ALGEBRA SCHEMES AND THEIR REPRESENTATIONS

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1 ALGEBRA SCHEMES AND THEIR REPRESENTATIONS AMELIA ÁLVAREZ, CARLOS SANCHO, AND PEDRO SANCHO Introducton The equvalence (Carter dualty) between the category of topologcally flat formal k-groups and the category of flat balgebras has been treated as a dualty of contnuous vector spaces (of functons) [G, Exposé VII B by P. Gabrel, 2.2.1]. Ths s owng to the fact that the reflexvty of vector spaces of nfnte dmenson does not hold f one does not provde them wth a certan topology and does not consder the contnuous dual. In ths paper we obtan ths dualty wthout provdng the vector spaces of functons wth a topology. Let R be a commutatve rng wth unt. It s natural to consder R-modules as R-module functors n the followng way: f E s an R-module, let E be the R- module functor defned by E(B) := E R B for every R-algebra B whch belongs to the category C R of R-algebras. Now, f F s a functor of R-modules, ts dual F can be defned n a natural way as the functor of R-modules defned F (B) := Hom B (F B, B). In ths work we wll prove that the functor defned by an R-module s reflexve: E E, even n the case of R beng a rng. We call the functors E R-module schemes and f they are R-algebra functors too, we wll say they are R-algebra schemes. In secton 2 we study and characterze the vector space schemes (2.3, 2.17) and we characterze when the module scheme closure of an R-module functor F s equal to F (2.8, 2.9). P. Gabrel [G, Exposé VII B, 1.3.5] proved that the category of topologcally flat formal R-varetes s equvalent to to the category of flat cocommutatve R- coalgebras, where R s a pseudocompact rng. We prove (4.2) that the category of R-algebra schemes s equvalent to the category of R-coalgebras, where R s a rng. From ths perspectve, on the theory of algebrac groups and ther representatons R-module schemes appear n a necessary way, as also do R-algebra schemes as lnear envelopes of groups. Let G = Spec A be an R-group and let G be the functor of ponts of G,.e., G (B) = Hom R schemes (Spec B, G) for all B C R, and let R[G ] be the lnear envelope of G (see secton 3). We prove that the R-algebra scheme closure of R[G ] s the R-algebra scheme A (3.3, 5.4) and the category of G-modules s equal to the category of A -modules (5.5). So, the theory of lnear representatons of a group G = Spec A s a partcular case of the theory of A - modules (5.7, 5.8, 6.4, etc). Moreover, there s a bjectve correspondence between the R-ratonal ponts of A and the multplcatve characters of G (5.6). When R s an algebracally closed feld and G s smooth we prove that the completon of R[G ] by ts deal functors of fnte codmenson s also A (3.5, 5.9). Date: October,

2 2 AMELIA ÁLVAREZ, CARLOS SANCHO, AND PEDRO SANCHO Fnally we prove that every R-algebra scheme A s an nverse lmt of fnte R- algebra schemes (4.12). We characterze the separable algebra schemes (7.4) and we prove the theorem of Wedderburn-Malcev (8.8) n the context of algebra schemes. Ths paper s essentally self-contaned. 1. R-module schemes. Reflexvty theorem. Let R be a commutatve rng wth unt, let C R be the category of commutatve R-algebras and let R : C R C R be the algebra functor that assgns the R-algebra R(A) := A to A. Let C Ab be the category of commutatve groups. Defnton 1.1. A functor F : C R C Ab wth a morphsm of functors R F θ F s sad to be an R-module functor f F (A) wth the morphsm A F (A) θ(a) F (A) s an A-module for each A C R. Gven an R-module E, the functor E defned by E(A) := E R A s an R-module functor. Unless otherwse stated, we assume that all functors consdered n ths artcle are functors from the category C R to another one. Defnton 1.2. Gven a par of R-module functors F and F, we denote by Hom R (F, F ) the functor of all R-lnear morphsms from F to F,.e., Hom R (F, F )(A) = Hom A (F A, F A ) where F A denotes the functor F restrcted to the category of commutatve A-algebras C A. An element of Hom A (F A, F A ) conssts of assgnng a morphsm of B-modules F (B) F (B) to each A-algebra B. We denote by F = Hom R (F, R) the dual functor of F 1. Proposton 1.3. For every R-module functor F and every R-module E, t holds that Hom R (E, F ) = Hom R (E, F (R)) Proof. Gven an R-lnear morphsm f : E F, we have for every R-algebra A a morphsm of A-modules f A : E R A F (A) and a commutatve dagram E R A E f A f R F (A) F (R) Hence, the morphsm of A-modules f A s determned by f R. Lemma 1.4. Let A be an R-algebra, let E be an R-module and let F, F be R- module functors. Then (1) E A s the functor assocated to E R A on C A. (2) Hom R (F, F ) A = Hom A (F A, F A ). 1 If F = E or F = E then Hom R (F, F )(A) s a set (see 1.3, 1.6 and Yoneda s lemma) and Hom R (F, F ) s a functor. When we wrte F or F we wll suppose that they are welldefned functors. However, for any F and F, n order for Hom R (F A, F A) to be a set (t wll be necessary n 2.2, 2.3 and 8.1), nstead of takng nto account the category of commutatve algebras, we consder an nfnte set X and the category of commutatve algebras whose cardnal s less than or equal to card(x N ). See [D, General conventons].

