Generic zeros and absolute primality

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

A prime ideal can describe a whole variety through one point, provided that point is allowed to have transcendental coordinates. Its specializations recover all the ordinary points. Changing the coefficient field asks a further question: does this variety remain integral, or does it split or acquire nilpotents? Generic zeros connect these questions to field extensions.

Let \(k\) be any field, \(R=k[x_1,\ldots,x_n]\), and let \(\Omega\supset k\) be algebraically closed. When generic points must belong to \(\Omega^n\), assume \(\operatorname{trdeg}_k\Omega\ge n\). For arbitrary points we may use different extension fields. We assume the Nullstellensatz, Theorems 2.1–2.2, and dimension and Noether normalization, Theorems 4.2–4.3 and 5.1. The primary-decomposition facts come from the preceding lessons and the associated-prime prerequisite. The field-extension criteria in Section 5 are proved here. Noether’s working English edition and the AI Integrated Stacks Project provide further reading.

1. A point that remembers exactly an ideal

For a tuple \(\xi\) in an extension field define

\[ I_k(\xi)=\{f\in R:f(\xi)=0\}. \]

It is prime because evaluation has values in a field.

Theorem 1.1 (generic zero). For a prime \(\mathfrak p\subset R\), put \(A=R/\mathfrak p\), \(L=\operatorname{Frac}(A)\), and let \(\xi_i\) be the image of \(x_i\) in \(L\). Then \(I_k(\xi)=\mathfrak p\).

Proof. Evaluation is the quotient map \(R\to A\) followed by the injective map \(A\to L\). Its kernel is precisely \(\mathfrak p\). \(\square\)

The tuple is called a generic zero. The field \(L=k(\xi_1,\ldots,\xi_n)\) is its function field. It can be embedded over \(k\) into the stated large \(\Omega\): send a transcendence basis of \(L/k\), of size at most \(n\), to algebraically independent elements of \(\Omega\), then extend over the finite algebraic extension to an embedding in the algebraically closed field. The relation ideal stays \(\mathfrak p\).

For \(\mathfrak p=(y-x^2)\), the generic zero is \((t,t^2)\), with \(t\) transcendental. Evaluation gives \(k[x,y]/\mathfrak p\cong k[t]\), so no additional relation is hidden.

2. Specialization acts on coordinate rings

Say that \(\eta\) is a specialization of \(\xi\) if \(I_k(\xi)\subset I_k(\eta)\). This definition preserves equations rather than prescribing a limiting process.

Theorem 2.1. The condition is equivalent to a \(k\)-algebra homomorphism

\[ k[\xi_1,\ldots,\xi_n]\longrightarrow k[\eta_1,\ldots,\eta_n], \qquad \xi_i\longmapsto\eta_i. \]

For a fixed target field \(F\supset k\), the specializations in \(F^n\) are exactly \(V_F(I_k(\xi))\).

Proof. Both coordinate algebras are quotients of \(R\) by their relation ideals. A map induced by the prescribed assignment exists exactly when every relation of \(\xi\) is a relation of \(\eta\). The last assertion restates this condition as evaluation of all polynomials in the relation ideal. \(\square\)

Such a map is not in general a map of function fields. For example, \(t\mapsto0\) defines \(k[t]\to k\) but cannot extend to \(k(t)\), since \(t^{-1}\) would require an inverse of zero. This is why specialization uses \(k[\xi]\).

The prime-inclusion picture of specialization in the spectrum is precisely the same order: a larger prime imposes more equations. Closed points have coordinates algebraic over \(k\), by the weak Nullstellensatz; generic zeros may require transcendental coordinates.

3. All zeros, including those of primary ideals

Proposition 3.1. For every ideal \(I\), \(V_F(I)=V_F(\sqrt I)\) in every extension field \(F\). If \(Q\) is \(\mathfrak p\)-primary, \(V_F(Q)=V_F(\mathfrak p)\).

