to S/J\) is integral for every ideal \(J\subset S\). If \(U\subset R\) is multiplicative, then \(U^{-1}R\to U^{-1}S\) is integral. If \(R\to S\) is finite, then \(R'\to S\otimes_R R'\) is finite for every \(R\)-algebra \(R'\).

Proof. Monic equations descend to a quotient. An equation for \(s\) becomes, after division by \(u^n\), a monic equation for \(s/u\) with degree-\(i\) coefficient \(a_i/u^i\). Every element of \(U^{-1}S\) has this form. Finally, the images of module generators of \(S\) generate its tensor product over \(R'\). \(\square\)

2. Integral closure survives localization

The normalization of a domain \(R\) is its integral closure in its fraction field \(K\). The domain is normal when this closure equals \(R\). We first prove a statement that allows zero divisors in the larger algebra.

Theorem 2.1 (localization of integral closure). Let \(C\subset S\) be the integral closure of \(R\) in an \(R\)-algebra \(S\), and let \(U\subset R\) be multiplicative. Inside \(U^{-1}S\), the integral closure of \(U^{-1}R\) is \(U^{-1}C\).

Proof. An element \(c/u\), with \(c\in C\), is integral over \(U^{-1}R\) by the same division of a monic equation used in Proposition 1.3. Conversely, suppose \(z=b/u\) satisfies

Put \(v=u\prod_i v_i\) and \(c=(\prod_i v_i)b\in S\). Then \(c/1=vz\). Multiplication of the equation by \(v^n\) gives

Here \(v^i/v_i\) means the product obtained by deleting one displayed factor \(v_i\); it involves no division in \(R\). Since this equation vanishes in the localization, some \(h\in U\) satisfies \(hP(c)=0\) in \(S\). Consequently \(d=hc\) satisfies the equation in the original algebra

It is monic, so \(d\in C\). Finally \(z=d/(hv)\), proving the reverse inclusion. The argument also covers a multiplicative set containing zero, when both localized algebras are zero. \(\square\)

The multiplier \(h\) matters. In \(S=k[u,v]/(uv)\), the element \(v\) becomes zero after inverting \(u\), although it is not integral over the embedded \(k[u]\). Indeed, reduction modulo \(u\) of a proposed monic equation would make the indeterminate \(v\) algebraic over \(k\). Clearing coefficient denominators in the equation \(v/1=0\) does not give \(v=0\) in \(S\). Multiplication by \(u\) supplies the missing equality.

Theorem 2.2 (normality is local). For a domain \(R\), the following are equivalent: \(R\) is normal; every \(R_{\mathfrak p}\), for \(\mathfrak p\in\operatorname{Spec}R\), is normal; every \(R_{\mathfrak m}\), for \(\mathfrak m\) maximal, is normal. Every localization of a normal domain by nonzero elements is normal.

Proof. All these localizations have the same fraction field \(K\). Theorem 2.1 proves that a localization of a normal domain is normal, giving the forward implications. Suppose all maximal localizations are normal, and \(\alpha\in K\) is integral over \(R\). It is integral over every \(R_{\mathfrak m}\), so belongs to each. Consider the denominator ideal

Membership \(\alpha\in R_{\mathfrak m}\) gives a denominator outside \(\mathfrak m\) in \(D\). Thus no maximal ideal contains \(D\). Every proper ideal is contained in a maximal ideal, so \(D=R\), and \(1\in D\) gives \(\alpha\in R\). \(\square\)

Proposition 2.3 (factorial domains are normal). Every unique factorization domain is normal. In particular, principal ideal domains and \(k[X_1,\ldots,X_r]\) for a field \(k\) are normal.

Proof. Write an integral fraction as \(a/b\) with coprime numerator and denominator. Clearing \(b^n\) from a monic equation shows \(b\mid a^n\). Unique factorization forces every irreducible factor of \(b\) to divide \(a\), so \(b\) has no irreducible factors and is a unit. Thus the fraction belongs to the domain.

For completeness, here is the polynomial factorization needed for the examples. In a unique factorization domain \(A\), the content of a polynomial is a greatest common divisor of its coefficients, up to a unit; a polynomial is primitive when its content is a unit. The product of two primitive polynomials is primitive. Otherwise an irreducible \(p\) would divide every coefficient of the product. Reduction modulo \(p\) gives a product of two nonzero polynomials equal to zero over the domain \(A/(p)\), which is impossible. Factoring out the contents then shows that content is multiplicative up to a unit.