3 ALGEBRA SCHEMES AND THEIR REPRESENTATIONS 3 Defnton 1.5. Gven a commutatve R-algebra A, we defne the functor (Spec A) to be (Spec A) (B) = Hom R algebras (A, B) for each commutatve R-algebra B. Ths functor wll be called the functor of ponts of Spec A. By Yoneda s lemma (see [E, Appendx A5.3]), Hom functors ((Spec A), F ) = F (A). Gven an R-module E, we wll denote by S R E the symmetrc algebra of E. Let us recall the next well-known lemma (see [D, II, 1, 2.1] or [G, Exposé VII B, 1.2.4]). Lemma 1.6. If E s an R-module, then E = (Spec S R E) as R-module functors. Proof. For every R-algebra A, t holds that E (A) = Hom A (E A, R A ) 1.4 = Hom A (E R A, A) 1.3 = Hom A (E R A, A) = = Hom R (E, A) = Hom R algebras (S RE, A) = (Spec S RE) (A) Defnton 1.7. The tensoral product of two functors F, G n the category of R-module functors s defned to be (F R G)(A) = F (A) A G(A). Proposton 1.8. Let E, E be R-modules. Then Hom R (E, E ) = E R E Proof. We know that E s represented by Spec S R E, therefore Hom R (E, E ) Hom functors (E, E ) = E (S RE) = S RE R E However, n order for w S R E R E to be a lnear applcaton, t must be w E R E. Hence, Hom R (E, E ) = E R E. For every R-algebra A, we have that Hom R (E, E )(A) = Hom A (E A, E A) = Hom A ((E R A), E R A) = (E R A) A (E R A) = (E R E )(A) Remark 1.9. If C R s the category of commutatve R-algebras whose cardnal s less than or equal to card (X N ) = m, then we have to suppose that S R E C R,.e., that card S R E m. If E s a free R-module of whchever cardnal, we obtan the proposton agan: Let {E } be the set of quotents of E whose cardnal s less than or equal to m. It s easly seen that E = lm E. Then Hom R (E, E ) = Hom R (lm E, E ) = lm (E R E ) = E R E where = s a consequence of the equalty E R E R S = lm (E R E R S) for every R-algebra S C R. Even more, we can assume E s a projectve R-module,.e., a drect sum of a free R-module. As a corollary we obtan the followng Theorem Let E be an R-module. Then E = E

4 4 AMELIA ÁLVAREZ, CARLOS SANCHO, AND PEDRO SANCHO Defnton Quas-coherent R-modules are defned to be R-module functors of the type E, where E s any R-module. We shall say that E s a coherent R-module f E s a fntely generated R-module. R-module schemes are defned to be R-module functors of the type E. If E s a free fntely generated R-module then E s a quas-coherent R-module and an R-module scheme. Theorem The category of quas-coherent modules over R s equvalent to the category of R-modules. The category of quas-coherent modules over R s antequvalent to the category of R-module schemes (the correspondence s establshed by takng the dual functor). In [G, Exposé VII B, 1.2.3], the ant-equvalence between the category of flat R- modules and the category of projectve pseudocompact R-modules s establshed, where R s a (commutatve) pseudocompact rng. Proposton The R-lnear morphsm E E s surjectve, n the category of R-module functors, f and only f the morphsm E E s njectve, n the category of R-module functors. Proof. It follows mmedately that f the morphsm E E s surjectve, then the morphsm E E s njectve. Inversely, let us suppose the morphsm E E s njectve. If V s the cokernel of the morphsm E E, we obtan V = 0. Hence V = V = 0 and the morphsm E E s surjectve. If a morphsm E E s surjectve then the assocated morphsm E E s njectve and t has a retracton. Let us consder the R-algebra A := R E, where e 1 e 2 = 0 for all e 1, e 2 E. Let w E (A) = Hom R (E, A) be defned by w(e) := e. Then, there exsts a w Hom R (E, A) such that w (e) = e for all e E. If π : A E s the natural projecton, then π w s a retracton of the morphsm E E. Let us recall the Formula of adjont functors. Defnton Let us consder the ncluson of categores C R = {commutatve R-algebras} C A = {commutatve A-algebras} where A s an R-algebra. Gven a functor G on C A we defne ( G)(B) := G(A R B) for each object B of C R. Gven a functor F on C R we defne ( F )(B) := F (B) for each object B of C A. Let us gve a drect proof of the followng theorem, although t can be obtaned from [B, 8.4,8.5] after many precsons and complcated techncal terms. Theorem 1.15 (Formula of adjont functors). Let F be an R-module functor and let G be an A-module functor. Then t holds that Hom A ( F, G) = Hom R (F, G) Proof. Gven a w Hom A ( F, G), we have morphsms w A B : F (A B ) G(A B ) for each R-algebra B. By composton wth the morphsms F (B ) F (A B ), we have the morphsms φ B : F (B ) G(A B ) = G(B ), whch n ther turn defne φ Hom R (F, G). Gven a φ Hom R (F, G), we have morphsms φ B : F (B ) G(B ) = G(A B ) for each A-algebra B. By composton wth the morphsms G(A B )