Proof. If \(f^r\in I\), then at a zero of \(I\) one has \(f(\eta)^r=0\); fields have no nonzero nilpotents, so \(f(\eta)=0\). The reverse inclusion follows from \(I\subset\sqrt I\). Apply this to \(\sqrt Q=\mathfrak p\). \(\square\)

If \(\mathfrak p_1,\ldots,\mathfrak p_s\) are the minimal primes over \(I\), then

\[ V_\Omega(I)=\bigcup_i V_\Omega(\mathfrak p_i). \]

Indeed \(\sqrt I=\bigcap_i\mathfrak p_i\). A point outside every \(V(\mathfrak p_i)\) admits \(f_i\in\mathfrak p_i\) nonzero there, and the product \(\prod f_i\in\sqrt I\) stays nonzero there, a contradiction. By Theorem 2.1 these sets are the specializations of the generic zeros, when those zeros are realized in \(\Omega\).

The strong Nullstellensatz implies

\[ I_k(V_\Omega(I))=\sqrt I. \]

For a general \(k\) this follows by applying it to \(I\Omega[x]\) and contracting: field extension is faithfully flat, and \(I\Omega[x]\cap k[x]=I\); the same holds for radicals, since powers detect them. This does not claim that \(\Omega\) need be large enough for a generic zero: an algebraic closure of \(k\) already suffices for the equality of ideals of zeros.

The primary ideal \((x^2,y)\) has only the origin as its geometric zero. The quotient still contains a nonzero nilpotent class \(x\). A set of field-valued zeros detects the radical, so multiplicity needs more algebra.

4. Counting parameters and excluding embedded dimensions

Krull dimension is the supremum of the lengths of strict prime chains. For a prime \(\mathfrak p\), the cited dimension theorem gives

\[ \dim(R/\mathfrak p)=\operatorname{trdeg}_k k(\xi). \]

Noether’s 1923 prime-chain formulation describes the same number as the parameter formulation. The catenarity and height formula belong to the cited commutative-algebra lesson. For the parabola, \((0)\subsetneq(t)\) in \(k[t]\) is a length-one chain, and its function field has one parameter.

An ideal in a polynomial ring is unmixed here if every associated prime of its quotient has the same height. This includes embedded associated primes in the test. Equidimensionality, which only tests minimal primes, is weaker. For \((x^2,xy)\), the associated heights are one and two, so it is not unmixed, although its reduced closed set is just one line.

Theorem 4.1 (Macaulay’s unmixedness theorem). In \(k[x_1,\ldots,x_n]\), an ideal generated by \(r\) elements and having height \(r\) is unmixed: every associated prime of its quotient has height \(r\).

Proof using the internal prerequisites. Polynomial localizations are regular by Proposition 3.3 of Regular local rings. They are Cohen–Macaulay by Theorem 6.1 of Regular sequences, depth and Cohen–Macaulay modules. Let \(\mathfrak q\) be associated to the quotient, and localize at \(\mathfrak q\). Every minimal prime \(\mathfrak p\) over \(I\) contained in \(\mathfrak q\) has height exactly \(r\): height is at least \(r\) by the assumed height of \(I\), and at most \(r\) by the generalized principal ideal theorem for \(r\) generators. The polynomial height and catenarity formulas in the dimension prerequisite give \(\dim (R_{\mathfrak q}/IR_{\mathfrak q})=\operatorname{ht}\mathfrak q-r\). Corollary 6.2 of the depth lesson makes this quotient Cohen–Macaulay with that dimension. Since its maximal ideal is associated, it has depth zero, by Lemma 2.2 of that same lesson. Its dimension is therefore zero, forcing \(\operatorname{ht}\mathfrak q=r\). This proves unmixedness. The generalized principal ideal theorem is proved in Dimension theory of Noetherian local rings, Section 3. \(\square\)

This argument uses regularity of polynomial localizations. The same number-of-generators condition in an arbitrary Noetherian ring is insufficient.

5. When the variety survives every field extension

The prime \(\mathfrak p\) is absolutely prime if \(\mathfrak p\bar k[x]\) is prime for an algebraic closure \(\bar k\) of \(k\). A \(k\)-algebra is geometrically integral when its tensor product with every extension field is a domain.

For a finitely generated field extension, separable over \(k\) means separably generated: it has a transcendence basis \(t_1,\ldots,t_d\) such that the finite algebraic extension over \(k(t_1,\ldots,t_d)\) is separable. A \(k\)-algebra is geometrically reduced when its tensor product with every extension field is reduced. The next four lemmas provide all field-extension facts used in the criterion, including the imperfect-field case.