Let \(F=\operatorname{Frac}(A)\). Every nonzero polynomial over \(F\) is a scalar in \(F\) times a primitive polynomial over \(A\), by clearing denominators and removing the content. Moreover, if two primitive polynomials over \(A\) differ by a scalar \(a/b\in F\), with \(a,b\) coprime, comparison of coefficients shows that \(b\) divides every coefficient of one of them and hence is a unit. Primitivity of the other then makes \(a\) a unit as well. Factorization over the Euclidean ring \(F[X]\) therefore gives, after these scalar adjustments, a factorization of every primitive polynomial over \(A\) into primitive factors corresponding to irreducibles over \(F\). Such factors are irreducible over \(A\): a factorization there would either give a nontrivial factorization over \(F\), or have a constant nonunit factor contrary to primitivity. Uniqueness follows from uniqueness over \(F\), the primitive-scalar observation, and uniqueness of contents in \(A\). This proves that \(A[X]\) is a unique factorization domain. Inducting from the field \(k\) proves the assertion for polynomial rings. A principal ideal domain has unique factorization: nonzero principal ideals satisfy ACC, which gives factorization into irreducibles, and Bezout's identity makes irreducibles prime, giving uniqueness. \(\square\)

The ACC used for a principal ideal domain follows because every ideal is finitely generated, by Theorem 1.1 of Noetherian and Artinian rings. More explicitly, in an ascending chain the union is an ideal; its single generator already belongs to one member, so the chain stabilizes there. Bezout's identity follows because the ideal generated by two elements is principal.

3. Lifting points and specializations

For an inclusion \(R\subset S\), a prime \(\mathfrak q\) lies over \(\mathfrak p\) when \(\mathfrak q\cap R=\mathfrak p\). Recall that inclusion of prime ideals is the specialization order on a spectrum.

Lemma 3.1 (integral domains and fields). If domains \(A\subset B\) form an integral extension, then \(A\) is a field if and only if \(B\) is a field.

Proof. If \(B\) is a field and \(0\ne a\in A\), its inverse satisfies a monic equation over \(A\). Multiplication of that equation by \(a^{n-1}\) expresses \(a^{-1}\) as an element of \(A\). Conversely, if \(A\) is a field and \(0\ne b\in B\), the algebra \(A[b]\) is a finite-dimensional \(A\)-vector space and a domain. Multiplication by \(b\) is injective, hence surjective on this vector space. Its image contains \(1\), giving an inverse for \(b\). \(\square\)

Theorem 3.2 (lying over and going up). If \(R\subset S\) is integral, every prime of \(R\) has a prime above it. For any integral map \(R\to S\), if \(\mathfrak q\) lies over \(\mathfrak p\) and \(\mathfrak p\subset\mathfrak p'\), there is \(\mathfrak q'\supset\mathfrak q\) lying over \(\mathfrak p'\).

Proof. For lying over, fix \(\mathfrak p\) and put \(U=R\setminus\mathfrak p\). Localization preserves the inclusion, so \(R_{\mathfrak p}\subset U^{-1}S\) is an integral inclusion of nonzero rings. Choose a maximal ideal \(\mathfrak n\) of \(U^{-1}S\). If \(\mathfrak m\) is its contraction, the inclusion

is integral, its source is a domain and its target a field. Lemma 3.1 makes \(\mathfrak m\) maximal. The only maximal ideal of \(R_{\mathfrak p}\) is \(\mathfrak pR_{\mathfrak p}\), so prime correspondence contracts \(\mathfrak n\) to a prime of \(S\) over \(\mathfrak p\).

For going up, the induced map \(R/\mathfrak p\to S/\mathfrak q\) is an injective integral map. Apply lying over to the prime \(\mathfrak p'/\mathfrak p\) and take the inverse image of the resulting prime of \(S/\mathfrak q\). It contains \(\mathfrak q\) and has the required contraction. \(\square\)

Lying over over the whole base spectrum requires injectivity. An arbitrary integral map has image \(V(\ker(R\to S))\), by applying lying over to its image ring.

Theorem 3.3 (incomparability and maximal ideals). For an integral map \(R\to S\), distinct primes with the same contraction are incomparable. A prime \(\mathfrak q\subset S\) is maximal if and only if its contraction \(\mathfrak p\subset R\) is maximal.

