Math 216A Solutions to Homework 1

1. Let R be any ring, and A ' R[X1, . . . , Xn]/I and B two R-algebras, with I a finitelygenerated ideal (one says A is ‘of finite presentation’ over R; when R is noetherian this is ofcourse equivalent to saying A is of finite type as an R-algebra). Let p and q be prime idealsin A and B respectively. For any a ∈ A \ p, b ∈ B \ q, and any R-algebra map f : Aa → Bb

with f−1(qb) = pa, there is a naturally induced R-algebra map fq : Ap → Bq which is a localmap. Prove the following:

Theorem. For every R-algebra map F : Ap → Bq which is a local map, there exist b ∈ B \qand an R-algebra map f : A→ Bb for which f−1(qb) = p and fq = F .

Solution. Choose bi ∈ B, β ∈ B \ q so that g : A → Ap → Bq takes Xi to bi/β. Leth1, . . . , hm ∈ I be generators. We get a well-defined R-algebra map ϕ : R[X1, . . . , Xn]→ Bβ

sending Xi to bi/β. The finitely many ϕ(hi) vanish in Bq because composing ϕ with Bβ → Bq

yields the composite of R[X1, . . . , Xn] � A and g. Thus, there exists β′ ∈ B \ q such thatβ′ϕ(hi) = 0 in Bβ (using Bq ' (Bβ)qβ). Define b = ββ′ and consider the composite mapR[X1, . . . , Xn] → Bββ′ induced by ϕ. This kills all hi’s, so we get a well-defined R-algebramap f : A→ Bb so that A→ Bb → Bq is equal to g. Thus, f−1(qb) = p, so f has the desiredproperties.

If f1 : A→ Bb1 and f2 : A→ Bb2 are both of the desired type, then by a similar argumentthere exists b ∈ B so that the maps A → Bb1b2b induced by f1 and f2 coincide (this is thesense in which f is ‘unique’).

2. Let A be a finite type k-algebra, where k is an algebraically closed field. Let k′/k be anextension field with k′ algebraically closed. Let A′ = k′ ⊗k A, a finite type k′-algebra. Itoften is the case that A has a property P if and only if A′ has a property P. Prove this forP equal to any of the following properties: non-zero, domain, reduced, of dimension d, hasa unique minimal prime, regular (i.e.., all localizations at maximal ideals are regular localrings). This sort of thing is essential in order to pass between algebraic geometry over Qand algebraic geometry over C (where analytic tools become available).

Also prove that if A1 and A2 are two finite type k-algebras, then A1⊗kA2 is reduced (resp.is a domain) if A1 and A2 are of this type. Give counterexamples if we drop the hypothesisthat k is algebraically closed.

Solution. Using k-bases (Zorn!) settles the case P = ‘non-zero’. By Noether normaliza-tion, there exists an injective k-algebra map i : k[T1, . . . , Td] ↪→ A which is finite as well.By dimension theory, this forces d = dimA. Since k → k′ is flat (Zorn!), applying k′⊗k toi yields an injection k′[T1, . . . , Td] ↪→ A′, and this is still finite too, so dimA = d = dimA′.Since A → A′ is injective, if A′ is reduced or a domain, so is A. For the other direction, itsuffices to consider the case where A is a domain, since for reduced A we have an injectionA ↪→

∏A/pi, where {pi} is the finite set of minimal primes in the noetherian ring A (and

all domains A/pi are of course finite type over k). Now assume that A is a domain, so thefraction field K of A makes sense. Since A′ injects into k′⊗kK, we will show that this latter(strange) ring is a domain (and so A′ is a domain).

