11. Schemes To defne schemes, just as wth algebrac varetes, the dea s to frst defne what an affne scheme s, and then realse an arbtrary scheme, as somethng whch s locally an affne scheme. The defnton of an affne scheme s motvated by the correspondence between affne varetes and fntely generated algebras over a feld, wthout nlpotents. The dea s that we should be able to assocate to any rng R, a topologcal space X, and a set of contnuous functons on X, whch s equal to R. In practce ths s too much to expect and we need to work wth a slghtly more general object than a contnuous functon. Now f X s an affne varety, the ponts of X are n correspondence wth the maxmal deals of the coordnate rng A = A(X). Unfortunately f we have two arbtrary rngs R and S, then the nverse mage of a maxmal deal won t be maxmal. However t s easy to see that the nverse mage of a prme deal s a prme deal. Defnton 11.1. Let R be a rng. X = Spec R denotes the set of prme deals of R. X s called the spectrum of R. Note that gven an element of R, we may thnk of t as a functon on X, by consderng t value n the quotent. Example 11.2. It s nterestng to see what these functons look lke n specfc cases. Suppose that we take X = Spec k[x, y]. Now any element f = f(x, y) k[x, y] defnes a functon on X. Suppose that we consder a maxmal deal of the form p = x a, x b. Then the value of f at p s equal to the class of f nsde the quotent k[x, y] R/p = x a, x b. If we dentfy the quotent wth k, under the obvous dentfcaton, then ths s the same as evaluatng f at (a, b). Now consder Z. Suppose that we choose an element n Z. Then the value of n at the prme deal p = p s equal to the value of n modulo p. For example, consder n = 60. Then the value of ths functon at the pont 7 s equal to 60 mod 7 = 4 mod 7. Moroever 60 has zeroes at 2, 3 and 5, where both 3 and 5 are ordnary zeroes, but 2 s a double zero. Suppose that we take the rng R = k[x]/ x 2. Then the spectrum contans only one element, the prme deal x. Consder the element x R. Then x s zero on the unque element of the spectrum, but t s not the zero element of the rng. Now we wsh to defne a topology on the spectrum of a rng. We want to make the functons above contnuous. So gven an element 1
f R, we want the set { p Spec R f(p) = 0 } = { p Spec R f p }, to be closed. Gven that any deal a s the unon of all the prncpal deals contaned n t, so that the set of prme deals whch contan a s equal to the ntersecton of prme deals whch contan every prncpal deal contaned n a and gven that the ntersecton of closed sets s closed, we have an obvous canddate for the closed sets: Defnton 11.3. The Zarsk topology on X s gven by takng the closed sets to be where a s any deal of R. V (a) = {p Spec R a p }, Lemma 11.4. Let R be a rng. Then X = Spec R s a topologcal space. Moreover the open sets form a base for the topology. Proof. Easy check. U f = { p R f / p }, By what we sad above, the Zarsk topology s the weakest topology so that the zero sets of f R are closed. Example 11.5. Let k be a feld. Then Spec k conssts of a sngle pont. Now consder Spec k[x]. If k s an algebracally closed feld, then by the Nullstellensatz, the maxmal deals are n correspondence wth the ponts of k. However, snce k[x] s an ntegral doman, the zero deal s a prme deal. Snce k[x] s a PID, the proper closed sets of X consst of fnte unons of maxmal deals. The closure of the pont ξ = 0 s then the whole of X. In partcular, not only s the Zarsk topology, for schemes, not Hausdorff or T 2, t s not even T 1. Now consder k[x, y], where k s an algebracally closed feld. Prme deals come n three types. The maxmal deals correspond to ponts of k 2. The zero deal, whose closure conssts of the whole of X. But there are also the prme deals whch correspond to prme elements f k[x, y]. The zero locus of f s then an rreducble curve C, and n fact the closure of the pont ξ = f s then the curve C. The proper closed sets thus consst of a fnte unon of maxmal deals, unon nfnte sets of the maxmal deals whch consst of all ponts belongng to an affne curve C, together wth the deal of each such curve. Now suppose that k s not algebracally closed. For example, consder Spec R[x]. As before the closure of the zero deal conssts of the whole 2
