MATH 8253 ALGEBRAIC GEOMETRY WEEK 10 2.5.6. Let a Be a Noetherian Ring, I a Proper Ideal of A

CIHAN˙ BAHRAN

2.5.6. Let A be a Noetherian ring, I a proper ideal of A. (a) Let x ∈ I. Show that ht(I/xA) + 1 ≥ ht(I), and that the equality holds if x does not belong to any minimal prime ideal of A. Show that any prime ideal minimal among those containing x has height equal to 1, if x is a regular element. (b) Show that ht(I) ≥ 1 if and only if I is not contained in the union of the minimal prime ideals of A (use (a) and Exercise 5.4). (c) Let r = ht(I). Show that I contains an ideal J generated by r elements such that ht(J) = r and that ht(I/J) = 0.

(a) Recall that by deﬁnition ht(I) = inf{ht(p): p ∈ V (I)} Fix p ∈ V (I). Since x ∈ I, the image of x in Ap, say y, is non-invertible. Now (Ap, pAp) is a Noetherian local ring and y ∈ pAp. Then by Theorem 5.15 we have

ht(p/xA) = dim((A/xA)p/xA) = dim(Ap/yAp) ≥ dim Ap − 1 = ht(p) − 1 . Now varying the p’s we get inf{ht(p/xA): p ∈ V (I)} ≥ inf{ht(p): p ∈ V (I)} − 1 inf{ht(q): q ∈ V (I/xA)} ≥ ht(I) − 1 ht(I/xA) ≥ ht(I) − 1 .

If x does not belong to any minimal prime ideal of A, then its image in Ap does not belong to any minimal prime ideal of Ap for every p ∈ V (I). Thus again by Theorem 5.15 the above inequalities degenerate into equalities. For the last part, assume x is regular and p is a prime ideal minimal among those containing x. Then we have ht(p) ≤ ht(p/xA) + 1 = 1 so it remains to show that p is not a minimal prime in A. This is true because every element of a minimal prime is a zero-divisor by Exercise 2.1.4. (b) Suppose that I is contained in the union of minimal prime ideals of A. Note that being Noetherian, A has finitely many minimal prime ideals; so by prime avoidance, I ⊆ p for some minimal prime p of A. Then ht(I) ≤ ht(p) = 0 so ht(I) = 0. Conversely, assume ht(I) = 0. Since heights are positive integers, we can replace the “infimum” in the definition of ht(I) with “minimum” and get 0 = ht(I) = min{ht(p): p ∈ V (I)} so there exists a prime ideal of height 0, that is, a minimal prime ideal containing I. (c) We employ induction on r. For the basis case, we are assuming ht(I) = 0. Note that I contains the zero ideal 0 which is generated by 0 elements; and ht(I/0) = ht(I) = 0. 1 MATH 8253 ALGEBRAIC GEOMETRY WEEK 10 2

Assume the claim is true for r − 1, where r ≥ 1. Let I be an ideal such that ht(I) = r. Since ht(I) ≥ 1, by (b) I contains an element x which does not belong to any minimal prime ideal of A. Hence by (a) we have ht(I) = ht(I/xA) + 1. Since I/xA is a proper ideal of the Noetherian ring A/xA and ht(I/xA) = r − 1, by the induction hypothesis I/xA contains an ideal J/xA (every ideal of A/xA is of this form where x ∈ J) generated by r − 1 elements such that ht(J/xA) = r − 1 and 0 = ht((I/xA)/(J/xA)) = ht(I/J).

If J/xA is generated by x1 + xA, . . . , xr−1 + xA in A/xA, then J is generated by x, x1, . . . xr−1 in A. So J is generated by r elements; moreover again by (a) we have ht(J) = ht(J/xA) + 1 = r. 2.5.7. Let f : X → Y be a morphism of algebraic varieties. We suppose that f(X) is everywhere dense in Y . Show that dim X ≥ dim Y (see Proposition 5.19). Note a counterexample below for schemes in general.