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Choosing a separating transcendence basis for K/k and using the primitive element theo-rem and a suitable scaling, we get a k-algebra isomorphism K ' k(x1, . . . , xr)[xr+1]/f wheref ∈ k[x1, . . . , xr+1] is irreducible and monic (!) over k(x1, . . . , xr). The usual UFD argument(via monicity) shows that f is irreducible in k[x1, . . . , xr+1] also. Since k′ ⊗k K is a local-ization of k′[x1, . . . , xr+1]/f (at the multiplicative set k[x1, . . . , xr] \ {0}), it suffices to showthat k′[x1, . . . , xr+1]/f is a domain. That is, we will show that f is irreducible over k′, givenits irreducibility over k. Suppose there is a non-trivial factorization f = gh over k′. Thismakes sense over a finitely generated k-algebra R inside of k′ (namely the one generated bythe coefficients of the g and h). If we consider monomials with the usual lexicographicalordering, the ‘leading’ coefficients a(f), a(g), a(h) satisfy a(f) = a(g)a(h) 6= 0. Since R isreduced of finite type over a field, we can choose a maximal ideal m not containing a(f). Bythe Nullstellensatz and the fact that k is algebraically closed(!), k → R/m is an isomorphism.Thus, passage to (R/m)[x1, . . . , xr+1] transforms the factorization of f with R coefficientsinto a factorization of our original f with R/m ' k coefficients. The choice of m ensures thisis non-trivial factorization, so we have a contradiction. This settles the case P = ‘domain’.

So far we have not needed that k′ is algebraically closed, so the above argument evenshows that for K ′/k any extension field, and A a domain of finite type over k, the ringK ′ ⊗k A is a domain. Thus, if A1 and A2 are domains of finite type over k, A1 ⊗k A2 is asubring of the domain K1 ⊗k A2, with K1 the fraction field of A1, so A1 ⊗k A2 is a domain.If the Ai are merely reduced, injecting them into the usual finite product of domains (givenby quotients by minimal primes) shows that A1 ⊗k A2 injects into a product of domainsand so is a reduced ring. For a counterexample when k is not assumed algebraically closed,choose k to be a non-perfect field of characteristic p and a ∈ k is not a pth power. ThenA1 = A2 = k(a1/p) = k[T ]/(T p − a) is a domain but A1 ⊗k A2 ' k(a1/p)/(T − a1/p)p is anon-reduced ring. Note that if k is perfect and not algebraically closed, then the ‘domain’condition can fail (look at A1 = A2 = k[T ]/f(T ) for f ∈ k[T ] an irreducible polynomialwith degree > 1), but the property of being reduced is still preserved under tensor products,essentially because of the existence of separating transcendence bases for finitely generatedextension fields of k.

The condition that A have a unique minimal prime is equivalent to the conditon thatthe quotient Ared by the nilradical is a domain (and this quotient is of course a finite typek-algebra). By the ‘reduced’ case considered above, the canonical map k′ ⊗k Ared → A′red isan isomorphism. Now by the ‘domain’ case, we conclude Ared is a domain if and only if A′red

is a domain. Observe that we still have not used that k′ is algebraically closed.Finally, we show that A has localizations at all maximal ideals regular if and only if A′ has

this property. In either case, one of A or A′ is reduced (by checking localizations at maximalideals), so we see that A and A′ are both reduced (by the case P = ‘reduced’ above). In fact,we can assume A and A′ are domains. To see this, let {pi} denote the finite set of minimalprimes of A. It is enough to check that ψA : A →

∏A/pi is an isomorphism (and then

work with A/pi in place of A). By the case P = ‘domain’, we see that {k′ ⊗k pi} is the setof minimal primes of A′ (without repetitions), so applying extension of scalars to ψA yieldsψA′ . Hence, ψA is an isomorphism if and only if ψA′ is an isomorphism. If A (resp. A′) isregular, then the A-linear map ψA (resp. A′-linear map ψA′) is an isomorphism, as this can

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be checked by localizing at maximal ideals, where it is clear (since regular local rings aredomains and so contain unique minimal primes).

So now we may assume that A and A′ are both domains, necessarily with the samedimension d, so all local rings are d-dimensional. Writing A = k[T1, . . . , Tn]/(f1, . . . , fm),regularity at a maximal ideal m means that the Jacobian matrix of the fi’s has rank exactlyn− d when ‘evaluated’ at m. Put another way, since this rank is always at most n− d, thepolynomial conditions for the rank to be ≤ n− d− 1 do not vanish at m. Thus, A is regularat all maximal ideals if and only if the ideal I generated by the n−d−1 by n−d−1 minorsof the Jacobian matrix of the fi’s does not lie inside of any maximal ideal of A, which is tosay that I = A. It is clear that I ′ = k′ ⊗k I is the corresponding ideal in A′ (so here weuse that k′ is algebraically closed, in order for the Jacobian criterion for regularity to makesense over k′). Since I = A if and only if I ′ = A′, we are done.