Lemma 5.A (finite separable extensions). A finite separable extension \(E/F\) has a primitive element. For any extension field \(H/F\), \(E\otimes_F H\) is a finite product of finite separable extensions of \(H\), hence is reduced. In characteristic \(p>0\), a finite extension \(E/F\) is separable if and only if \(E=FE^p\), where \(FE^p\) denotes the subfield generated by \(F\) and all \(p\)-th powers in \(E\).

Proof. First recall the root-count argument. In a finite tower of simple algebraic extensions, each embedding of an intermediate field into an algebraic closure extends by choosing a root of the transported minimal polynomial. The number of choices is its degree when that polynomial is separable and is at most its degree in general. Thus a separable tower has exactly its total degree many embeddings. Every element of its top field is separable: an inseparable element would have fewer roots than its degree, and counting embeddings by restriction to the field it generates would give fewer embeddings than the total degree. This also proves transitivity of finite algebraic separability and that the separable elements in a finite extension form a subfield.

If \(F\) is infinite and \(E=F(a,b)\) is separable, there are \([E:F]\) embeddings. Two distinct embeddings disagree on \(a\) or \(b\), so their images of \(a+cb\) coincide for at most one \(c\in F\). Avoiding the finitely many forbidden values gives an element with \([E:F]\) distinct conjugates. Its degree is therefore \([E:F]\), so it generates \(E\). Induct on a finite list of generators. If \(F\) is finite, then \(E\) is finite. Its multiplicative group is cyclic: for its exponent \(m\), combining commuting elements of maximal prime-power orders produces an element of order \(m\); all group elements are roots of \(T^m-1\), so the group has at most \(m\) elements, and it has at least \(m\) because of this element. A generator of that group generates the field. This proves the primitive-element assertion in both cases.

Write \(E=F(a)\) with separable minimal polynomial \(f\). Bézout’s identity for \(f,f'\) remains valid over \(H\), so the irreducible factors of \(f\) in \(H[T]\) are distinct and separable. The Chinese remainder theorem gives

\[ E\otimes_F H=H[T]/(f)\cong\prod_j H[T]/(g_j), \]

with each factor a finite separable field. The Chinese remainder theorem here follows directly from the Bézout identities between distinct irreducible polynomials: their ideals are comaximal, and the usual remainder maps have kernel their product.

For the last assertion, if \(a\in E\) is separable over \(F\), its minimal polynomial over \(F(a^p)\) is separable and divides \(T^p-a^p\), which has only one distinct root. That minimal polynomial has degree one, so \(a\in F(a^p)\). Consequently separability implies \(E=FE^p\).

Conversely let \(E_s\) be the subfield of elements of \(E\) separable over \(F\). Every irreducible polynomial in characteristic \(p\) can be written \(g(T^{p^e})\) with \(g'\ne0\); \(g\) is irreducible and hence separable. It follows that some \(p\)-power of each element of \(E\) belongs to \(E_s\). A finite generating list supplies one exponent \(N\) such that \(E^{p^N}\subset E_s\). If \(E=FE^p\), then \(E=E_sE^p\), and iteration gives \(E=E_sE^{p^N}=E_s\). This is separability. \(\square\)

Lemma 5.B (separability and geometric reducedness). For a finitely generated field extension \(L/k\), the following are equivalent:

  1. \(L/k\) is separably generated.
  2. \(L\otimes_k K\) is reduced for every extension field \(K/k\).
  3. \(L\otimes_k\bar k\) is reduced.

In characteristic \(p>0\), they are also equivalent to reducedness of \(L\otimes_k k^{1/p}\), where \(k^{1/p}=\{c\in\bar k:c^p\in k\}\).