Consider the unique closed subscheme Xred of X which has the same underlying topological space as X, so in particular dim Xred = dim X. Since the structure sheaf OXred is defined by dividing OX by a sheaf of ideals, if {Spec Aα} is an affine chart for OX then {Spec(Aα/Iα)} will yield an affine chart for OXred for some Iα’s. Hence Xred is also an algebraic variety over the same field with X, say k. Note that the composition Xred → X → Y is topologically the same with the original morphism f. So we may assume that X is reduced. Now we have f : X → Y where X is reduced. Then by Proposition 4.2 f uniquely factors into a morphism X → Yred, which is again same as f as a continuous map. So we may assume Y is also reduced. Being an algebraic variety, Y is a Noetherian scheme. So by Proposition 4.9 it has finitely many irreducible components, say Y1,...,Yn. Since dim Y = max{dim Yj}j, it −1 suffices to show that dim X ≥ dim Yj for every j. So fix j and let Xj = f (Yj) for each j = 1, . . . , n. On one hand, since f(Xj) ⊆ Yj and Yj is closed in Yj we have f(Xj) ⊆ Yj. On the other hand, we have

n ! n n [ [ [ Y = f(X) = f Xi = f(Xi) = f(Xi) i=1 i=1 i=1 so intersecting with Yj we get n [ [ Yj = f(Xi) ∩ Yj = f(Xj) ∪ (f(Xi) ∩ Yj) . i=1 i6=j

Suppose Yj 6= f(Xj). Then the irreducibility of Yj forces Yj = f(Xi) ∩ Yj for some i 6= j. This yields Yj ⊆ f(XÄi) ⊆ Yi, a contradiction.ä

So we conclude f(Xj) = Yj. Since Yj is a closed subset of Y , by Proposition 4.2 (d), Yj has a unique structure of reduced closed subscheme. Observe that since closed subschemes of affine varieties are themselves affine varieties (because if we quotient out a finitely generated k-algebra by an ideal, we again get a finitely generated k-algebra), it follows that closed subschemes of algebraic varieties are themselves algebraic varieties. So Yj is a sub-variety of Y . Similarly, Xj is a sub-variety of X. So if we knew that dim Xj ≥ dim Yj, then we would conclude dim X ≥ dim Xj ≥ dim Yj, as desired.

By construction, f restricts to a continuous map between the topological spaces Xj and Yj. It also restricts to a scheme morphism from Xj to Yj (this is possible due to the MATH 8253 ALGEBRAIC GEOMETRY WEEK 10 3 reduced-ness, see Proposition 1 after the solution), we reduce tho the case where Y is irreducible.

Let X1,...Xm be the irreducible components of X (note that they are diﬀerent from the above Xj’s). As above, each Xj has the structure of a closed and reduced subscheme of X and are themselves varieties. Then we have m m m [ [ [ Y = f(X) = f Xj = f(Xj) = f(Xj) . j=1 j=1 j=1

Since Y is irreducible, Y = f(Xj) for some j, so the composition Xj ,→ X → Y maps Ñ é Xj to a dense subset of Y . So we can assume X is also irreducible. Finally, we are in the case where both X and Y are integral algebraic varieties. Let ξ be the generic point of X. By the continuity of f, we have f(X) = f({ξ}) ⊆ f({ξ}) . But f(X) = Y , so {f(ξ)} = Y . So by Proposition 4.12 (a), f(ξ) is the generic point of # Y . The sheaf morphism f : OY → OX which induces a ring homomorphism # fξ : OY,f(ξ) → OX,ξ between the stalks. But since X,Y are integral with generic points ξ and f(ξ), these # stalk rings are ﬁelds; thus fξ is injective. Also since X,Y are algebraic varieties over k, # both OY,f(ξ) and OX,ξ are ﬁeld extensions of k and fξ is compatible with this structure (more precisely it is a k-algebra homomorphism).

Thus by making identiﬁcations we get ﬁeld extensions k ⊆ OY,f(ξ) ⊆ OX,ξ. Using Proposition 5.19, we get dim X = trdegk(OX,ξ) ≥ trdegk(OY,f(ξ)) = dim Y . We are done. Here is the proposition we promised: Proposition 1. Let f : X → Y be a morphism of schemes where X is reduced. Suppose that Z is a closed subset of Y such that as a continuous map, f maps X into Z, that is, f(X) ⊆ Z. Then putting the unique reduced closed subscheme structure on Z where i : Z → Y is the closed immersion, then f uniquely factors through i as a scheme morphism.