Proof. The fibre ring

is integral over \(\kappa(\mathfrak p)\), by Proposition 1.3 and the monic equations. Every prime quotient of \(C\) is a domain integral over the embedded field \(\kappa(\mathfrak p)\), hence a field by Lemma 3.1. Thus every prime of \(C\) is maximal. Primes of \(S\) over \(\mathfrak p\) correspond, with their inclusions, to primes of \(C\). Two comparable such primes are therefore equal. For the second assertion apply Lemma 3.1 directly to the injective integral map \(R/\mathfrak p\subset S/\mathfrak q\). Its two domains are fields simultaneously. \(\square\)

Theorem 3.4 (closed maps and finite fibres). An integral map \(R\to S\) induces a closed map of spectra. More precisely, for every ideal \(J\subset S\),

A finite ring map has finite fibres. Each fibre, with its induced topology, is a finite discrete space.

Proof. The quotient inclusion \(R/\varphi^{-1}(J)\subset S/J\) is integral. Lying over proves precisely the displayed equality, which gives closedness.

For finiteness, the fibre algebra \(C=S\otimes_R\kappa(\mathfrak p)\) is finite-dimensional over \(\kappa(\mathfrak p)\). A chain of ideals is a chain of vector subspaces, so \(C\) is Artinian. The Artinian structure theorem gives finitely many primes, all maximal. Every point of \(\operatorname{Spec}C\) is consequently closed, and a finite space with closed points is discrete, since the complement of any one point is a finite union of closed points. The prime correspondence identifies this spectrum homeomorphically with the fibre subspace. One can check the topology directly: a basic open of the fibre algebra, defined by a fraction \(s/u\), corresponds to the intersection of the fibre with \(D_S(s)\). A zero fibre algebra gives the empty fibre. \(\square\)

4. Going down: why the normal base matters

Going up starts with a point upstairs and follows a specialization of its image. Going down asks for a point below a prescribed point upstairs. Quotienting alone cannot solve this: the proof must control denominators at that prescribed point. Normality makes the coefficients of a minimal polynomial available in the base ring.

Lemma 4.1 (monic factors over a normal domain). Let \(R\) be normal, with fraction field \(K\). If a monic \(Q\in K[X]\) divides a monic \(P\in R[X]\), then \(Q\in R[X]\). If all nonleading coefficients of \(P\) belong to an ideal \(I\subset R\), all nonleading coefficients of \(Q\) belong to \(\sqrt I\).

Proof. Obtain a splitting field of \(P\) by repeatedly adjoining a root of an irreducible factor: the quotient by that factor is a field, and division by the new linear factor decreases the remaining degree. In this field every root of \(P\) is integral over \(R\), since \(P\) is monic. The roots of \(Q\), counted with multiplicities, form a sublist of these roots. The coefficients of \(Q\) are signed elementary symmetric expressions in that sublist, hence integral over \(R\) by Theorem 1.2. They lie in \(K\), so normality puts them in \(R\). This argument allows inseparable polynomials and repeated roots.

Monic division in \(R[X]\) now shows \(P=QH\) with monic \(H\in R[X]\): the remainder is zero because division over \(K\) has zero remainder. For any prime \(\mathfrak p\supset I\), reduction modulo \(\mathfrak p\) gives

Over the fraction field of this domain, the only monic divisors of a power of \(X\) are powers of \(X\). Thus \(\overline Q=X^{\deg Q}\). Every other coefficient of \(Q\) belongs to every prime over \(I\), whose intersection is \(\sqrt I\). \(\square\)

In particular, an element integral over a normal domain has its monic minimal polynomial over \(K\) in \(R[X]\). Indeed, that minimal polynomial divides any polynomial over \(K\) vanishing at the element, by Euclidean division. The intersection-of-primes description of radicals used here is Theorem 1.2 of Spectra of rings.

An element \(y\) of an \(R\)-algebra is integral over an ideal \(I\) if it satisfies an equation

The powers on the coefficients record how strongly the element is controlled by \(I\).