An important generalization. In case you are interested, here is an argument forhow to obtain the regularity assertion when k′ is allowed to be any extension of k, withno other hypotheses on k′, but we will need to use much more sophisticated commutativealgebra in order to handle the possibility that k′ might not be perfect. (There is a moregeometric approach using the theory of smooth and etale morphisms, but that requires mucheffort to set up.)

The essential philosophical point is that because A → A′ is faithfully flat (since k → k′

is), the complications should arise in the ‘fibers’, and these can be controlled because k isalgebraically closed (or even just perfect). Now for the proof (with references to hard corestuff in Matsumura). The first point is the fundamental (but non-trivial) fact that if B → Cis a flat local map of local noetherian rings, with the ‘closed fiber’ ring C/mBC a regular ring,then B is regular if and only if C is regular. This follows from Theorem 23.7 in Matsumura,and ultimately relies on Serre’s homological characterization of regularity (Theorem 19.2in Matsumura). Another important consequence of Serre’s theorem is Theorem 19.3 inMatsumura, which implies that if A or A′ is regular at all maximal ideals, then one getsregularity of local rings at all prime ideals (since a localization at a prime is a localization ofthe local ring at a maximal over the prime). This is critical, since a maximal ideal in A′ willtypically contract to a non-maximal prime ideal in A, since k′/k is usually not algebraic.

Pick a prime q in A′, with contraction p in A, and consider the resulting flat local mapAp → A′q. Observe that every p in A arises in this way from some q, since the ‘fiber ring’κ(p) ⊗A A′ ' κ(p) ⊗k k′ is a non-zero ring and so contains prime ideals! The ‘fiber ring’k′ ⊗k κ(p) is noetherian, since we can view κ(p) as a localization of a finite type k-algebraA/p, so the fiber ring is a localization of a finite type k′-algebra, hence is noetherian. Wewill show that the fiber ring k′ ⊗k κ(p) has all of its local rings regular. In particular, A′q/pwill be regular, so by Theorem 23.7 in Matsumura, regularity of Ap and A′q are equivalent(this clearly settles the original question).

In order to study the fiber ring, we operate more generally, and will prove the regularityof the local rings of the noetherian ring k′ ⊗k K for K any finitely generated extension fieldof k. By the perfectness of k (!), we may choose a separating transcendence basis, givingK ' k(X1, . . . , Xn)[Xn+1]/f as k-algebras, for an irreducible f ∈ k[X1, . . . , Xn+1] which

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is monic and separable in k(X1, . . . , Xn)[Xn+1]. Thus, if ∆f ∈ k[X1, . . . , Xn] is the non-zero discriminant of f , then k′ ⊗k K is a localization of (k′[X1, . . . , Xn+1]/f)∆f

where f ∈k′[X1, . . . , Xn+1] is monic in k′(X1, . . . , Xn)[Xn+1] and separable too, since the discriminantis the non-zero ∆f ∈ k[X1, . . . , Xn] ⊆ k′[X1, . . . , Xn].

It now suffices to show that if k′ is any field and f ∈ k′[X1, . . . , Xn+1] is monic andseparable in k′(X1, . . . , Xn)[Xn+1], then the noetherian ring (k′[X1, . . . , Xn+1]/f)∆f

has alllocal rings regular. This ring is to be thought of as a ‘covering space’ of affine n-space overk′ (in coordinates X1, . . . , Xn), which should reduce us to checking the regularity of localrings on affine space. Now for the algebra. Consider the flat map

ψ : R = k′[X1, . . . , Xn]→ (k′[X1, . . . , Xn+1]/f)∆f= T.