5 ALGEBRA SCHEMES AND THEIR REPRESENTATIONS 5 G(B ), we have the morphsms w B : F (B ) G(B ), whch n ther turn defne w Hom A ( F, G). Now we shall show that w φ and φ w are mutually nverse. Gven w Hom A ( F, G) we have φ Hom R (F, G). Let us prove that the latter defnes w agan. We have the followng dagram, where B s an A-algebra, F (B ) F (A B ) F (B ) w A B G(A B ) w B p G(B ) The composte morphsm p w A B = p φ B s that assgned to φ, and concdes wth w B snce the whole dagram s commutatve. Gven φ Hom k (F, G) we have w Hom A ( F, G). Let us see that the latter defnes φ. We have the followng dagram, where B s an k-algebra, F (B ) φ B ( G)(B ) r j F (A B ) φ A B w A B G(A B ) ( G)(A B ) G(A A B ) The composte morphsm w A B r assgned to w agrees wth φ B, snce p j = Id and the whole dagram s commutatve. For smplcty of notaton, gven a functor F we wll sometmes wrte w F nstead of w F (A). Proposton Let F be k-vector space functors and let E be a k-vector space. It holds that Hom k ( F, E) = Hom k (F, E) p Proof. From the njectve morphsm F j F one obtans the morphsm j : Hom k ( F, E) Hom k ( F, E) = Hom k (F, E) The am s to prove that ths morphsm s njectve and ts mage s Hom k (F, E). For the frst queston, let w Hom k ( F, E) be a lnear form such that w 0 but w F = 0. Then there exsts a k-algebra A and elements f F (A) such that w((f ) ) 0, and composng wth the morphsms φ : A F A, φ((a ) ) = (a f ) we get a lnear form w φ Hom A ( A, E A) that s not null but s null on A, whch s mpossble snce Hom A ( A) A (E A) = E A. ( A, E A) = Hom A (( A), E A) 1.8 =

6 6 AMELIA ÁLVAREZ, CARLOS SANCHO, AND PEDRO SANCHO To prove that Imj = Hom k (F, E) t s enough to prove that Hom k ( F, E) = Hom k (F, E), because n that case we wll have Hom A ( F A, E A ) 1.15 = Hom k ( 1.15 = F, E A) = Hom k (F, E A) Hom A (F A, E A ) Gven a lnear form w Hom k ( F, E) we have to prove that there exsts at most a fnte subset of ndces such that w F 0. Let us suppose that ths s not true,.e., that there exsts a set of ndces n, where n N, and k-algebras A n such that w(f n ) 0 for some f n F n (A n ). Let A = A n and denote by h m : A m n A n the natural njectons and by f m the mage of f m by the nduced morphsm n F m (h m ) : F m (A m ) F m ( n A n ). It s easy to see that w( f m ) = h m (w(f m )) 0. Therefore, we get a lnear form w : n A E A, w((a n ) n ) := w((a n fn ) n ), whch s not null on any factor A n A. Agan ths contradcts the fact that Hom A ( n A, E A) = n E A. 2. Characterzatons of vector space schemes. Let R be a commutatve rng wth unt and let k be a commutatve feld. Defnton 2.1. An R-module functor F s sad to be reflexve f F = F. Theorem 2.2. If F s a reflexve functor of k-vector spaces such that Hom k (F, ) commutes wth drect sums,.e., Hom k (F, F ) = Hom k (F, F ) I I for all k-vector space functors F, then F s a k-vector space scheme. Proof. From the adjont functor formula, we have that Hom k (F, A) = Hom k (F, k) = Hom A (F A, A) = F (A) However, A = k and the property that F satsfes by hypothess means that I Hom k (F, A) = F (k) A. Hence, F (A) = F (k) k A and F = E, where E = F (k), and therefore F = F = E. We can now rephrase ths result n terms of drect lmts. The defnton that we work wth s taken from [E, Appendx 6]. Theorem 2.3. Let F be a reflexve k-vector space functor. The functor on the category of quas-coherent k-vector spaces, Hom k (F, ), commutes wth drect lmts f and only f F s a k-vector space scheme. Proof. The necessary condton s a consequence of the prevous theorem, snce t was only necessary that Hom k (F, ) commuted wth drect sums of quas-coherent vector spaces for F to be a k-vector space scheme.