Lemma 4.2 (integrality over an ideal). If \(S\) is integral over \(R\), every element of \(IS\) is integral over \(I\). For an arbitrary \(R\)-algebra \(S\), an element \(y\in S\) is integral over \(I\) if and only if \(yT\in S[T]\) is integral over the subalgebra

Proof. Write \(y=\sum_{j=1}^r a_js_j\), with \(a_j\in I\). The subalgebra \(C=R[s_1,\ldots,s_r]\subset S\) is finite over \(R\). Choose module generators \(m_1,\ldots,m_e\). Since \(yC\subset IC\), write \(ym_i=\sum_j b_{ij}m_j\) with all \(b_{ij}\in I\). In \(\det(XI-(b_{ij}))\), the degree-\(i\) coefficient below the leading term is a sum of products of \(i\) entries, hence belongs to \(I^i\). The adjugate argument makes this polynomial evaluated at \(y\) kill \(C\). It therefore kills \(1\) and vanishes in \(S\), giving the required equation.

For the second assertion, multiplying such an equation by \(T^n\) immediately gives a monic equation for \(yT\) over \(\mathcal R(I)\). Conversely, suppose

Taking the coefficient of \(T^n\) gives \(y^n+\sum_i c_i y^{n-i}=0\), where \(c_i\) is the coefficient of \(T^i\) in \(A_i(T)\). By the definition of \(\mathcal R(I)\), it belongs to \(I^i\). \(\square\)

Theorem 4.3 (going down). Let \(R\subset S\) be an integral inclusion of domains, with \(R\) normal. Given primes \(\mathfrak p\subset\mathfrak p'\) of \(R\) and a prime \(\mathfrak q'\) of \(S\) over \(\mathfrak p'\), there is a prime \(\mathfrak q\subset\mathfrak q'\) over \(\mathfrak p\).

Proof. We will show

The inclusion from right to left is immediate. If \(z\in R\) belongs to the left side, combine denominators to write \(z=y/g\) in \(S_{\mathfrak q'}\), with \(y\in\mathfrak pS\) and \(g\in S\setminus\mathfrak q'\). Since \(S\) is a domain, its localization embeds in its fraction field, and \(zg=y\) holds in \(S\). If \(z=0\), it belongs to \(\mathfrak p\); assume otherwise.

Let \(K=\operatorname{Frac}(R)\). The element \(g\) is algebraic over \(K\), with monic minimal polynomial

by Lemma 4.1. At least one \(a_i\) is outside \(\mathfrak p\). Otherwise the equation would put \(g^n\) in \(\mathfrak pS\subset\mathfrak q'\), forcing \(g\in\mathfrak q'\).

Because \(z\ne0\) lies in \(K\), the monic minimal polynomial of \(y=zg\) is

To verify minimality, a smaller-degree polynomial vanishing at \(zg\), after substitution \(X\mapsto zX\), would give a smaller-degree polynomial vanishing at \(g\). Lemma 4.2 gives a monic equation \(P(y)=0\) whose nonleading coefficients belong to \(\mathfrak p^i\), hence to \(\mathfrak p\). The minimal polynomial \(Q_z\) divides \(P\), so Lemma 4.1 gives

Choose \(i\) with \(a_i\notin\mathfrak p\). Primality forces \(z\in\mathfrak p\), proving (1).

The ideal \(\mathfrak pS_{\mathfrak q'}\) is therefore disjoint from the multiplicative set given by the image of \(R\setminus\mathfrak p\). Prime separation, proved in Lemma 1.1 of Spectra of rings, supplies a prime \(P\) of \(S_{\mathfrak q'}\) containing this ideal and disjoint from that set. Its contraction to \(R\) is exactly \(\mathfrak p\). Prime correspondence for \(S\to S_{\mathfrak q'}\) contracts \(P\) to a prime \(\mathfrak q\subset\mathfrak q'\), with \(\mathfrak q\cap R=\mathfrak p\). \(\square\)

The proof used neither finite generation of \(S\) as an algebra nor Noetherianity of either domain. Normality was used in the coefficient lemma, and the domain hypothesis on \(S\) justified the equality \(zg=y\) before forming minimal polynomials.

5. Chains and dimension

The length of a finite chain of prime ideals

is \(n\). The Krull dimension \(\dim R\) is the supremum of these lengths. We set the supremum of the empty set to \(-\infty\), so the zero ring has dimension \(-\infty\); unbounded lengths give dimension \(+\infty\). The height of a prime \(\mathfrak p\) is the supremum of lengths of chains ending at \(\mathfrak p\), equivalently \(\dim R_{\mathfrak p}\) by prime correspondence. The later dimension lessons will develop further tools; here finite chains suffice.