of X. The maxmal deals come n two flavours. Frst there are the deals x a, where a R. But n addton there are also the deals x 2 + ax + b = (x α β)(x α + β), where a, b, α and β > 0 are real numbers, so that b 2 4a < 0. There s a very smlar (but more complcated) pcture nsde Spec R[x, y]. The set V (x 2 + y 2 = 1) does not contan any deals of the frst knd, but t contans many deals of the second knd. In the classcal pcture, the conc does x 2 + y 2 = 1 does not contan any ponts but t does contan many ponts f you nclude all prme deals. Now suppose that we take Z. In ths case the maxmal deals correspond to the prme numbers, and n addton there s one pont whose closure s the whole spectrum. In ths respect Spec Z s very smlar to Spec k[t]. We wll need one very useful fact from commutatve algebra: Lemma 11.6. If a R s an deal n a rng R then the radcal of a s the ntersecton of all prme deals contanng a. Proof. One ncluson s clear; every prme deal p s radcal (that s equal to ts own radcal) and so the ntersecton of all prme deals contanng a s radcal. Now suppose that r does not belong to the radcal of a. Let b be the deal generated by the mage of a nsde the rng R r. Then the mage of r nsde the quotent rng R r /b s non-zero. Pck an deal n ths rng, maxmal wth respect to the property that t does not contan the mage of r. Then the nverse mage p of ths deal s a prme deal whch does not contan r. Lemma 11.7. Let X be the spectrum of the rng R and let f R. If U f = U g then f n = b g, where b 1, b 2,..., b k R. In partcular U f s compact. Proof. Takng complements, we see that V ( f ) = V ( g ) = V ( g ). Now V (a) conssts of all prme deals that contan a, and the radcal of a s the ntersecton of all the prme deals that contan a. Thus f = g. But then, n partcular, f n s a fnte lnear combnaton of the g and the correspondng open sets cover U f. 3
As ponted out above, we need a slghtly more general noton of a functon than the one gven above: Defnton 11.8. Let R be a rng. We defne a sheaf of rngs O X on the spectrum of R as follows. Let U be any open set of X. A secton σ of O X (U) s by defnton any functon s: U p U R p, where s(p) R p, whch s locally represented by a quotent. More precsely, gven a pont q U, there s an element f R such that U f U and such that the secton σ U s represented by a/f n, for some a R and n N. An affne scheme s then any locally rnged space somorphc to the spectrum of a rng wth ts assocated sheaf. A scheme s a locally rnged space, whch s locally somorphc, as locally rnged space, to an affne scheme. It s not hard to see that O X (U) s a rng (sums and products are defned n the obvous way) and that we do n fact have a sheaf rather than just a presheaf. The key result s the followng: Lemma 11.9. Let X be an affne scheme, somorphc to the the spectrum of R and let f R. (1) For any p X, the stalk O X,p s somorphc to the local rng R p. (2) The rng O X (U f ) s somorphc to R f. In partcular O X (X) R. Proof. We frst prove (1). There s an obvous rng homomorphsm O X,p R p, whch just sends a germ (g, U) to ts value g(p) at p. On the other hand, there s an obvous rng homomorphsm, R O X,p, whch sends an element r R to the par (r, X). Suppose that f / p. Then (1/f, U f ) defnes an element of O X,p, and ths element s an nverse of (f, X). It follows, by the unversal property of the localsaton, that there s a rng homomorphsm, R p O X,p, whch s the nverse map. Hence (1). 4
Now we turn to the proof of (2). As before there s an obvous rng homomorphsm, R O X (U f ), whch nduces a rng homomorphsm R f O X (U f ). We have to show that ths map s an somorphsm. We frst consder njectvty. Suppose that a/f n R f s sent to zero. Then for every p Spec R, f / p, the mage of a/f n s equal to zero n R p. For each such prme p there s an element h / p such that ha = 0 n R. Let a be the annhlator of a n R. Then h a and h / p, so that a s not a subset of p. Snce ths holds for every p U f, t follows that V (a) U f =. But then f a so that f l a, for some l. It follows that f l a = 0, so that a/f n s zero n R f. Thus the map s njectve. Now consder surjectvty. Pck s O X (U f ). By assumpton, we may cover U f by open sets V such that s s represented by a /g n on V. Replacng g by g n we may assume that n = 1. By defnton g / p, for every p V, so that V U g. Now