For every prime ideal p of T , with contraction q = ψ−1(p), the induced local map Rq → Tpis not only flat, but has a ‘fiber ring’ Tp/q which is a field. To see this latter claim, note

that the fiber ring is a local ring of κ(q)[Xn+1]/f with f ∈ κ(q)[Xn+1] a monic polynomialwith discriminant ∆f ∈ κ(q) non-zero (since p lies over q and p does not contain ∆f ). That

is, f ∈ κ(q)[Xn+1] is a separable polynomial, so κ(q)[Xn+1]/f is a finite product of finiteseparable field extensions of κ(q). In particular, the local factor ring Tp/q is a field.

Because Rq → Tp is a local flat map of local noetherian rings and the ‘fiber ring’ isregular, it follows from Theorem 23.7 that regularity of Tp for arbitrary p is a consequenceof regularity of Rq for arbitrary q. In other words, we are reduced to checking the regularityof the local rings of F [Z1, . . . , Zn] for any field F , and it is enough to check the localizationsat maximal ideals. The case n = 0 is clear. Since any maximal ideal contracts to a maximalideal in F [Z1, . . . , Zn−1] (by the Nullstellensatz!), one can carry out an induction argumenton n by going through the proof of Theorem 19.5 in Matsumura.

3. Let f ∈ K[T1, . . . , Tn] be a polynomial, K a field. Explain how the condition that fbe irreducible over K is equivalent to the non-solvability over K of a suitable system ofpolynomial equations. Using the Nullstellensatz over K (!), give a formulation in terms ofyour polynomial constraints for what it means to say that f is irreducible over an algebraicclosure of K. Conclude that if f ∈ Z[T1, . . . , Tn] is irreducible over an algebraic closure of Q(over equivalently (!), over C), then for all but finitely many primes p, f is irreducible whenviewed with coefficients over an algebraically closed field of characteristic p.

Solution. Giving monomials the usual lexicographical ordering, scale f so that its highestdegree term is monic of degree d > 0. Any non-trivial factorization f = gh can always beassumed to have g and h similarly ‘monic’ with degrees a and d−a, where 0 < a < d. Givingg and h ‘indeterminate’ coefficients, the condition that f be reducible as a product of degreea and d − a ‘monics’ amounts to the solvability in K of a suitable ‘universal’ finite systemof polynomial equations, depending only on the integers a and d, and whose coefficientsare universal polynomials over Z (!) in the coefficients of f . Denote these polynomials by{pi,a(f)}, and let Ii,a(f) be the ideal they generate in the appropriate polynomial ring overK (in a number of indeterminates depending only on d, n, and the ‘leading’ monomial of f).

Thus, the condition that f be reducible over K is that the product ideal Ii(f) =∏Ii,a(f)

have a K-rational zero. Because of the ‘universal’ nature of this construction, the formation

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of Ii(f) is compatible with extension of the ground field, so the condition that f be geo-metrically irreducible is that Ii(f) have no geometric zeros. But every maximal ideal m ina polynomial ring K[Z1, . . . , Zm] has a geometric solution, since K[Z1, . . . , Zm]/m is a finiteextension of K (and so admits a K-embedding into an algebraic closure of K). Thus, lack ofa geometric zero of Ii(f) is equivalent to Ii(f) lying in no maximal ideal, which is to say thatIi(f) = (1). Observe how this latter condition holds over K and makes no reference to analgebraic closure of K (it is this step that would fail if we dropped the ‘geometric’ condition,since for non-algebraically closed fields K, there are plenty of ideals in polynomial rings overK which are proper and have no K-rational zero).