7 ALGEBRA SCHEMES AND THEIR REPRESENTATIONS 7 The suffcent condton s obtaned as an mmedate consequence of Proposton 1.8, snce the functor lm E s agan a quas-coherent k-vector space and I Hom k (F, lm E ) 1.8 = F (lm I I E ) = lm (F E ) 1.8 = lm Hom k (F, E ) I I Defnton 2.4. Gven an R-module functor F, we shall say that F s the R-module scheme closure of F f F s an R-module scheme and Hom R (F, V ) = Hom R ( F, V ) for every R-module V. As F s defned to be the representant on the category of R-module schemes of the functor Hom R (F, ) t s unque up to somorphsms, and there exsts a canoncal morphsm F F correspondng to the dentty morphsm F F. Notaton 2.5. We mean by F (R) the quas-coherent R-module correspondng to the R-module F (R),.e., F (R)(A) = F (R) R A. Lemma 2.6. Let F, G be functors of R-modules. Then Proof. Hom R (F, G ) = Hom R (G, F ) Hom R (F, G ) = Hom R (F R G, R) = Hom R (G, F ) Proposton 2.7. Let F be an R-module functor. It holds that F = F (R). Proof. Hom R (F, V ) 2.6 = Hom R (V, F ) 1.3 = Hom R (V, F (R)) 1.3 = Hom R (V, F (R)) 2.6 = Hom R (F (R), V ) Unfortunately, the R-module scheme closure of an R-module functor F s not stable under base change. Proposton 2.8. Let F be an R-module functor. If F s a quas-coherent R- module then F = F and F = F. Proof. If F s a quas-coherent R-module then Hom R (F, E ) = Hom R (E, F ) = Hom R (F, E ) Therefore F = F. Moreover, F = (F ) = (F ) = F. Proposton 2.9. The R-module scheme closure of an R-module functor F s stable under base change f and only f F s a quas-coherent R-module. Proof. If F A = F A, then takng Hom A (, A) we obtan that F (A) = F (R) R A. Inversely, f F s quas-coherent then F A = F A = F A = F A.

8 8 AMELIA ÁLVAREZ, CARLOS SANCHO, AND PEDRO SANCHO Example If F 1,..., F n are R-module functors whose duals are quas-coherent R-modules, then (F 1... F n ) = F 1... F n, whch n partcular s a quascoherent R-module: Hom R (F 1... F n, R) = Hom R (F 1... F n 1, F n) = Hom R (F 1... F n 2, Hom R (F n 1, F n)) = Hom R (F 1... F n 2, Hom R (F n, F n 1)) 1.8 = Hom R (F 1... F n 2, F n 1 F n) =... = F 1... F n Hence, F 1... F n = (F 1... F n ) = (F 1... F n) and F 1... F n = F 1... F n. If we denote by the tensoral product n the category of R-module schemes then E 1 E 2 = E 1 E 2 = (E 1 E 2 ). Moreover, commutes wth nverse lmts: (lm E ) E = (lm E ) E = ((lm E ) E) = (lm (E E)) = lm (E E) = lm (E E ) Henceforward, we shall only work wth functors of k-vector spaces. Proposton The morphsm F F s njectve f and only f the morphsm F F s njectve. Proof. Let us prove the necessary condton. Gven s F (A) such that s = 0 n F (A) = F (k) (A) = Hom A (F (k) A, A) 1.15 = Hom k (F (k), A) 1.3 = Hom k (F (k), A), then s(w) := w(s) = 0 for all w F (k). Gven a k-algebra B, f one wrtes B = k e, one notces that F (B) = Hom B (F B, B) 1.15 = Hom k (F, B) = Hom k (F, k) Hom k (F, k) whch assgns to every w B F (B) a (w ) F (k). Explctly, gven s F (B), then w B (s) = w (s) e. Therefore w B (s) = 0 for all w B F (B). Snce the morphsm F F s njectve, ths means that s = 0,.e., the morphsm F F s njectve. For the suffcent condton, we consder the morphsm F (k) F, whch, by takng duals, becomes F F (k). Snce the composte morphsm F F F = F (k) s njectve, so s the morphsm F F. Lastly, we wll show another characterzaton of k-vector space schemes by means of complete reflexve functors. Frst, we need some techncal results before Defnton Proposton (1) The morphsm F (k) F (k) s njectve f and only f the morphsm F F(k) s njectve. (2) The morphsm F (k) F (k) s njectve f and only f for every quascoherent vector space E the mage of any k-lnear morphsm F E s a quas-coherent subspace of E. Proof.