snce sets of the form U h form a base for the topology, we may assume that V = U h. As U h U g t follows that V (g ) V (h ) so that h g. But then h n g, so that h n a = c g. In partcular = c a g h n. Replacng h by h n and a by c a, we may assume that U f s covered by U h, and that s s represented by a /h on U h. Now observe that by (11.7), f n = b h, where b 1, b 2,..., b k R and U f can be covered by fntely many of the sets U h. Thus we may assume that we have only fntely many h. Now on U h h j = U h U hj, there are two ways to represent s, one way by a /h and the other by a j /h j. By njectvty, we have a /h = a j /h j n R h h j so that for some n, (h h j ) n (h j a h a j ) = 0. Snce there are only fntely many and j, we may assume that n s ndependent of and j. We may rewrte ths equaton as If we replace h by h n+1 a /h and moreover h n+1 j (h n a ) h n+1 (h n j a j ) = 0. and a by h n a, then s s stll represented by h j a = h a j. 5
Let a = b a, where f n = b h. Then for each j, h j a = b a h j = b h a j = f n a j. But then a/f n = a j /h j on U hj. But then a/f n represents s on the whole of U f. Note that by (2) of (11.9), we have acheved our am of constructng a topologcal space from an arbtrary rng R, whch realses R as a natural subset of the contnuous functons. Defnton 11.10. A morphsm of schemes s smply a morphsm between two locally rnged spaces whch are schemes. The gves us a category, the category of schemes. Note that the category of schemes contans the category of affne schemes as a full subcategory and that the category of schemes s a full subcategory of the category of locally rnged spaces. Theorem 11.11. There s an equvalence of categores between the category of affne schemes and the category of commutatve rngs wth unty. Proof. Let F be the functor that assocates to an affne scheme, the global sectons of the structure sheaf. Gven a morphsm (f, f # ): (X, O X ) = Spec B (Y, O Y ) = Spec A, of locally rnged spaces then let φ: A B, be the nduced map on global sectons. It s clear that F s then a contravarant functor and F s essentally surjectve by (11.9). Now suppose that φ: A B s a rng homomorphsm. We are gong to construct a morphsm (f, f # ): (X, O X ) (Y, O Y ), of locally rnged spaces. Suppose that we are gven p X. Then p s a prme deal of B. But then q = φ 1 (p) s a prme deal of A. Thus we get a functon f : X Y. Now f a s an deal of A, then f 1 (V (a)) = V ( φ(a) ), so that f s certanly contnuous. For each prme deal p of B, there s an nduced morphsm φ p : A φ 1 (p) B p, 6
of local rngs. Now suppose that V Y s an open set. We want to defne a rng homomorphsm f # (V ): O Y (V ) O X (f 1 (V )). Suppose frst that V = U g, where g A. Then O Y (V ) = A g and f 1 (V ) U φ(g). But then there s a restrcton map O X (U φ(g) ) B φ(g) O X (f 1 (V )). On the other hand, composng there s a rng homomorphsm A B φ(g). Snce the mage of g s nvertble, by the unversal property of the localsaton, there s an nduced rng homomorphsm A g B φ(g). Puttng all of ths together, we have defned f # (V ) when V = U g. Snce the sets U g form a base for the topology, and these maps are compatble n the obvous sense, ths defnes a morphsm f # : O Y f O X, of sheaves. Clearly the nduced map on local rngs s gven by φ p, and so (f, f # ) s a morphsm of local rngs. Fnally t suffces to prove that these two assgnments are nverse. The composton one way s clear. If we start wth φ and construct (f, f # ) then we get back φ on global sectons. Conversely suppose that we start wth (f, f # ), and let φ be the map on global sectons. Gven p X, we get a morphsm of local rngs on stalks, whch s compatble wth φ and localsaton, so that we get a commutatve dagram A φ B f # p A f(p) B p. But snce f p # s a morphsm of local rngs, t follows that φ 1 (p) = f(p), so that f concdes wth the map nduced by φ. But then f # s also the map nduced by φ. Defnton 11.12. Let X be a scheme and let x X be a pont of X. The resdue feld of X at x s the quotent of O X,x by ts maxmal deal. We recall some basc facts about valuatons and valuaton rngs. 7