Theorem 5.1 (dimension under integral inclusions). If \(R\subset S\) is integral, then \(\dim R=\dim S\). For \(\mathfrak q\in\operatorname{Spec}S\), with \(\mathfrak p=\mathfrak q\cap R\),

If the extension has going down, equality holds for every such pair.

Proof. Contraction turns a strict chain in \(S\) into a strict chain in \(R\): adjacent contractions cannot agree, by incomparability. This proves \(\dim S\leq\dim R\), and the same argument for chains ending at \(\mathfrak q\) proves the height inequality. Conversely, start with any finite strict chain in \(R\). Lying over lifts its first prime; repeated going up lifts the rest. The lifted inclusions are strict because their contractions are distinct. Thus \(\dim S\geq\dim R\).

If going down holds and a chain ends at \(\mathfrak p\), start at the prescribed \(\mathfrak q\) and lift the chain backward, one step at a time. This gives a chain of the same length ending at \(\mathfrak q\), so the reverse height inequality follows. These arguments compare all finite lengths, even when the suprema are infinite. If \(R=0\), the unital inclusion forces \(S=0\), giving equality with the empty-spectrum convention. \(\square\)

Equality of dimensions is a statement about integral inclusions. For a general integral map, the same argument instead gives \(\dim S=\dim(R/\ker\varphi)\).

6. Arithmetic and normalization examples

Three fibres of the Gaussian integers. The inclusion \(\mathbb Z\subset\mathbb Z[i]\) is finite, with module basis \(1,i\). Its fibre at \((p)\) is

At \(p=2\), the polynomial is \((T+1)^2\), so there is one prime above \((2)\), namely \((2,i+1)=(1+i)\). For the equality, \(2=(1+i)(1-i)\), so both generators on the left lie in \((1+i)\); conversely \(1+i\) is a generator on the left. The fibre is nonreduced: the class of \(T+1\) is nonzero and square-zero. At \(p=3\), the squares are \(0,1\), so \(T^2+1\) is irreducible; the unique prime is \((3)\), with residue field of nine elements. At \(p=5\), the roots are \(2,-2\), giving the two primes \((5,i-2)\) and \((5,i+2)\), each with residue field \(\mathbb F_5\). Finite fibres need not have constant cardinality or be reduced.

The cusp. Let \(C=k[t^2,t^3]\subset k[t]\). Its elements are exactly polynomials with no \(t\)-term: every exponent at least two is a nonnegative combination of \(2\) and \(3\). Thus \(t\notin C\), while \(t\) is integral over \(C\) by the equation \(X^2-t^2=0\). Also \(t=t^3/t^2\) in \(\operatorname{Frac}(C)\), so that fraction field is \(k(t)\). The polynomial ring \(k[t]\) is normal and integral over \(C\). Any element of \(k(t)\) integral over \(C\) is integral over \(k[t]\), using the same monic equation, hence belongs to \(k[t]\). Therefore \(k[t]\) is the normalization of \(C\). The inclusion is finite with generators \(1,t\).

The node. Assume \(\operatorname{char}k\ne2\) and set

The parametrization

embeds \(A_0\) in \(k[t]\). To prove injectivity, monic division in \(y\) gives a unique representative \(f(x)+yg(x)\). Under (2), the first term has only even powers of \(t\), and the second only odd powers. If their sum vanishes, both vanish. Substitution \(k[x]\to k[t]\), \(x\mapsto t^2-1\), is injective by comparing highest degrees, so \(f=0\); the domain property of \(k[t]\) then gives \(g=0\).

The element \(t\) satisfies \(t^2=x+1\), so \(k[t]=A_0+A_0t\) is finite over \(A_0\). Since \(t=y/x\) in the fraction field, \(\operatorname{Frac}(A_0)=k(t)\). Normality of \(k[t]\) proves that it is the normalization, exactly as for the cusp. It is a proper extension: every element of \(A_0\) has the same value at \(t=1\) and \(t=-1\), whereas \(t\) has different values there. The fibre over \((x,y)\) is \(k[t]/(t^2-1)\), with two primes \((t-1)\) and \((t+1)\). They separate the two branches meeting at the node. Away from this point, \(x\) is invertible and \(t=y/x\), so \((A_0)_x=k[t]_x\).