9 ALGEBRA SCHEMES AND THEIR REPRESENTATIONS 9 (1) If the morphsm F F(k) s njectve then takng sectons on k the morphsm F (k) F (k) s njectve. Inversely, from the commutatve dagram Hom k (F, k) Hom k (F (k), k) Homk (F, k) Hom k (F (k), k) one has that Hom k (F, k) Hom k (F (k), k). Snce A = k, then F (A) = Hom A (F A, A) 1.15 = Hom k (F, k) Hom k (F (k), k) = Hom A (F (k) k A, A) 1.3 = Hom A (F(k) k A, A) 1.4 = Hom A (F(k) A, A) = Hom k (F(k), k)(a) = F(k) (A).e., the morphsm F F(k) s njectve. (2) Let us suppose that the mage of any morphsm F E s a quas-coherent subspace of E. Gven w F (k),.e., a morphsm w : F k, Im w s equal to the quas-coherent vector space assocated to w(f (k)). Hence f w(f (k)) = 0 then w = 0. Inversely, let E be the mage of the morphsm F (k) E and consder F V := E/E. The morphsm V F(k) s null. Hence the morphsm V F s null, and the composte morphsm F F V = V s null. Therefore, the mage of the morphsm F E s E. Corollary Let F be a reflexve functor and let E be a k-vector space. Then the mage of any k-lnear morphsm F E s a quas-coherent subspace of E. Proof. If F = F, by Proposton 2.11 the morphsm F F = F(k) s njectve. Then by Proposton 2.12 the proof s complete. Defnton Gven a k-vector space functor F such that the mage of any k-lnear morphsm F E s a quas-coherent subspace of E, let us consder the k-vector space subfunctors F F such that F/F are coherent k-vector spaces. Then we defne ˆF := lm F/F. The drect lmt of quas-coherent vector spaces, n the category of k-vector space functors, s a quas-coherent vector space. Therefore, the nverse lmt of k-vector space schemes s a k-vector space scheme. Hence ˆF s a k-vector space scheme, namely, ˆF := (lm (F/F ) ). Proposton Let E be a k-vector space. Then, E s complete and separate,.e., Ê = E. Proof. By the reflexvty theorem, the coherent cokernels of E correspond to the subspaces E E of fnte dmenson. Hence, Ê = lm (E ) = ( lm E ) = E dm k E < dm k E <

10 10 AMELIA ÁLVAREZ, CARLOS SANCHO, AND PEDRO SANCHO Proposton Let F be a k-vector space functor such that the mage of any k-lnear morphsm F E s a quas-coherent subspace of E. Then the vector space closure of F s equal to the completon of F,.e., ˆF = F. In partcular, ˆF = E, where E = F (k), and ˆF s complete, separate, and reflexve. Proof. Frst, let us suppose that E s a fnte-dmensonal space. Observe that the dual of an nverse lmt of k-vector space schemes s equal to the drect lmt of the quas-coherent dual vector spaces, (lm V ) = (lm V ) = lm V, then Hom k ( ˆF, E ) = Hom k (lm F/F, E ) = lm Hom k (F/F, E ) = Hom k (F, E ) In general, E = lm E, where dm ke <. Then Hom k ( ˆF, E ) = Hom k ( ˆF, lm E ) = lm Hom k ( ˆF, E ) = lm Hom k (F, E ) Therefore, ˆF = F. = Hom k (F, E ) Theorem Let F be a reflexve k-vector space functor. Then F s a k-vector space scheme f and only f F s complete and separate. 3. Lnearzatons of varetes. Closure of lnearzatons. Defnton 3.1. Let X = Spec A be an affne R-scheme and let us denote by X the functor of ponts of X,.e., X (B) = Hom R algebras (A, B). Let R[X ] be the R-module functor defned by R[X ](B) := B = { the formal fnte B-lnear combnatons of ponts of X n B }. X (B) It s clear that for every R-module functor F t holds that Hom R (R[X ], F ) = Hom functors (X, F ) Snce every morphsm of R-algebras A B s n partcular R-lnear, we have a morphsm of functors φ : X A, where the morphsm between schemes s gven by the natural epmorphsm of R-algebras S R A A. Then we have a morphsm R[X ] A. Notaton 3.2. It s usual the notaton X B = Spec A R Spec B = Spec(A R B) and A B = A R B. Theorem 3.3. Let X = Spec A be an affne R-scheme. It holds that (1) R[X ] = A (2) R[X ] = R[X ] = A Proof. R[X ] (R) = Hom R (R[X ], R) = Hom functors (X, R) = A and lkewse R[X ] (B) = Hom B (R[X ] B, B) = Hom B (B[X B], B) = A B = A(B) Hence, R[X ] = A and takng duals A 2.8 = R[X ] = R[X ].