Defnton 11.13. Let K be a feld and let G be a totally ordered abelan group. A valuaton of K wth values n G, s a map ν : K {0} G, such that for all x and y K {0} we have: (1) ν(xy) = ν(x) + ν(y). (2) ν(x + y) mn(ν(x), ν(y)). Defnton-Lemma 11.14. If ν s a valuaton, then the set R = { x K ν(x) 0 } {0}, s a subrng of K, whch s called the valuaton rng of ν. The set m = { x K ν(x) > 0 } {0}, s an deal n R and the par (R, m) s a local rng. Proof. Easy check. Defnton 11.15. A valuaton s called a dscrete valuaton f G = Z and ν s surjectve. The correspondng valuaton rng s called a dscrete valuaton rng. Any element t R such that ν(t) = 1 s called a unformsng parameter. Lemma 11.16. Let R be an ntegral doman, whch s not a feld. The followng are equvalent: R s a DVR. R s a local rng and a PID. Proof. Suppose that R s a DVR. Then R s certanly a local rng. Suppose that a and b R and ν(a) = ν(b). Then ν(b/a) = ν(b) ν(a) = 0 and so a = b. It follows that the deals of R are of the form I k = { a R ν(a) k }. As ν s surjectve, there s an element t R such that ν(t) = 1. Then I k = t k = m k. Thus R s a PID. Now suppose that R s a local rng and a PID. Let m be the unque maxmal deal. As R s a PID, m = t, for some t R. Defne a map ν : K Z, by sendng a to k, where a m k m k+1 and extendng ths to any fracton a/b n the obvous way. It s easy to check that ν s a valuaton and that R s the valuaton rng. 8
There are two key examples of a DVR. Frst let k be feld and let R = k[t] t. Then R s a local rng and a PID so that R s a DVR. t s a unformsng parameter. Note that R s the stalk of the struture shear of the affne lne at the orgn. Now let = { z C z < 1 }, be the unt dsc n the complex plane. Then the stalk O,0 of the sheaf of holomorphc functons s a local rng. The order of vanshng realses ths rng as a DVR. z s a unformsng parameter. In fact f C s a smooth algebrac curve, an algebrac varety of dmenson one, then O C,p s a DVR. Example 11.17. Let R be the local rng of a curve over an algebracally closed feld (or more generally a dscrete valuaton rng). Then Spec R conssts of two ponts; the maxmal deal, and the zero deal. The frst t 0 s closed and has resdue feld the groundfeld k of C, the second t 0 has resdue feld the quotent rng K of R, and ts closure s the whole of X. The ncluson map R K corresponds to a morphsm whch sends the unque pont of Spec K to t 1. There s another morphsm of rnged spaces whch sends the unque pont of Spec K to t 0 and uses the ncluson above to defne the map on structure sheaves. Snce there s only one way to map R to K, ths does not come from a map on rngs. In fact the second map s not a morphsm of locally rnged spaces, and so t s not a morphsm of schemes. Example 11.18. It s nterestng to see an example of an affne scheme, n a seemngly esoterc case. Consder the case of a number feld k (that s a fnte extenson of Q, wth ts rng of ntegers A k (that s the ntegral closure of Z nsde k). As a partcular example, take k = Q( 3). Then A = Z Z 3. The pcture s very smlar to the case of Z. There are nfntely many maxmal deals, and only one pont whch s not closed, the zero deal. Moreover, as there s a natural rng homomorphsm Z A, by our equvalence of categores, there s an nduced morphsm of schemes Spec A Spec Z. We nvestgate ths map. Consder the fbre over a pont p Spec Z. Ths s just the set of prmes n A contanng the deal pa. It s well known by number theorsts, that three thngs can happen: (1) If p dvdes the dscrmnant of k/q (whch n ths case s 12), that s p = 2 or 3, then the deal p s a square n A. 2 A = 1 + 3 2, 9
and 3 A = 3 2. (2) If 3 s a square modulo p, the prme p factors nto a product of dstnct prmes, 11 A = 4 + 3 3 4 3 3, or 13 A = 4 + 3 4 3, (3) If p > 3 and 3 s not a square mod p (e.g p = 5 and 7), the deal p s prme n A. Defnton 11.19. Let C be a category and let X be an object of C. Let D = C X be the category whose objects consst of pars f : Y X, where f s a morphsm of C, and whose morphsms, consst of commutatve dagrams Y f φ Z X. D s known as the category over X. If X s a scheme, then a scheme over X s exactly an object of the category of schemes over X. Let R be a rng. Affne n-space over R, denoted A n R, s the spectrum of the polynomal rng R[x 1, x 2,..., x n ]. One of the key deas of schemes, s to work over arbtrary bases. Note that snce there s an ncluson R R[x 1, x 2,..., x n ] of rngs, affne space over R s a scheme over Spec R. Thus we may defne affne space over any affne scheme. g 10