Proof. Suppose \(F=k(t_1,\ldots,t_d)\subset L\) is a separating transcendence basis. For any \(K/k\),

\[ B=F\otimes_k K=(k[t_1,\ldots,t_d]\setminus\{0\})^{-1}K[t_1,\ldots,t_d] \]

is a domain. If \(H=\operatorname{Frac}(B)\), the injection \(B\hookrightarrow H\) remains injective after tensoring with the finite-dimensional \(F\)-vector space \(L\): choose an \(F\)-basis to see this coordinate by coordinate. Hence

\[ L\otimes_k K=L\otimes_F B\hookrightarrow L\otimes_F H. \]

The ring on the right is reduced by Lemma 5.A, so its subring on the left is reduced. Thus (1) implies (2), which implies (3). In characteristic zero, every finitely generated field extension has a separating transcendence basis: choose any transcendence basis, and its remaining finite extension is separable because an irreducible nonconstant polynomial has nonzero derivative. This proves the equivalences in that characteristic.

Assume now that the characteristic is \(p>0\) and that \(L\otimes_k k^{1/p}\) is reduced. Condition (3) implies this hypothesis, since tensoring the injection \(k^{1/p}\subset\bar k\) with the \(k\)-vector space \(L\) is injective. Place the compositum \(Lk^{1/p}\) in an algebraic closure of \(L\). The multiplication map

\[ L\otimes_k k^{1/p}\longrightarrow Lk^{1/p} \]

is injective. Indeed, for \(z=\sum_i a_i\otimes c_i\), \(z^p=(\sum_i a_i^pc_i^p)\otimes1\); if the multiplication image of \(z\) vanishes, this power is zero, and reducedness gives \(z=0\). Frobenius gives an isomorphism from this tensor ring to \(L^p\otimes_{k^p}k\). Therefore its multiplication map into \(L\) is also injective. Explicitly, any elements of \(k\) linearly independent over \(k^p\) remain linearly independent over \(L^p\).

Take any transcendence basis \(t_1,\ldots,t_d\), put \(F=k(t_1,\ldots,t_d)\), and write \(n=[L:F]\). Frobenius is an isomorphism of the extensions \(L/F\) and \(L^p/F^p\), so \([L^p:F^p]=n\). The injective tensor product \(k\otimes_{k^p}L^p\) is free of rank \(n\) over its subring \(k\otimes_{k^p}F^p\). The fraction field of the latter subring is \(kF^p=k(t_1^p,\ldots,t_d^p)\). Localizing at its nonzero elements consequently gives a domain of dimension \(n\) over \(kF^p\). A finite-dimensional domain over a field is a field: multiplication by a nonzero element is an injective endomorphism of a finite-dimensional vector space, hence surjective. This field is the compositum \(kL^p\). Thus

\[ [kL^p:kF^p]=n,\qquad [L:kL^p]=\frac{[L:F][F:kF^p]}{[kL^p:kF^p]}=p^d. \tag{5.1} \]

Here \([F:kF^p]=p^d\): the monomials \(t_1^{e_1}\cdots t_d^{e_d}\), \(0\le e_i<p\), are independent by grouping polynomial exponents modulo \(p\) after clearing denominators, and they span because a polynomial denominator \(q(t)\) has inverse \(q(t)^{p-1}/q(t)^p\), with \(q(t)^p\in kF^p\).

Put \(M=kL^p\). Each element of \(L\) has its \(p\)-th power in \(M\). Whenever an element \(u\) is outside an intermediate field containing \(M\), adjoining it has degree \(p\): its minimal polynomial divides \((T-u)^p\), and a monic factor of degree \(e\) with \(0<e<p\) would have coefficient \(-eu\), forcing \(u\) into that field. By (5.1), adjoining such elements successively stops after exactly \(d\) steps. We obtain

\[ L=M(u_1,\ldots,u_d), \qquad \{u_1^{e_1}\cdots u_d^{e_d}:0\le e_i<p\} \text{ is an }M\text{-basis of }L. \tag{5.2} \]

These \(u_i\) are algebraically independent over \(k\). If not, choose a nonzero relation \(f(u)=0\) of least total degree. Group its exponents to write

\[ f(X)=\sum_{0\le e_i<p}X^e f_e(X_1^p,\ldots,X_d^p),\qquad f_e\in k[Y_1,\ldots,Y_d]. \]

The basis (5.2) forces every \(f_e(u_1^p,\ldots,u_d^p)\) to vanish. Choose a nonzero \(f_e\), and express its finitely many coefficients using elements \(b_1,\ldots,b_r\in k\) independent over \(k^p\): \(f_e(Y)=\sum_j b_jg_j(Y)\), with \(g_j\in k^p[Y]\). The injectivity of \(k\otimes_{k^p}L^p\to L\) forces each \(g_j(u_1^p,\ldots,u_d^p)=0\). Taking \(p\)-th roots of its coefficients gives a polynomial \(h_j\in k[X]\) with \(g_j(u_1^p,\ldots,u_d^p)=h_j(u)^p\). At least one \(h_j\) is nonzero, and its total degree is at most \(\deg f_e\le\deg f/p<\deg f\). Its vanishing contradicts minimality. If \(d=0\), the asserted independence is empty and needs no argument.