In the setting over Z, we pass to working over Z[1/M ] wth M the (non-zero) ‘leading’coefficient of f , so we may assume f is ‘monic’ as above. We have the polynomials pi,a(f)which have coefficients in Z[1/M ] (!), so we let Ii,a(f) and Ii(f) denote the correspondingideals in the appropriate polynomial ring over Z[1/M ]. When we extend scalars to any of theresidue fields of Z[1/M ] (i.e., Q or Fp for p not dividing M), the resulting ideal generatedby Ii(f) is the ideal constructed above for detecting geometric irreducibility of f over theresidue field. Geometric irreducibility over Q thereby implies that Ii(f) contains 1 if weallow Q-coefficients. This expression for 1 involves the introduction of only finitely manydenominators, so for a suitable non-zero integer N , Ii(f) generates 1 over Z[1/NM ]. Thus,for all prime ideals of Z not containing NM , we can pass to residue fields of Z[1/NM ] andthereby get that Ii(f) generates 1 over Fp for all p not dividing NM (so f is irreducible overall algebraically closed fields with characteristic equal to such p, since Exercise 2 reduces thisissue to the case of Fp).

This argument carries over to the case where Z is replaced by any integral domain A andbecomes the statement that for a polynomial f over A which is geometrically irreducible overthe fraction field, there exists a non-zero a ∈ A such that f is also geometric irreducible overall residue fields of all prime ideals of A which do not contain a. The same assertion is truefor ‘geometric reducibility’, and is much easier to prove. This very down-to-earth result liesat the bottom of a vast array of useful theorems in EGA IV of the type “if X → Y is a ‘nice’map of schemes, then the set of points of Y over which the ‘geometric’ fibers have propertyP is topologically ‘decent’,”, where the meanings of ‘nice’ and ‘decent’ can be made preciseand the property P can be many things. Another illustration of the tremendous importanceof the Nullstellensatz. Thank you, Hilbert!

Note thatX2−2 ∈ Z[X] is irreducible in Q[X], but is reducible in Fp[X] for infinitely manyp (and is irreducible in Fp[X] for infinitely many p also). Thus, the ‘geometric’ condition ofirreducibility over an algebraic closure of Q is essential (of course, X2 − 2 is not irreduciblein Q[X]).

4. First, give a direct proof of the following special (but important) case of the CohenStruture Theorem (Thm 5.5A, Ch. 1) which includes the cases of interest in classical algebraicgeometry:

Theorem. Let K be a field and (A,m) be a complete local noetherian K-algebra. Assumethere exists a subfield k ⊆ A containing K such that k → A/m is an isomorphism. Showthat A is regular if and only if there is an isomorphism of K-algebras A ' k[[T1, . . . , Tn]], forsome n, and that in such cases there is even an isomorphism as k-algebras.

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Suppose now that K is not a perfect field and a ∈ K is not a pth power, with p =

char(K) > 0. Define p = (T p − a) ⊆ K[T ] and A = K[T ]p. Show that A is a completediscrete valuation ring (in particular, it is regular) with residue field isomorphic to K(a1/p)(as a K-algebra). However, show that there does not exist a subfield k ⊆ A containing Kfor which k → A/m is an isomorphism (hint: look at A ⊗K K(a1/p)). Explain why this isnot inconsistent with the Cohen Structure Theorem.

Solution. Formal power series rings over fields are certainly regular, so we need to prove theconverse aspect of the Theorem. Assuming A is regular with dimension n, we may assumen > 0. Pick x1, . . . , xn ∈ m lifting a k-basis of m/m2, so by completeness of A there is aunique local k-algebra map

f : k[[T1, . . . , Tn]]→ A

satisfying f(Tj) = xj. We just need to show that f is surjective, since then the local domainA is the quotient of k[[T1, . . . , Tn]] modulo a prime ideal P , and if P 6= 0 then then quotient Amodulo P must have dimension strictly smaller than dim k[[T1, . . . , Tn]] = n, yet n = dimAby design. Hence, P would have to be 0, so f is an isomorphism if it is surjective.

To prove surjectivity, we will use the method of “successive approximation”. Let M bethe maximal ideal of k[[T1, . . . , Tn]], so the local f induces a k-algebra map

k[[T1, . . . , Tn]]→ A/m2

that is clearly surjective with M � m/m2 due to how the xj’s were chosen because Tj modM2 7→ xj mod m2. Assume f mod me is surjective for some e ≥ 2 with M � m/me, so forany 2 ≤ j ≤ e− 1 we have M j � mj/me.