11 ALGEBRA SCHEMES AND THEIR REPRESENTATIONS 11 Theorem 3.4. If X = Spec A s an R-scheme and F s a reflexve R-module functor, then the morphsm Hom R (A, F ) Hom functors (X, F ) A φ F X A F s an somorphsm. Moreover, f A s a free R-module such lnear applcatons of functors are determned each by ts value on global sectons,.e., Proof. Frstly, we have Hom R (A, F ) Hom R (A, F (R)) Hom R (A, F ) 2.6 = Hom R (F, A) 3.3 = Hom R (F, R[X ] ) whch s the somorphsm to compose wth φ. Secondly, snce A = R R we get 2.6 = Hom R (R[X ], F ) = Hom functors (X, F ) Hom R (A, F ) 2.6 = Hom R (F, A) Hom R (F, R) 2.6 = Hom R ( R, F ) 1.3 = Hom R ( R, F (R)) Snce the njectve morphsm Hom R (A, F ) Hom R ( R, F (R)) factorzes through Hom R (A, F (R)), the morphsm Hom R (A, F ) Hom R (A, F (R)) s njectve. Theorem 3.5. Let us suppose that the only functon a A of the k-scheme X = Spec A that s null on every k-ratonal pont s the zero functon a = 0. Then, t holds that k[x ] = A. Proof. By hypothess the morphsm k[x ] (k) = A k[x ](k) s njectve, hence we are under the hypothess of Defnton 2.14 and Proposton Therefore, by Proposton 2.16 k[x ] = k[x ] 3.3 = A. Maybe t s more natural the defnton R[X ] (B) := < Hom R algebras (A, B) > B Hom R (A, B).e., R[X ] s the mage of R[X ] n A. Proposton 3.6. It holds Proof. (1) Hom R (R[X ], F ) = Hom functors (X, F ) for every reflexve functor F. (2) R[X ] = A. (3) R[X ] = A. (4) The mnmum reflexve subfunctor of A that contans R[X ] s A. (1) It s a consequence of the equaltes Hom R (A, F ) 3.4 = Hom functors (X, F ) = Hom R (R[X ], F ) (2),(3) They are consequences of (1).

12 12 AMELIA ÁLVAREZ, CARLOS SANCHO, AND PEDRO SANCHO (4) Let us suppose we have morphsms R[X ] F A, where F s a reflexve functor. Takng double duals, we obtan that the composte morphsm A F A s the dentty morphsm. Therefore, the morphsm F A s surjectve and (4) follows. 4. Algebra schemes. Defnton 4.1. We call an R-module scheme A an R-algebra scheme f t s also an R-algebra functor (.e., A (B) s a B-algebra and the morphsms A (B) A (B ) are morphsms of B-algebras for every morphsm B B of R-algebras). Proposton 4.2. The category of coalgebras wth count, C coalg, s ant-equvalent to the category of algebra schemes, C alg. The functors whch gve the equvalence are C coalg C alg, B B and C alg C coalg, A A. Proof. Observe that Hom R (E 1... E n, V ) 2.6 = Hom R (V, (E 1... E n) ) 2.10 = Hom R (V, E 1... E n ). Gvng an R-algebra functor structure on a scheme A s equvalent to gvng the morphsm of multplcaton A A A and the unt R A, so that the dagrams that state dstrbutve, assocatve and the lke propertes are commutatve. Ths s equvalent to gvng morphsms A A A, A R whch endow A wth a coalgebra structure wth count. Notaton 4.3. From now on, n ths and next sectons, A denotes an R-algebra scheme. Defnton 4.4. Let F be an R-module functor and let A be an R-algebra functor. We say that F s an A-module f there exsts a morphsm of R-algebra functors A End R (F ). We wll say that an R-module E s an A-module f E s an A-module. Gvng a structure of A -module on E s equvalent to the exstence of a morphsm A E E verfyng the obvous propertes, whch s equvalent to the exstence of a morphsm E A E verfyng the obvous propertes, snce Hom R (A, Hom R (E, E)) = Hom R (A E, E) = Hom R (E, Hom R (A, E)) = Hom R (E, A E) = Hom R (E, A E) By these equvalences, f we have the morphsm E A E, e a e, then w e = w(a )e gven w A. If w s the general lnear form,.e., w = Id A (A) = Hom R (A, A), then w e = a e. If A s an algebra scheme, then A s n a natural way a rght and left A -module as t follows: (w a)(w ) := a(w w) (a w)(w ) := a(w w ) where a A, w, w A. We shall say that A s the regular A -module of A. Gven an R-submodule E E we wll say that E E s a cuascoherent submodule.