The \(u_i\) therefore form a transcendence basis. The extension \(L/k(u_1,\ldots,u_d)\) is finite, and (5.2) says \(L=k(u_1,\ldots,u_d)L^p\). Lemma 5.A makes that finite extension separable. This proves (1) from the \(k^{1/p}\) test, completing all equivalences. \(\square\)

Lemma 5.C (algebraic-closure tests). For any \(k\)-algebra \(S\),

\[ S\text{ is geometrically integral} \Longleftrightarrow S\otimes_k\bar k\text{ is a domain}, \]

and

\[ S\text{ is geometrically reduced} \Longleftrightarrow S\otimes_k\bar k\text{ is reduced}. \]

No finite-generation hypothesis on \(S\) or on the test extension is required.

Proof. Only the reverse implications need proof. Suppose \(u,v\in S\otimes_k K\) are nonzero with \(uv=0\). Express them as \(u=\sum_i a_i\otimes c_i\), \(v=\sum_j b_j\otimes d_j\), choosing the \(a_i\) linearly independent over \(k\), and likewise the \(b_j\). Choose \(c_{i_0},d_{j_0}\ne0\). Form the nonzero finite-type \(k\)-algebra

\[ C=k[c_i,d_j,(c_{i_0}d_{j_0})^{-1}]\subset K. \]

Expand the finitely many products \(a_ib_j\) in a \(k\)-basis of their span. The equation \(uv=0\) says that the corresponding finite list of polynomial expressions in the \(c_i,d_j\) vanishes in \(C\). Choose a maximal ideal of \(C\). Its residue field is finite algebraic over \(k\), by the proved weak Nullstellensatz, and embeds over \(k\) in \(\bar k\). The resulting map \(C\to\bar k\) preserves all these equations and keeps \(c_{i_0},d_{j_0}\) nonzero, since their product is invertible in \(C\). Linear independence of the chosen lists then gives nonzero images \(\bar u,\bar v\in S\otimes_k\bar k\) with \(\bar u\bar v=0\), contradicting the domain hypothesis. The zero ring cannot intervene: if \(S\otimes_k\bar k\) is a domain, then \(S\ne0\), and tensoring its nonzero vector space with any field extension remains nonzero.

For reducedness, take a nonzero \(u=\sum_i a_i\otimes c_i\) with \(u^m=0\), choose \(c_{i_0}\ne0\), and use \(C=k[c_i,c_{i_0}^{-1}]\). Expand the finitely many products arising in \(u^m\) in a basis of their span. The same specialization produces a nonzero nilpotent in \(S\otimes_k\bar k\), a contradiction. This proves the second test as well. \(\square\)

Lemma 5.D (purely inseparable scalar extension). Let \(F/E\) be a field extension and \(P/E\) a purely inseparable algebraic extension. Then \(F\otimes_E P\) has exactly one prime ideal. It is a domain whenever it is reduced.

Proof. In characteristic zero \(P=E\), so this is immediate. In characteristic \(p>0\), choose compatible embeddings in an algebraic closure of \(F\), and multiply tensors into the compositum \(FP\). For any \(z=\sum_i a_i\otimes c_i\), there is \(N\) with every \(c_i^{p^N}\in E\); hence \(z^{p^N}\) belongs to the embedded field \(F\). If the multiplication image of \(z\) is zero, this power is zero; conversely a nilpotent has zero image in a field. Thus the kernel is exactly the nilradical. If the image of \(z\) is nonzero, its scalar power is nonzero, and \(z^{p^N-1}/z^{p^N}\) is already an inverse in the tensor ring. The image is therefore a field. Every prime contains the nilradical, and its quotient is this field, proving uniqueness. If the ring is reduced, that unique prime is zero, so the ring is a domain. \(\square\)

Theorem 5.1 (regular-extension criterion). For \(A=R/\mathfrak p\) and \(L=\operatorname{Frac}(A)\), the following are equivalent:

  1. \(\mathfrak p\) is absolutely prime.
  2. \(A\) is geometrically integral.
  3. \(L/k\) is separable and every element of \(L\) algebraic over \(k\) belongs to \(k\).