Proof. The reduced closed subscheme structure (Z, OZ ) on Z is given by the sheaf of ideals J of OY deﬁned by

J (V ) = {s ∈ OY (V ): sy ∈ my for every y ∈ Z ∩ V } so that OZ (Z ∩ V ) = OY (V )/J (V ) for every open subset V of Y . And the sheaf part # i : OY → i∗OZ of the closed immersion i : Z → Y is given by the projections # iV : OY (V ) → (i∗OZ )(V ) = OZ (Z ∩ V ) = OY (V )/J (V ) . Note that as a continuous map, f obviously factors through i since f(X) ⊆ Z, so say f = i◦g where g : X → Z is a continuous map. We want to construct a sheaf morphism # # # # g : OZ → g∗OX which satisﬁes i∗(g ) ◦ i = f :

f # OY / f∗OX 8 i# # i∗(g ) i∗OZ MATH 8253 ALGEBRAIC GEOMETRY WEEK 10 4

# Let’s observe that the diagram forces what i∗(g ) has to be. Indeed for every open subset V of Y , evaluating the sheafs we get

f # V −1 OY (V ) / OX (f (V ))

# iV OY (V )/J (V )

# # # # Since iV is the projection, fV factors through iV if and only if fV (J (V )) = 0 and in this case the factorization is unique. The uniqueness of the factorization forces the maps when V varies to be compatible with the restrictions so they uniquely determine a sheaf morphism from i∗OZ to f∗OX . Let s ∈ J (V ). For x ∈ f −1(V ) we have f(x) ∈ Z ∩ V since f(X) ⊆ Z by assumption. # # −1 Then sf(x) ∈ mf(x) and hence [fV (s)]x ∈ mx. So writing t = fV (s) ∈ OX (f (V )), we −1 obtained that tx ∈ mx for every x ∈ f (V ). Let U = Spec A be an aﬃne open subset −1 contained in f (V ) and consider t|U ∈ A. Then (t|U )p ∈ pAp for every p ∈ Spec A, which means t|U ∈ p for every prime ideal p of A. Then t|U is nilpotent in A, so t|U = 0 −1 because A = OX (W ) is a reduced ring by Proposition 4.2 (a). Since f (V ) can be # covered by aﬃne open subsets of X and OX is a sheaf, we get fV (s) = t = 0. # # Thus there indeed is a unique sheaf morphism ϕ : i∗OZ → f∗OX such that ϕ ◦ i = f . Note that the map

HomShfZ (OZ , g∗OX ) → HomShfY (i∗OZ , f∗OX )

ψ 7→ i∗(ψ) is a bijection since i is a closed immersion. Therefore there exists a unique ψ = g# : # OZ → g∗OX such that i∗(g ) = ϕ.

2.5.10 Let k be an uncountable ﬁeld. Let X be an algebraic variety over k with dim X ≥ 1, (Yn)n a sequence of closed subsets of X with dim Yn < dim X. We want S to show that n Yn 6= X. 1 m (a) Show the result for X = Ak and then for Ak . (b) By using Noether’s normalization lemma, show the result for an arbitrary aﬃne variety. Deduce the general case from this.

1 (a) First, assume A = k[T ] and X = Spec A = Ak. For every n the closed subset Yn of X is necessarily of the form V (In) where In is an ideal of A. Since dim X = 1 < dim V (In), each In must be a nonzero ideal. Clearly we may assume that In 6= A since that implies Yn = ∅. Observe that if I is a nonzero proper ideal of A, since A is a PID, I is generated by a nonzero non-unit element a and V (I) consists of ﬁnitely many primes which are precisely the ideals generated by the irreducible factors of a.

Thus each Yn is a ﬁnite set, hence their countable union is countable. But there is an injective map k → Spec A = X α 7→ (T − α) S so X is uncountable since k is uncountable by hypothesis. Therefore X 6= n Yn. MATH 8253 ALGEBRAIC GEOMETRY WEEK 10 5

m Second, assume A = k[T1,...,Tm] and X = Spec A = Ak . We employ induction on m. We have just completed the base case. So assume m ≥ 2 and that the claim holds for S m − 1. For the sake of contradiction, suppose X = n Yn. For every prime ideal p ???????????? ???????????? (b) Let A be a finitely-generated k-algebra and X = Spec A. By Noether normalization, there exists a finite injective homomorphism k[T1,...,Tm] ,→ A for some m ≥ 0. Then m by Proposition 2.5.10(b), the induced scheme morphism f : X → Ak is surjective and dim X = m. Theeeeeeen???? n 2.5.12. Let k be a field. We will determine the affine open subsets of Ak and of n Pk . See also Exercise 4.1.15. n n n (a) Show that the principal open subsets of Ak and Pk (not equal to Pk ) are affine. S n (b) Let X = i D(fi) be a finite union of principal open subsets of Ak . Show that O n (X) = k[T ,...T ] , where f = gcd{f } . Show that every affine Ak 1 n f i i n open subset of Ak is principal. n (c) Show that every irreducible closed subset of Pk of dimension n − 1 is principal. n (d) Let X be an affine open subset of Pk . Show that the irreducible components n of Pk r X are of dimension n − 1, and that X is a principal open subset of n Pk .