Going down can fail. Add a variable \(z\): put \(A=A_0[z]\subset B=k[t,z]\), with the same parametrization. The inclusion is still finite, and \(A_x=B_x\). Let

The ideal \(\mathfrak q_D\) is prime, since its quotient is \(k[t]\). Evaluation at \(t=1,z=0\) annihilates it and, on \(A\), is evaluation at \(x=y=z=0\). Thus \(\mathfrak p\subset\mathfrak p'\). This inclusion is strict: the image of \(z\) in \(B/\mathfrak q_D\) is \(t-1\ne0\), so \(z\notin\mathfrak p\), whereas \(z\in\mathfrak p'\). The prime \(\mathfrak q'\) contracts to \(\mathfrak p'\), by evaluation at \(t=-1,z=0\).

There is only one prime of \(B\) over \(\mathfrak p\). Indeed, \(x\notin\mathfrak p\), because its image modulo \(\mathfrak q_D\) is \(t^2-1\ne0\). All primes over \(\mathfrak p\) therefore lie in \(D_B(x)\), where \(A_x=B_x\); prime correspondence gives a unique one, namely \(\mathfrak q_D\). But the generator \(z-t+1\) has value \(2\ne0\) at \(t=-1,z=0\), so \(\mathfrak q_D\not\subset\mathfrak q'\). There is no prime below \(\mathfrak q'\) over \(\mathfrak p\), as going down would require. The base \(A\) is not normal: \(t=y/x\) is integral and is absent from \(A\), as evaluation at the two points again shows.

7. Exercises

Exercise 7.1 (easy: quadratic integers). Let \(d\ne0,1\) be a squarefree integer, allowing negative \(d\). Determine the integral closure of \(\mathbb Z\) in \(\mathbb Q(\sqrt d)\). Show that it is \(\mathbb Z[\sqrt d]\) if \(d\not\equiv1\pmod4\), and \(\mathbb Z[(1+\sqrt d)/2]\) if \(d\equiv1\pmod4\). Explain separately the case \(d=1\).

Exercise 7.2 (easy: the missing linear term). For a field \(k\), prove that \(k[t^2,t^3]\) is not normal and determine its normalization. Give a finite set of module generators for the normalization over the original ring.

Exercise 7.3 (medium: the topology of a finite fibre). For a finite map \(R\to S\), prove directly that the fibre over each \(\mathfrak p\in\operatorname{Spec}R\) is finite and discrete. Give an upper bound for its cardinality using \(\dim_{\kappa(\mathfrak p)}(S\otimes_R\kappa(\mathfrak p))\), and include the empty case.

Exercise 7.4 (medium: heights at a prescribed point). Let \(R\subset S\) be integral. For a prime \(\mathfrak q\) with contraction \(\mathfrak p\), prove \(\operatorname{ht}(\mathfrak q)\leq\operatorname{ht}(\mathfrak p)\), with equality if going down holds. Your argument should also handle infinite heights.

Exercise 7.5 (medium: separating the wrong branch). Verify the failure of going down for \(A\subset B\) in the final example of Section 6. Prove that \(\mathfrak p\subsetneq\mathfrak p'\), identify the contraction of \(\mathfrak q'\), and justify uniqueness of the prime above \(\mathfrak p\).

Exercise 7.6 (hard: orbits over the invariant ring). Let a finite group \(G\) act by automorphisms on an arbitrary ring \(S\). Put \(R=S^G\). Prove that \(S\) is integral over \(R\) and that \(G\) acts transitively on the primes of \(S\) above any given prime of \(R\). Do not assume that \(S\) is of finite type over \(R\).

8. Solutions

Solution 7.1. Put \(K=\mathbb Q(\sqrt d)\). Since \(d\) is squarefree and is neither zero nor one, it is not a rational square, so \(1,\sqrt d\) form a \(\mathbb Q\)-basis. Write an integral element as \(\alpha=a+b\sqrt d\), with \(a,b\in\mathbb Q\). If \(b=0\), normality of \(\mathbb Z\) makes \(a\) an integer. Suppose \(b\ne0\). The minimal polynomial is

and its coefficients are integers by Lemma 4.1 applied to the normal domain \(\mathbb Z\). Thus \(u=2a\in\mathbb Z\) and \(N=a^2-db^2\in\mathbb Z\). The identity

forces \(v=2b\in\mathbb Z\). To see this, express \(2b=r/s\) in lowest terms with \(s>0\). Then \(s^2\mid dr^2\), and coprimality gives \(s^2\mid d\). Squarefreeness of \(|d|\) forces \(s=1\).