A field extension satisfying the third condition is called regular.

Proof. The first two conditions are equivalent by Lemma 5.C, since \(A\otimes_k\bar k=\bar k[x_1,\ldots,x_n]/\mathfrak p\bar k[x]\). For every extension \(K/k\), tensoring the injection \(A\hookrightarrow L\) over the field \(k\) gives an injection \(A\otimes_k K\hookrightarrow L\otimes_k K\). Also \(L\otimes_k K\) is the localization of \(A\otimes_k K\) at the images of \(A\setminus\{0\}\). These images are nonzero because field extension preserves the injection of \(A\). Hence these two rings are domains simultaneously: one implication uses the injection, the other uses localization of a domain at nonzero elements. It suffices to establish the criterion for \(L\).

If \(L\otimes_k\bar k\) is a domain, it is reduced, so Lemma 5.B gives separability of \(L/k\). If \(\alpha\in L\) is algebraic over \(k\), tensoring the subfield injection gives \(k[\alpha]\otimes_k\bar k\hookrightarrow L\otimes_k\bar k\). The source is \(\bar k[T]/(f)\), where \(f\) is the minimal polynomial of \(\alpha\). If \(\deg f>1\), write \(f=(T-a)g\) in \(\bar k[T]\). Both factors have positive degree smaller than \(\deg f\), so both give nonzero classes whose product is zero. This is impossible in a subring of a domain, including when all roots of \(f\) coincide. Hence \(\deg f=1\), giving \(\alpha\in k\).

Conversely assume the third condition. Any irreducible polynomial \(f\in k[T]\) remains irreducible over \(L\). Indeed, the coefficients of a monic factor in \(L[T]\) are elementary symmetric expressions in a selection of roots in an algebraic closure of \(L\), so are algebraic over \(k\). The constants hypothesis puts those coefficients in \(k\); monic division then puts the complementary factor in \(k[T]\) too, contradicting irreducibility. For a finite separable extension \(E/k\), Lemma 5.A supplies a primitive element, and this irreducibility makes \(L\otimes_k E\) a field. The separable closure \(k^{\mathrm{sep}}\) is the directed union of its finite separable subextensions. Their tensor rings inject into one another and are fields, so their union \(F=L\otimes_k k^{\mathrm{sep}}\) is a field.

The extension \(\bar k/k^{\mathrm{sep}}\) is purely inseparable: in characteristic \(p\), the minimal polynomial of any algebraic element has the form \(g(T^{p^e})\) with \(g\) separable, so its \(p^e\)-th power belongs to \(k^{\mathrm{sep}}\). In characteristic zero this stage is absent. Consequently

\[ L\otimes_k\bar k=F\otimes_{k^{\mathrm{sep}}}\bar k \]

has one prime by Lemma 5.D. Separability of \(L/k\) makes it reduced by Lemma 5.B. Lemma 5.D therefore makes it a domain, proving absolute primality and the equivalence of all three conditions. \(\square\)

6. Two distinct failures

Over \(\mathbb R\), the polynomial \(x^2+y^2\) is irreducible: a degree-two factorization would be into homogeneous linear factors, but a real line cannot lie in its zero set, which is only the origin. Since \(\mathbb R[x,y]\) is a unique factorization domain, the generated ideal is prime. Over \(\mathbb C\),

\[ x^2+y^2=(x+iy)(x-iy), \]

so it is not absolutely prime. In its function field, \(y\ne0\) and \((x/y)^2=-1\), yielding a new algebraic constant. Separability holds in characteristic zero; algebraic closedness of the constants fails.