2.5.13. Let k be a field. A function field in n variables over k is a field K that is finitely generated over k (i.e., there exists f1, . . . fr ∈ K such that K = k(f1, . . . , fr)), of transcendence degree trdegk K = n. Show that an extension K of k is a function field in n variables over k if and only if K = K(X), where X is an integral algebraic variety over k, of dimension n. If this is the case, we can take X projective.

3.1.3. Let f : X → Y be a dominant morphism of irreducible schemes. Show that the generic ﬁber of f is dense in X.

3.1.4. Let f : X → Y be a morphism of affine schemes. Show that for any affine open subscheme V of Y , f −1(V ) is affine (see also Exercise 3.8).

Note that

f −1(V ) / X

V / Y is a fiber square. It is easy to see that the diagram is a fiber square of topological spaces, but in addition it is a fiber square of schemes; this is because V and f −1(V ) are open in X and Y , respectively. The key observation is, if Z → Y is a morphism of schemes whose image lies in V , then it uniquely factors through V as a scheme morphism. MATH 8253 ALGEBRAIC GEOMETRY WEEK 10 6

On the other hand, since V,X,Y are all affine by the first part of Proposition 1.2 their fiber product, which is necessarily isomorphic to f −1(V ) by the universal property, is affine. 3.1.5. Let X,Y be algebraic varieties over a field k. Show that

dim(X ×k Y ) = dim X + dim Y (use Proposition 2.1.9).

Let {Uα}α and {Vβ}β be open coverings of affine varieties for X and Y , respectively. Writing p : X ×k Y → X and q : X ×k Y → Y for the canonical maps, by Proposition −1 −1 1.4(c), each fiber Uα ×k Vβ can be identified with the open subscheme p (Uα)∩q (Vβ) of X ×k Y . Observe that [ −1 −1 [ [ −1 −1 p (Uα) ∩ q (Vβ) = p (Uα) ∩ q (Vβ) α,β α β

[ −1 [ −1 = p (Uα) ∩ q (Vβ) Ä ä α Ä β ä

[ −1 −1 [ = Ñp (Uα) ∩ q ( Vβ)é α β [ −1 −1 = p (Uα) ∩ q (Y ) α Ñ é [ −1 = p (Uα) ∩ (X ×k Y ) α [ −1 = Äp (Uα) = X ×k Y.ä α

Thus {Uα ×k Vβ}α,β is an open covering of XÄ×k Y . Assuming thatä the claim is valid for aﬃne varieties, we obtain

dim(X ×k Y ) = sup{dim(Uα ×k Vβ)}α,β

= sup{dim(Uα) + dim(Vβ)}α,β

= sup{dim(Uα)}α + sup{dim(Vβ)}β = dim X + dim Y. Thus we reduce to the affine case. If X = Spec A and Y = Spec B where A, B are finitely generated k-algebras, by Noether normalization there exists m, n ≥ 0 and finite injective homomorphisms ϕ : k[T1,...,Tm] ,→ A and ψ : k[T1,...,Tn] ,→ B. Since tensoring over k is exact, the finite homomorphism ∼ ϕ ⊗ ψ : k[T1,...,Tm+n] = k[T1,...,Tm] ⊗k k[T1,...,Tn] → A ⊗k B is injective. In particular, each of ϕ, ψ and ϕ ⊗ ψ are integral and injective. Thus by Proposition 5.10,

dim A = m , dim B = n , dim(A ⊗k B) = m + n . ∼ Since X ×k Y = Spec(A ⊗k B), we get

dim(X ×k Y ) = dim(A ⊗k B) = dim A + dim B = dim X + dim Y. MATH 8253 ALGEBRAIC GEOMETRY WEEK 10 7

3.1.6. Let π : T → S be a morphism of schemes. Let us suppose that π : T → S is an open or closed immersion, or that S = Spec A, and π is induced by a localization A → F −1A for a multiplicative set F ⊆ A. (a) Let f, g : Z → T be two morphisms of schemes such that π ◦ f = π ◦ g. Show that f = g. (b) Let X,Y be T -schemes. Show that the canonical morphism X ×T Y → X ×S Y is an isomorphism. (c) Show that (b) can be false if the hypothesis on T → S is not veriﬁed.