For any ξ ∈ A/me+1 we can find h ∈ k[[T1, . . . , Tn]] such that f(h) ≡ ξ mod me, soξ − f(h) ∈ me/me+1. Since m =

∑kxi + m2, by passing to e-fold products we see that

me/me+1 is spanned by e-fold products among the xj’s. Hence, we can find a homogeneouspolynomial ge in the T ’s of degree e such that ge(x1, . . . , xn) ≡ ξ − f(h) mod me+1. Inother words, ξ = f(h + ge) mod me+1. By repeating this process, we can find homogenouspolynomials g1, g2, . . . in the T ’s with deg gi = i so that

ξ ≡∑i

gi(x1, . . . , xn) mod me

for all e ≥ 1 (note that gi’s don’t matter here for i ≥ e). Thus, for g :=∑

i gi(T1, . . . , Tn) ∈k[[T1, . . . , Tn]] we have f(g) ≡ ξ mod me for all e ≥ 1 (where clearly the gi’s for i ≥ e don’tmatter since

∑i≥e gi ∈ M e and f(M e) ⊂ me). Hence, f(g) − ξ

⋂em

e = 0 due to the KrullIntersection Theorem, so ξ = f(g).

Now consider imperfect K of characteristic p > 0 and A = K[T ]p for p = (T p − a) fora ∈ K −Kp. Since K[T ] is a PID, its local ring at any maximal ideal is clearly a local PIDand hence a discrete valuation ring. Thus, its completion is regular of dimension 1, withmaximal ideal principal by regularity and Nakayama’s Lemma, so A is a discrete valuationring. The residue field is the same as before completion, so it is K[T ]/p = K(a1/p) as aK-algebra.

Assume there exists a subfield k ⊂ A mapping isomorphically onto the residue fieldK(a1/p)and containing K, so A ' K(a1/p)[[T ]] as K-algebras by what we have shown above. Applying

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(·)⊗K K(a1/p), the right side becomes

(K(a1/p)⊗K K(a1/p))[[T ]]

since we can pass a finite-dimensional vector space tensor product through a formal powerseries ring (why?). This is non-reduced, since

K(a1/p)⊗K (K[T ]/(T p − a)) = K(a1/p)[T ]/(T p − a) = K(a1/p)[T ]/(T − a1/p)p.

We claim that in contrast, the left side A⊗KK(a1/p) is reduced, and even a discrete valuationring. More specifically, since K(a1/p) is a finite free K-module, it is easy to check thatA ⊗K K(a1/p) is identified with the p-adic completion of K[T ] ⊗K K(a1/p) = K(a1/p)[T ](why?). But pK(a1/p)[T ] is the pth power of (T − a1/p), so the latter completion coincideswith the (T − a1/p)-adic completion of K(a1/p)[T ]. This is the completion of a PID at amaximal ideal, so it is a discrete valuation ring (hence reduced); explicitly, for T ′ = T − a1/p

we have K(a1/p)[T ] = K(a1/p)[T ′], so this completion is K(a1/p)[[T ′]].The Cohen Structure Theorem for A says that A ' F [[T ] for some field F , and passing

to residue fields shows F ' K(a1/p) as fields, so A ' K(a1/p)[[T ]] as abstract rings. Butthere is no guarantee in the Structure Theorem that this should exist as an isomorphism ofK-algebras, and the preceding shows that in fact no such isomorphism can exist.

A reality check. You might wonder how the existence of a “coefficient field” is provedfor an equicharacteristic complete local noetherian ring A (i.e., if A is an algebra over theprime field F of its residue field k, how do we find a ring-theoretic section to A � k), sincein the above explicit counterexample one doesn’t readily see such a subfield inside the givencomplete discrete valuation ring. Here is the construction.

First, consider a transcendence basis {xi}i∈I for k/F, so k is an algebraic extension ofF(xi)i∈I . Let ai ∈ A× be a lift of xi ∈ k×. The ai’s are algebraically independent over Finside A since if f a nonzero polynomial in a finite subset of the variables {Yi} over F thenf(a) ∈ A reduces to f(x) 6= 0 (using the algebraic independence of the xi’s over F). Thus,f(a) 6= 0 and in fact f(a) ∈ A×. This shows that the map F[Yi]i∈I → A defined by Yi 7→ aiis injective with all nonzero elements mapping to units, so it uniquely factors through a ringhomomorphism

F(Yi)i∈I → A.