The remaining norm condition is \(u^2\equiv dv^2\pmod4\). If \(v\) is even, then \(u\) is even. If \(v\) is odd, then \(v^2\equiv1\pmod4\), and the congruence requires \(u\) odd and \(d\equiv1\pmod4\): a squarefree even \(d\) is \(2\pmod4\), which is not a square modulo four. Hence, when \(d\not\equiv1\pmod4\), both \(u,v\) are even and \(\alpha\in\mathbb Z[\sqrt d]\). When \(d\equiv1\pmod4\), they have the same parity and

Conversely, \(\sqrt d\) satisfies \(X^2-d=0\). If \(d\equiv1\pmod4\), the element \(\omega=(1+\sqrt d)/2\) satisfies \(X^2-X+(1-d)/4=0\), a monic integer polynomial. The integral elements form a subring, so each proposed ring consists of integral elements. This proves both equalities. For \(d=1\), the field is just \(\mathbb Q\) and its integral closure of \(\mathbb Z\) is \(\mathbb Z\).

Solution 7.2. The monomials occurring in \(C=k[t^2,t^3]\) have exponents \(2a+3b\) with \(a,b\geq0\). These exponents are zero and all integers at least two: an even integer uses only \(2\), and an odd integer at least three is \(3\) plus an even integer. Linear independence of polynomial monomials therefore shows \(t\notin C\). But \(t=t^3/t^2\) belongs to \(\operatorname{Frac}(C)\) and satisfies the monic equation \(X^2-t^2=0\), proving nonnormality. The same identity shows \(\operatorname{Frac}(C)=k(t)\). Reduction of powers using \(t^2\in C\) gives \(k[t]=C+Ct\), a finite integral extension. By Proposition 2.3, \(k[t]\) is normal. An element of \(k(t)\) integral over \(C\) is integral over \(k[t]\), since its monic equation has coefficients there, so it lies in \(k[t]\). Conversely every element of \(k[t]\) is integral over \(C\). The normalization is exactly \(k[t]\), generated as a \(C\)-module by \(1,t\).

Solution 7.3. Let \(F=\kappa(\mathfrak p)\), \(C=S\otimes_R F\), and \(n=\dim_F C\), which is finite because module generators of \(S\) descend to generators of \(C\). Prime correspondence identifies the fibre with \(\operatorname{Spec}C\). For any prime \(\mathfrak n\subset C\), its quotient is a finite-dimensional domain over the embedded field \(F\). Multiplication by a nonzero element is injective and hence surjective, so this quotient is a field. Thus all primes of \(C\) are maximal.

Any \(r\) distinct maximal ideals are pairwise comaximal. The Chinese remainder theorem gives a surjection

Each factor is a nonzero \(F\)-vector space, so \(r\leq n\). There can therefore be at most \(n\) primes. They are closed points; since there are finitely many, the complement of any point is closed, and the spectrum is discrete. The homeomorphism with the fibre is the one verified in Theorem 3.4. If \(n=0\), then \(C=0\) and the fibre is empty. This argument does not impose any finiteness assumption on \(R\) itself.

Solution 7.4. A strict finite chain ending at \(\mathfrak q\) contracts to a chain ending at \(\mathfrak p\). If any adjacent contractions agreed, the corresponding comparable primes upstairs would agree by Theorem 3.3, contradicting strictness. Hence every length occurring in the height of \(\mathfrak q\) occurs in that of \(\mathfrak p\), which proves the inequality. Under going down, start with a finite chain ending at \(\mathfrak p\). Prescribe \(\mathfrak q\) as its last lift and lift each preceding prime backward. The inclusions are strict because their contractions are distinct, so the resulting chain ends at \(\mathfrak q\) and has the same length. This proves the reverse inequality. We have compared all finite lengths, so the argument also proves equality when arbitrarily long chains exist; it does not require a chain attaining an infinite supremum.