(a) Suppose π is an open or closed immersion. Then since π is injective, we have f = g as continuous maps. To check they are equal as scheme morphisms, we need to check # # that the sheaf morphisms f , g : OT → f∗OZ = g∗OZ coincide. This can be checked on stalks. Indeed, given z ∈ Z the stalk maps OS,π(f(z)) → OZ,z induced by π ◦ f and OS,π(g(z)) → OZ,z induced by π ◦ g coincide. This map factors through OT,f(z), sitting in a commutative diagram

OS,π(f(z)) / OZ,z 8

# fz OT,f(z) The vertical map above is surjective if π is an open or closed immersion. Since compos- # # # # ing this vertical map either with fz or gz gives the same result, it follows that fz = gz for every z ∈ Z. This completes the case of immersions. For the second part, note that the functor Spec : Ring → Schop is left adjoint (this follows from Proposition 2.3.25) to the global sections functor, hence is cocontinuous1. Therefore it preserves epimorphisms, that is, if ϕ is an epimorphism in Ring, then Spec(ϕ) is an epimorphism in Schop, hence a monomorphism in Sch. Localizations are epimorphisms. Let’s verify the two important assertions we have made in the above paragraph: • Cocontinuous functors preserve epimorphisms. Observe that a morphism ϕ : A → B in a category C is an epimorphism if and only if the diagram

ϕ A / B

ϕ idB

idB B / B is a push-out square in C. In that case if F : C → D is a cocontinuous functor, then the diagram

F (ϕ) F (A) / F (B)

F (ϕ) idF (B)

idF (B) F (B) / F (B) is a push-out square in D, hence F (ϕ) is an epimorphism in D.

1I’ve always wanted to cancel the “co”s and write ntinuous instead. MATH 8253 ALGEBRAIC GEOMETRY WEEK 10 8

• Localizations are epimorphisms. Let f : A → S−1A be the localization map a 7→ a/1 where A is a ring and S is a multiplicative set. Suppose g1, g2 : −1 S A → B are ring homomorphisms such that g1 ◦ f = g2 ◦ f. Then for every a ∈ A, we have g1(a/1) = g2(a/1). In particular, for every s ∈ S we have −1 g1(s/1) = g2(s/1). But in S A, the element s/1 is a unit and its inverse is 1/s. Therefore both g1(1/s) and g2(1/s) are inverses of the element g1(s/1) = g2(s/1) in B. Thus g1(1/s) = g2(1/s) for every s ∈ S and so for every a ∈ A and s ∈ S, we have

g1(a/s) = g1(a/1)g1(1/s) = g2(a/1)g2(1/s) = g2(a/s) ,

thus g1 = g2. (b) This can be checked in abstract nonsense terms. If we add a monomorphism to the lower right corner of a pull-back square, the big diagram is also a pull-back. Indeed, if we have a pull-back diagram p P / X

q f Y g / T in a category and i : T → S is a monomorphism, we claim that the commutative square p P / X

q i◦f Y / S i◦g is also a pull-back. Indeed, if there is an object Z equipped with morphisms k : Z → X and l : Z → Y such that (i ◦ f) ◦ k = (i ◦ g) ◦ l , since i is a monomorphism we get f ◦ k = g ◦ l. Then because of the initial pull-back diagram, there exists a unique morphism ϕ : Z → P making the diagram

Z k ϕ # " P p / X l q f $ Y g / T commute. So it makes

Z k ϕ # " P p / X l q i◦f $ Y / S i◦g MATH 8253 ALGEBRAIC GEOMETRY WEEK 10 9 commute and it is the unique morphism from Z to P that does so. We veriﬁed the desired universal property of the pull-back. (c) We have seen above that the conclusion of (b) holds whenever π is an epimorphism in the category of schemes. Let’s see what happens when we drop this assumption. ∼ Consider the ring homomorphism i : R ,→ C. Then C ⊗C C = C while C ⊗R C is a 4- dimensional R-algebra whereas C is a 2-dimensional R-algebra. Hence C⊗C C C⊗R C. Thus

Spec(C ⊗C C) Spec(C ⊗R C)

Spec C ×Spec C Spec C Spec ×Spec R Spec C . And the reason is Spec i : Spec C → Spec R is not a monomorphism of schemes.

3.1.7. Let X,Y be S-schemes, p and q the projection morphisms from X ×S Y to X,Y . Let us ﬁx s ∈ S. Show that for any x ∈ Xs, and any y ∈ Ys, there exists a natural homeomorphism

Spec(k(x) ⊗k(s) k(y)) → {z ∈ X ×S Y : p(z) = x, q(z) = y} .