13 ALGEBRA SCHEMES AND THEIR REPRESENTATIONS 13 Lemma 4.5. Let E 1,..., E n be projectve R-modules and let E 0 be an R-module. The R-lnear morphsm T : E 1 R... R E n 1 R E n E 0 factorzes va an epmorphsm onto a coherent submodule of E 0. Proof. As E 1,..., E n are projectve R-modules, they are drect summands of free modules L 1,..., L n. Then, E s a quotent of L and we can assume that E = L are free modules. By Proposton 1.8, Hom R (E 1 R... R E n, E 0 ) = E 1 R... R E n E 0. Let {e j } be a bass for E j, for every j. Then for every T Hom R (E 1 R... R E n, E 0 ) we can wrte T = e e nn e 1... n 1,..., n where only a fnte number of the elements e 1... n E 0 are not null. It s easy to check that T factorzes va an epmorphsm onto the coherent R-module assocated to E =< e 1... n > 1... n. Proposton 4.6. Let A be an R-algebra scheme, let E be an A -module and let E E be an R-submodule. Let us suppose that A s a projectve R-module. Then, E s an A -submodule of E f and only f E s an A -submodule of E. Proof. Obvously, f E s an A -submodule of E then E s an A -submodule of E. Inversely, let us suppose E s an A -submodule of E and let us consder the natural morphsm of multplcaton A E E. By the prevous lemma wth consderng a free R-module domnatng E, ths morphsm factorzes va a quas-coherent R- submodule of E and t concdes wth the quas-coherent R-module assgned to A E = E, whch proves that E s an A -submodule of E. Proposton 4.7. Let A be an R-algebra scheme and let E be an A -module (respectvely a rght and left A -module). Let us suppose A s a projectve R-module. Every fntely generated R-submodule of E s ncluded n an A -submodule of E (respectvely a rght and left A -module) that s a fntely generated R-submodule. Proof. Gven a fntely generated R-module E E then A E (respectvely A E A ), the obvous mage of the morphsm A E E (respectvely A E A E), s an A -submodule (respectvely a rght and left A -submodule) of E that s a fntely generated R-module. Remark 4.8. In partcular, an A -module E s a k-vector space of fnte dmenson f and only f s a fntely-generated A -module,.e., there exsts an epmorphsm of A -modules A. n.. A E. Defnton 4.9. Let A be an R-algebra scheme. We wll say that a submodule scheme I A s an deal scheme f t s an deal subfunctor. We wll say that I A s a blateral deal scheme f t s a blateral deal subfunctor. The kernel of a morphsm of algebra schemes s a blateral deal scheme. Defnton Gven a fnte R-algebra B, we wll say that B s a coherent R-algebra. Remark Owng to the categoral equvalence between the category of R- modules and the category of quas-coherent R-modules, there s an obvous categoral equvalence between fnte R-algebras and coherent R-algebras.

14 14 AMELIA ÁLVAREZ, CARLOS SANCHO, AND PEDRO SANCHO Proposton Let A be an R-algebra scheme. Let us suppose A s a projectve R-module. Then A s an nverse lmt of quotents B, whch are coherent R-algebras. Proof. A s a drect lmt of ts fntely generated R-submodules M A. Then by Proposton 4.7 t s a drect lmt of ts rght and left A -submodules N that are fntely generated R-modules. The kernels of the morphsms A N are blateral deal schemes I = (A/N ) of A. Let R n N an epmorphsm of R-modules. The composte morphsm A N (Rn ) = R n factorzes va the epmorphsm A B, where B = A /I and t s a fnte R-algebra, by Lemma 4.5. Dually, we obtan the morphsms N B A. Takng drect lmt we obtan the sequence A lm B A. Hence, lm B = A. Dually, lm B = A. 5. Closure of an algebra functor. Defnton 5.1. Let F be an R-algebra functor. We defne F to be the representant on the category of R-algebra schemes, f t exsts, of the functor Hom R alg (F, ). I.e., Hom R alg (F, A ) = Hom R alg ( F, A ) Notaton 5.2. We wll denote by the tensoral product on the category of R- algebra schemes. Then, we have that Hom R alg ( F G, A ) = Hom R alg (F G, A ) Therefore, F G = F G. = Hom R alg (F, A ) Hom R alg (G, A ) = Hom R alg ( F, A ) Hom R alg ( G, A ) = Hom R alg ( F G, A ) Proposton 5.3. If F s an R-algebra functor such that F s a quas-coherent R-module, then F 2.8 = F = F. Moreover, f H s an R-module functor such that G := H s an R-algebra functor, then Hom R alg (F, G) = Hom R alg ( F, G) Proof. By Lemma 2.6, Example 2.10 and Proposton 2.8 t holds for every R- module functor T := S that Hom R (F... F, T ) = Hom R (S, F... F ) = Hom R ( F... F, T ) If we consder T = F, t follows easly that the structure of algebra of F defne a structure of algebra on F. Fnally, f we consder T = G, we see at once that Hom R alg (F, G) = Hom R alg ( F, G). Remark 5.4. In partcular, (1) If G = Spec A s an R-group, then R[G ] = R[G ]. (2) If A and B are R-algebra schemes, then A B = A B.