Solution 7.5. The embedding \(A_0\subset k[t]\) in Section 6 extends coefficientwise to \(A=A_0[z]\subset k[t,z]=B\). The equation \(t^2=x+1\) makes it finite, with generators \(1,t\), and \(t=y/x\) gives \(A_x=B_x\). The ideal \(\mathfrak q_D=(z-t+1)\) is prime, with quotient obtained by putting \(z=t-1\). Evaluation at \(t=1\) on this quotient corresponds to evaluation at the origin on \(A\); thus its contracted prime \(\mathfrak p\) is contained in \(\mathfrak p'=(x,y,z)\). The image of \(z\) is \(t-1\), which is not zero, whereas \(z\in\mathfrak p'\); this proves strict containment. Evaluation at \(t=-1,z=0\) maps \(x,y,z\) to zero and its restriction to \(A\) has kernel precisely \((x,y,z)\). Thus \(\mathfrak q'=(t+1,z)\) lies over \(\mathfrak p'\).

The image of \(x\) modulo \(\mathfrak q_D\) is \(t^2-1\ne0\), so \(x\notin\mathfrak p\). Every prime above \(\mathfrak p\) avoids \(x\) and corresponds to a prime of \(B_x=A_x\) with contraction \(\mathfrak pA_x\). Since this is an equality of rings, the only such localized prime is \(\mathfrak pA_x\). Contracting back gives the already known prime \(\mathfrak q_D\). Finally, \(z-t+1\) evaluates to \(2\ne0\) modulo \(\mathfrak q'\), so \(\mathfrak q_D\) is not contained in \(\mathfrak q'\). No prime over \(\mathfrak p\) lies below that prescribed prime, and going down fails.

Solution 7.6. For each \(s\in S\), the polynomial

is monic and vanishes at \(s\), since the identity element is among the factors. Every group element permutes the factors, so every coefficient is fixed by \(G\) and lies in \(R\). Thus \(S\) is integral over \(R\). In particular, lying over gives a nonempty fibre above every prime of \(R\).

Let \(\mathfrak q,\mathfrak q'\) have the same contraction \(\mathfrak p\). Each \(g(\mathfrak q')\) also contracts to \(\mathfrak p\), since \(g\) fixes \(R\). Suppose \(\mathfrak q\) is not among these orbit primes. Incomparability then says \(\mathfrak q\) is contained in none of them. After deleting repetitions, they are a finite family of pairwise incomparable primes \(\mathfrak r_1,\ldots,\mathfrak r_e\). Choose \(a_i\in\mathfrak q\setminus\mathfrak r_i\), and for \(j\ne i\) choose \(b_{ij}\in\mathfrak r_j\setminus\mathfrak r_i\). Put \(c_i=\prod_{j\ne i}b_{ij}\), with empty product \(1\). Then

lies outside every \(\mathfrak r_i\): modulo \(\mathfrak r_i\), all terms except \(a_ic_i\) vanish, and that term is nonzero by primality. This is the finite prime-avoidance argument, valid in arbitrary rings.

Consider \(N(x)=\prod_{g\in G}g(x)\). It belongs to \(R\), since the action permutes the factors, and belongs to \(\mathfrak q\), since one factor is \(x\). Thus \(N(x)\in\mathfrak p\subset\mathfrak q'\). But no factor \(g(x)\) lies in \(\mathfrak q'\): otherwise \(x\in g^{-1}(\mathfrak q')\), contrary to its choice. Primality says their product cannot lie in \(\mathfrak q'\), a contradiction. Hence \(\mathfrak q=g(\mathfrak q')\) for some \(g\), proving transitivity. Nothing here asserts that \(S\) is finite over its invariant ring; the monic equations and finite orbit of primes are sufficient.

References

• The Stacks project authors, The Stacks project, Commutative Algebra. Integral-element tests and closure: Tag 052I, Tag 00GO, Tag 00GN, Tag 02JJ. Localization and normality: Tag 0307, Tag 030B, Tag 0AFV.
• The same work, prime lifting and fibres: Tag 00GQ, Tag 00GU, Tag 00GT, Tag 00GR, Tag 00HZ, Tag 05DR. Going down and dimension: Tag 00H3, Tag 00H7, Tag 00H8, Tag 00OK. The links use the AI Integrated Stacks Project English reader described in the course introduction.
• Ravi Vakil, The Rising Sea: Foundations of Algebraic Geometry, public draft of 27 July 2024, Section 5.4 on normality, Section 8.2 on lying over and going up, and the node and finite-fibre passages in Section 8.3. Author’s public draft.