3.1.9. Let k be a ﬁeld. We want to study Spec A, where A = k(u) ⊗k k(v) is the tensor product of two purely transcendental extensions of k, of transcendence degree 1. (a) We have k(u) = T −1k[u], where T is the multiplicative set made up of the non-zero elements of k[u]. Deduce from this that A is the localization of k[u, v] with respect to the multiplicative set T 0 made up of the non-zero elements of the form P (u)Q(v) ∈ k[u, v]. (b) Let m be a maximal ideal of k[u, v]. Show that there exists a P (u) ∈ mr{0}. Deduce from this that T 0 ∩ m 6= ∅. (c) Show that the maximal ideals of A are of the form gA, with g ∈ k[u, v]r (k[u] ∪ k[v]) irreducible in k[u, v]. (d) Show that Spec A is an inﬁnite set and dim A = 1.

(a) Consider the k-algebra homomorphism

ϕ : k[u, v] → k(u) ⊗k k(v) = A u 7→ u ⊗ 1 , v 7→ 1 ⊗ v . Clearly ϕ maps the subrings k[u] and k[v] of k[u, v] to A∗. Thus ϕ maps T 0 to A∗. Suppose ψ : k[u, v] → B is another k-algebra homomorphism such that ψ maps T 0 to B∗. Mapping the subsets k[u]r {0} and k[v]r {0} to B∗, ψ uniquely factors through k(u) and k(v), and hence through A = k(u) ⊗k k(v) via ϕ since A is the coproduct of k(u) and k(v) in the category of k-algebras. (b) By Corollary 2.1.12, k[u, v]/m is a ﬁnite extension of k. Consider the coset u ∈ k[u, v]/m, If u = 0, then u ∈ m and we can choose P (u) = u. If u 6= 0, then there is a nonzero polynomial P [t] ∈ k[t] such that 0 = P (u) = P (u) in k[u, v]/m. Thus P (u) ∈ mr {0}. So P (u) ∈ T 0 ∩ m. (c) By (a), ϕ induces is an inclusion preserving bijection 0 Spec A ↔ {p ∈ Spec k[u, v]: p ∩ T = ∅} . MATH 8253 ALGEBRAIC GEOMETRY WEEK 10 10

So by (b), none of the maximal ideals of k[u, v] lie in the above set, hence dim A ≤ dim k[u, v] − 1 = 1. Let g ∈ k[u, v]r(k[u] ∪ k[v]) be an irreducible element in k[u, v]. Since k[u, v] is a UFD, 0 pg := (g) is a nonzero prime ideal of k[u, v], hence ht(pg) ≥ 1. Suppose f ∈ pg ∩ T . Then f = PQ where P ∈ k[u]r {0} and Q ∈ k[v]r {0}. Since pg is prime, WLOG P ∈ pg. So there exists h ∈ k[u, v] such that gh = P . Since h 6= 0, we get

0 < degv(g) ≤ degv(g) + degv(h) = degv(gh) = degv(P ) = 0 , 0 −1 a contradiction. Thus pg ∩T = ∅. Thus there exists q ∈ Spec A such that ϕ (q) = pg. Viewing A as a k[u, v]-algebra via ϕ, since pg = g · k[u, v], we have q = g · A. Since ht(q) = ht(pg) ≥ 1 ≥ dim A, the prime ideal q = gA of A is actually maximal. Conversely, suppose n is a maximal ideal of A. Then n = ϕ−1(p) for some p ∈ Spec k[u, v] with p ∩ T 0 = ∅ and p is maximal with this property. Since there are irreducible elements inside k[u, v]r T 0 (u + v for instance), p cannot be 0. So p contains a nonzero element g. Since p is prime and k[u, v] is a UFD, we may assume that g is irreducible. Considering the inclusion of prime ideals 0 ( (g) ⊆ p in k[u, v], the fact that ht(p) = ht(n) ≤ 1 forces (g) = p (and ht(n) = 1, consequently dim A = 1). We get n = gA and since g ∈ p, we have g∈ / k[u] ∪ k[v]. (d) We obtained in (c) that dim A = 1. By what we have shown in (c), there is a bijection {g · k[u, v]: g ∈ k[u, v]r (k[u] ∪ k[v]) irreducible} ↔ Spec A. So it is enough to show that k[u, v]r(k[u] ∪ k[v]) contains inﬁnitely many non-associate n irreducible elements. By Eisenstein’s criterion, the family (u + v)n∈N provides this.