Over \(k=\mathbb F_p(t)\), the polynomial \(X^p-t\) is irreducible. The element \(t\) is not a \(p\)-th power, as its order of vanishing at \(t=0\) is one. If \(a^p=t\) in an algebraic closure, a monic proper factor of \((X-a)^p\) would be \((X-a)^e\), \(0<e<p\); its coefficient \(-ea\) would force \(a\in k\). This proves irreducibility. Its quotient is the field \(k(t^{1/p})\). Over \(\bar k\) it becomes \((X-t^{1/p})^p\); the class of \(X-t^{1/p}\) is nonzero by monic division and is nilpotent. The field extension is inseparable and has new algebraic constants, so both regularity conditions fail.

Separability cannot be dropped even if the constants condition holds. Here is a complete example. Let \(s,x,y\) be independent indeterminates over \(\mathbb F_p\), put

\[ L=\mathbb F_p(s,x,y),\qquad t=y^p-sx^p,\qquad k=\mathbb F_p(s,t). \tag{6.1} \]

The elements \(s,x,t\) are algebraically independent: in a polynomial in \(t\), substitution of \(y^p-sx^p\) makes its highest \(t\)-degree term the unique highest \(y\)-degree term. Thus \(x\) is transcendental over \(k\). The element \(sx^p+t\) is not a \(p\)-th power in \(k(x)\), since its formal derivative with respect to \(t\) is one, whereas every \(p\)-th power has derivative zero. The preceding monic-factor argument proves that

\[ A=k[x,y]/(y^p-sx^p-t) \]

is a domain with fraction field \(L\). Over \(\bar k\), its equation is \((y-s^{1/p}x-t^{1/p})^p\). Write \(h=y-s^{1/p}x-t^{1/p}\). The ring \(A\otimes_k\bar k=\bar k[x,y]/(h^p)\) has nilradical \((h)\), quotient \(\bar k[x]\), and a nonzero nilpotent \(h\). Reduction modulo \(h\) embeds \(A\) in \(\bar k[x]\): the displayed value of \(y\) has the irreducible degree-\(p\) polynomial just proved as its minimal polynomial over \(k(x)\). Thus the elements of \(A\setminus\{0\}\) stay outside \((h)\). Localize at those elements to obtain \(L\otimes_k\bar k\). Its reduction is a subring of \(\bar k(x)\) containing \(k(x)\), because every nonzero polynomial in \(k[x]\) is among the denominators. Every element of this subring is algebraic over \(k(x)\): only finitely many coefficients in the algebraic extension \(\bar k/k\) occur in each rational function. A nonzero such element has a minimal polynomial with nonzero constant coefficient, which expresses its inverse as a polynomial in it with coefficients in \(k(x)\). Hence the reduction is a field. Since \(h\) is nilpotent, this makes \((h)\) the unique prime after localization. Also \(h\ne0\) there: its annihilator before localization is \((h^{p-1})\subset(h)\), disjoint from the denominators. Thus \(L\otimes_k\bar k\) is nonreduced, so \(L/k\) is not separably generated by Lemma 5.B.

Nevertheless \(k\) is algebraically closed in \(L\). First there are no new separable algebraic constants. A proper finite separable subfield \(E/k\) would give an injection of \(E\otimes_k\bar k\), a product of at least two copies of \(\bar k\), into \(L\otimes_k\bar k\). A ring with one prime has no idempotents except zero and one: its unique prime is its nilradical, so for an idempotent \(e\), either \(e\) or \(1-e\) is both nilpotent and idempotent and must be zero. This rules out that injection.

To exclude purely inseparable constants, put \(U=x^p,V=y^p\). Then \(L^p=\mathbb F_p(s^p,U,V)\). If \(f\in k\cap L^p\), differentiate in \(\mathbb F_p(s,U,V)\), holding \(U,V\) fixed. Since \(t=V-sU\), the result is

\[ 0=\frac{\partial f}{\partial s}-U\frac{\partial f}{\partial t}, \]

where the two derivatives on the right are computed in \(k=\mathbb F_p(s,t)\). The element \(U\) is transcendental over \(k\), so both derivatives vanish. The monomials \(s^it^j\), \(0\le i,j<p\), are a basis of \(k\) over \(k^p\), by the same rational monomial argument used in (5.1). Differentiating their expansion first in \(s\) and then in \(t\) forces all coefficients except the constant one to vanish. Hence \(f\in k^p\), proving \(k\cap L^p=k^p\). If \(a\in L\) has \(a^p\in k\), then \(a^p=c^p\) for some \(c\in k\), and uniqueness of \(p\)-th roots in a field gives \(a=c\). Repetition excludes every nontrivial purely inseparable constant. Finally, an arbitrary element algebraic over \(k\) has a suitable \(p\)-power separable over \(k\), by the polynomial argument in Lemma 5.A. That power is in \(k\), and the purely inseparable argument puts the element itself in \(k\). Thus (6.1) satisfies the constants condition but fails separability, as claimed.

