Reid 5.2. Describe the irreducible components of V (J) for J = (y2− x4, x2−2x3− x2y + 2xy + y2− y) in k[x, y, z]. Here k is algebraically closed.Answer: Note that the first generator factors as (y−x2)(y+x2), while the secondfactors as (y − x2)(−1 − 2x + y). Thus J = (y − x2)(y + x2, −1 − 2x + y), soV (J) = V (y − x2) ∪ V (y + x2, −1 − 2x + y).I claim that each of V (y − x2), V (y + x2, −1 − 2x + y) is irreducible. We check thefirst by verifying that (y − x2) ⊆ k[x, y, z] is prime. Indeed,k[x, y, z]/(y − x2)∼−→ k[x, z]via x 7→ x, y 7→ x2, and z 7→ z, with inverse given by x 7→ x, z 7→ z. For the second,note that, by high school algebra,V (y + x2, −1 − 2x + y) = V (x + 1, y − 1),andk[x, y, z]/(x + 1, y − 1)∼−→ k[z]via x 7→ −1, y 7→ 1, z 7→ z, with inverse z 7→ z. Thus each of V (y − x2),V (y +x2, −1 − 2x + y) = V (x + 1, y − 1) is irreducible; to check they are the compo-nents, we just need to see that neither contains the other. Again this is high schoolalgebra: we find that if the characteristic of k is not two, there are no containments,whereas if the characteristic is two, the second zero set is redundant, and we havejust one component, the first.Remark: The isomorphisms used to check that the above ideals are prime couldalso been seen geometrically in terms of the zero sets, the first as projecting theparabola-sheet to a plane, the second as recognizing that V (x + 1, y − 1) is just aline. See the algebra-geometry correspondence sketched in the next exercise.5.16. Let k be a field and A 6= 0 a finitely generated k-algebra. Then thereexist elements y1, . . . , yr∈ A which are algebraically independent over k and suchthat A is integral over k[y1, . . . , yr]. Moreover, if k is infinite and x1, . . . , xnis anyset of generators for A, we can choose the yias linear combinations of the xj. Thishas the geometric consequence that if k is algebraically closed and V is a closedsubvariety of kn, then there is a linear map kn→ krwhich, when restricted to V ,is [finite-to-one and] surjective.Proof: The first part was done in lecture, so assume k infinite and let’s provethe refined statement. Let’s go by induction on n. The case n = 0 being trivial,assume n ≥ 1. There are two cases:First suppose that xnis algebraic over k[x1, . . . , xn−1]. Then we have a nonzeropolynomial in n + 1 variables f with f(x1, . . . , xn) = 0. Let F denote its highestdegree homogeneous part of f; so F 6= 0. Then its dehomogenization with respect tothe last variable is nonzero, hence nonzero as a function since k is infinite; thus thereare λ1, . . . , λn−1∈ k such that F (λ1, . . . , λn−1, 1) 6= 0. Let x0i= xi− λixn. Applythe inductive hypothesis to the x0iand get y1, . . . , yr; these are linear combinationsof the x0i, hence of the xi. I claim that A is integral over k[y1, . . . , yr], which wouldmean we’re done. By transitivity of integrality, it suffices to see that A is integral12over k[x01, . . . x0n−1]; since A = k[x01, . . . , x0n−1][xn], though, it suffices to see that xnis integral over k[x01, . . . , x0n−1]. And indeed, we havef(x01+ λ1xn, x02+ λ2xn, . . . , x0n−1+ λn−1xn, xn) = 0,which, when expanded out as a polynomial in xn, has leading coefficient F (λ1, . . . , λn−1, 1) 6=0; dividing through by this we have a monic polynomial which does the job.Now suppose that xnis transcendental over k[x1, . . . , xn−1]. Apply the induc-tive hypothesis to x1, . . . , xn−1, and add xnto the yjthat you get; this works forx1, . . . , xn.Now for the geometric interpretation.Lemma: Let Aφ−→ B be a map of rings, and f : Spec(B) → Spec(A) the mapgiven by pulling back prime ideals along φ. Then for p ∈ Spec(A), we have acanonical bijectionSpec(Bp/pBp)∼−→ f−1({p}) ⊆ Spec(B),induced by pulling back prime ideals along the natural map B → Bp→ Bp/pBp.Proof: Here by BpI mean the localization of B at the multiplicative set φ(A\p),and by pBpI mean φ(p)Bp; this is indeed an ideal, as you should check. Actually,I’ll leave this whole thing for you to check; you know what the induced map on Specis for localizations, and you know what the induced map on Spec is for moddingout by an ideal; it suffices to put these two together.Corollary: Let Aφ−→ B be a map of rings which is finite, i.e. for which B isa finite A-module, and f its induced map on Spec. Then im(f ) = V (ker(φ)) ⊆Spec(A). In particular, if φ is injective, f is surjective.Proof: Recall that Spec of a ring is empty if and only if the ring is zero. Thusthe lemma givesim(f) = {p | Bp6= pBp}.However, Bpis a finite Ap-module, so Nakayama shows that in factim(f) = {p | Bp6= 0}.But Bp= 0 if and only if 1/1 = 0 in Bp, which means by definition that there issome s ∈ A \ p such that φ(s) = 0, which is equivalent to saying that p 6⊇ ker(φ),as desired.Remark 1: The result holds more generally if φ is just assumed integral; the proofthen just has one tricky step. Recall that finite is the same as integral plus finitelygenerated, so for k-algebra maps between coordinate rings of varieties they’re thesame thing.Remark 2: For a general φ : A → B, the closure of the image of f is alwaysV (ker(φ)) (nice exercise), so the corollary is equivalent to just saying that the imageof f is closed.3Remark 3: We can also show that, under the hypotheses of the corollary, thefibers of f are all finite. This is because Bp/pBp, being a finite dimensional vectorspace over the field Ap/pAp, is Artinian, and thus has only finitely many primeideals.Remark 4: Also in this situation, if f (P) = p, then P is maximal if and onlyif p is. Indeed, φ induces an integral injection A/p → B/P, from which the claimfollows by a result from class. We’ll probably see a more refined version of thiswhen we get to dimension theory.OK, back to the show. Let V be a closed subvariety of kn; let A = k[x1, . . . , xn]/I(V ).Use the first part of the problem to get linear forms L1, . . . , Lrsuch that withyi= Li(x1, . . . , xn) the yiare algebraically independent and A is integral (finite)over k[y1, . . . , yr]. Define L : kn→ krby L = (L1, . . . , Lr). The claim is that Lrestricted to V is surjective, so let p ∈ …