This gives a subfield k0 ⊂ A mapping isomorphically onto the subfield F(xi)i∈I ⊂ k.It would now suffice to build a k0-algebra section to A � k, where k/k0 is algebraic.

Zorn’s Lemma provides a “maximal” pair (k′0, s) consisting of a subfield k′0 ⊂ k equippedwith a k0-algebra section s : k′0 → A, and if k/k0 is separable then by maximality we musthave k′0 = k (as otherwise we could choose ξ ∈ k − k′0 and then apply Hensel’s Lemma tothe k′0-algebra A to lift into A the root ξ ∈ k of the separable minimal polynomial of ξ overk′0, contradicting the maximality of (k′0, s)). In particular, that settles the case when F ischaracteristic 0, so now assume the equicharacteristic A is an Fp-algebra for a prime p.

For a general choice of (xi)i∈I (when k is not algebraic over Fp) the algebraic extensionk/k0 is not separable, and when k isn’t finitely generated over Fp it can happen that k/k0

is inseparable for every choice of transcendence basis (even if the transcendence degree isfinite); an example is k equal to the perfect closure of Fp(X). And when k/k0 is not

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separable, typically no k0-algebra section exists, as we saw in the counterexample above(take A = K[T ]∧(T p−a) for a ∈ K −Kp and use a transcendence basis for k = K(a1/p) over

Fp that is one for k/Fp containing a, in which case we cannot succeed because we saw thatthe element a ∈ k ⊂ A has no pth root in A).

But it turns out that there is a special class of transcendence bases for which these problemsdo not occur, so-called “p-bases”. For a field extension k/k0 in characteristic p > 0, a subset{ai}i∈I ⊂ k is called a p-basis over k0 if

[k0kp(ai1 , . . . , air) : k0k

p] = pr

(the maximal possible degree) for any finite set of distinct elements i1, . . . , ir ∈ I. In otherwords, monomials in the ai’s with exponents < p are k0k

p-linearly independent. From thisviewpoint, by considering the extension k/k0k

p it is elementary to construct p-bases (useZorn’s Lemma) and to prove that for perfect k0 they are algebraically independent over k0

(see the proof of Theorem 26.8 in Matsumura).We take our transcendence basis {xi} above to be a p-basis for k/Fp, so k = k0k

p. In fact,by considering linear dependence relations over kp, we see that the natural map k0⊗kp0 k

p → kis an isomorphism (check!), so in particular k0⊗kp0 k

p is a field. Hence, since A is a k0-algebra,

to make an k0-algebra section to A � k it is equivalent to make a kp0-algebra section overthe subfield kp ⊂ k containing kp0.

Letting m ⊂ A be the maximal ideal, we shall proved inductively on n > 0 that anygiven k0-algebra section sn : k → A/mn (as we have for n = 1!) lifts to a k0-algebra sectionsn+1 : k → A/mn+1. Upon choosing a compatible sequence of such sections, we would obtain

s : k → lim←−A/mn = A

(using completeness!), as desired. Letting R be the k0-subalgebra preimage in A/mn+1 of thek0-subalgebra sn(k) ⊂ A/mn, we have R � k with kernel J = mn/mn+1 that is a square-zeromaximal ideal.

Finally, our task comes down to showing that if R is a k0-algebra containining a square-zero ideal J such that R/J = k as k0-algebras then there exists a k0-algebra section k → R.But recall that k = k0⊗kp0 k

p, so it is the same to make a kp0-algebra section over the subfieldkp ⊂ k. For any c ∈ k and lift c ∈ R, the power cp depends only on c and not on c since forany t ∈ J clearly (c + t)p = cp + tp = cp (as Jp ⊂ J2 = 0). Hence, cp 7→ cp is a well-definedmap kp → R that is easily seen to be a kp0-algebra section over kp ⊂ k.