7. Exercises

  1. Basic. Find a generic zero for \((y^2-x^3)\), and describe its specializations over an algebraically closed extension.
  2. Intermediate. Show that a point whose coordinates are algebraic over \(k\) specializes the generic zero of \(\mathfrak p\) exactly when it lies in \(V(\mathfrak p)\). Explain why a map on coordinate rings suffices.
  3. Intermediate. Prove that \((x^2+y^2)\) is prime over \(\mathbb R\) and compute its complex extension as an intersection of prime ideals.
  4. Intermediate. For \((X^p-t,Y)\subset\mathbb F_p(t)[X,Y]\), identify the quotient field and its base change to an algebraic closure.
  5. Advanced. Prove the regular-extension criterion in characteristic zero, identifying exactly where the characteristic-zero hypothesis removes a step.

8. Solutions

1. Use \((t^2,t^3)\), with \(t\) transcendental. Dividing by the monic relation in \(y\) leaves \(a(x)+yb(x)\). Its evaluation has disjoint even powers from \(a(t^2)\) and odd powers at least three from \(t^3b(t^2)\), so vanishing forces both polynomials zero. The kernel is exactly the given ideal, which is therefore prime. Every cusp point is parametrized: if \(x\ne0\), set \(t=y/x\), giving \(t^2=x,t^3=y\); if \(x=0\), then \(y=0\) and set \(t=0\). These are all specializations by Theorem 2.1.

2. Containment of relation ideals is exactly the condition that every polynomial in \(\mathfrak p\) vanish at the point; this is membership in \(V(\mathfrak p)\). The quotient map therefore descends to the coordinate algebra. Algebraicity ensures the point’s coordinate algebra is a finite field extension, by adjoining finitely many algebraic elements, but it does not make a specialization map extend to the generic function field.

3. The real irreducibility argument in Section 6 gives primality. In \(\mathbb C[x,y]\) the distinct irreducible linear factors are relatively prime as elements, so their principal ideals intersect in their product:

\[ (x^2+y^2)=(x+iy)\cap(x-iy). \]

These two ideals are not comaximal: their sum is \((x,y)\). The complex variety is two lines meeting at the origin.

4. The quotient is \(\mathbb F_p(t)(t^{1/p})\), a purely inseparable degree-\(p\) extension. The base-changed quotient is \(\bar k[\epsilon]/(\epsilon^p)\) with \(\epsilon=X-t^{1/p}\) and \(Y=0\). It is nonreduced and not a domain. The extension is not separable and \(k\) is not algebraically closed in it.

5. In characteristic zero every finitely generated field extension is separably generated. If \(k\) is algebraically closed in \(L\), the monic-factor argument in Theorem 5.1 keeps every finite separable minimal polynomial irreducible over \(L\), so \(L\otimes_k\bar k\) is the directed union of fields and is a domain. There is no purely inseparable stage. Conversely any algebraic \(\alpha\in L\setminus k\) would give a degree-greater-than-one separable polynomial whose tensor quotient over \(\bar k\) is a product of fields, injecting into \(L\otimes_k\bar k\); this contradicts its being a domain. The passage between \(A\) and \(L\) is the injective localization argument already proved.

In Noether’s words

Elimination Theory and General Ideal Theory, work 24 announces the arithmetic dimension in its introduction and develops zeros in Sections 2–3, dimension in Section 4, and absolute prime ideals in Section 7. Thus absolute primality occurs later than the initial zero theory. The short announcements are works 18 and 25. The German authority gives the original text.

References

Editable sources

Complete source archive · Complete course in LaTeX · This lesson in LaTeX. The archive includes all five lesson texts, the figures and their reproducible source.