15 ALGEBRA SCHEMES AND THEIR REPRESENTATIONS 15 Theorem 5.5. Let G = Spec A be an R-group scheme. The category of G-modules s equal to the category of A -modules. Proof. Let E be an R-module. Let us observe that End R (E) = (E E). Therefore, by Proposton 5.3 and Theorem 3.3, (2), Hom R alg (R[G ], End R (E)) = Hom R alg (A, End R (E)). In concluson, endowng E wth a structure of G-module s equvalent to endowng E wth structure of A -module. Defnng a morphsm R[G ] E E s equvalent to defnng a morphsm A E E, because Hom R (R[G ] E, E) = Hom R (A E, E) by Lemma 2.6, snce (R[G ] E) = (A E). Now t s easy to check that Hom G mod (E, E ) = Hom A (E, E ). Proposton 5.6. Let G = Spec A be an R-group and let G m = Aut R (R) End R (R). It holds that Hom R groups (G, G m ) = Hom R alg (A, R) Proposton 5.7. [W, 3.3] Let E be a G-module. Every vector subspace of E of fnte dmenson s ncluded n a G-submodule of E of fnte dmenson. Proof. It s a consequence of Proposton 4.7 Proposton 5.8. [W, 3.4] If G = Spec A s an algebrac group then t s a subgroup of a lnear group Gl n. Proof. Let us consder the natural ncluson G A. By Proposton 4.12 we know that A = lm A s an nverse lmt of fnte quotent k-algebras. By the noetheranty of G, there exsts an ndex such that the morphsm G A s njectve. However, we have the natural njecton A End k (A ), then an njecton G End k (A ). In ths secton, from now on, F wll be an algebra functor such that the mage of any k-lnear morphsm F E s a quas-coherent subspace of E, for example, f F s a reflexve k-vector space functor. Theorem 5.9. F = lm F/F, where {F } s the set of blateral deal subfunctors of F such that F/F s a coherent k-vector space. Proof. Let us denote F = lm F/F. We must proof the functoral expresson Hom k alg (F, A ) = Hom k alg (F, A ) Frst let us suppose that A s a fnte k-algebra scheme. Every morphsm of k-algebra functors F A has as kernel an F, then t factorzes through F/F, then through F. Inversely, let us see that every morphsm F A factorzes through F/F : Hom k (F, A ) = Hom k (lm F/F, A ) = Hom k (A, lm (F/F ) ) = lm Hom k (A, (F/F ) ) = lm Hom k (F/F, A ) where = holds because A s a fnte-dmensonal k-vector space.

16 16 AMELIA ÁLVAREZ, CARLOS SANCHO, AND PEDRO SANCHO In the general case, Hom k alg (F, A ) 4.12 = Hom k alg (F, lm A ) = lm Hom k alg (F, A ) = lm Hom k alg (F, A ) = Hom k alg (F, lm A ) = Hom k alg (F, A ) Proposton Let F be a k-algebra functor. Then, (1) The category of k-coherent F -modules s the same as the category of k- coherent F -modules. (2) The natural morphsm F F s surjectve. Proof. (1) If F F s a blateral deal functor such that F/F s a coherent k-vector space, then the epmorphsm F F/F factorzes through F and hence the morphsm F F/F s surjectve. If F F s a blateral deal functor such that F / F s a coherent k- vector space, then the mage of the morphsm F F / F s an algebra scheme, therefore the nduced morphsm F F / F values on that mage. In concluson, the morphsm F F / F s surjectve. Now (1) follows easly. (2) By the last argument, the composte morphsm F F F / F s surjectve. The nverse lmt of such surjectons s surjectve, because dually the drect lmt of njectons of quas-coherent vector spaces s an njecton. Then the morphsm F F s surjectve. Theorem Let F be a k-algebra functor such that F s a k-algebra functor and F F s a morphsm of k-algebra functors. Then F = F. Proof. The morphsm of k-algebra functors F F factorzes through a morphsm : F F. The morphsm of k-algebra functors F F s a k-lnear morphsm, then t factorzes through a morphsm j : F F. As j : F F s the dentty morphsm on F, j = Id. Then the morphsm j s njectve and, snce t s surjectve by the prevous proposton, ths proves that F = F. Defnton Let F be a k-algebra functor. We call the k-vector space of dstrbutons of fnte support of F, and we denote t by D, the vector subspace D F (k) consstng of lnear 1-forms of F that are null on some blateral deal of F whose cokernel s a coherent k-vector space. By Theorem 5.9, F = D, then F = D. It holds that Hom coalg (B, D) = Hom k alg (D, B ) = Hom k alg (F, B ) for every coalgebra B. Gven a commutatve k-algebra A and a closed pont x Spec A, f we consder t as an deal of A we wll wrte m x for t. Proposton Let A be a commutatve k-algebra of fnte type. It holds that

17 Proof. (1) Ã = ALGEBRA SCHEMES AND THEIR REPRESENTATIONS 17 x Spec max A Â x, where Âx := lm n A/m n x. (2) The natural morphsm D A s surjectve, where D s the k-vector space of the dstrbutons of fnte support of A. (1) If A/I s a fnte k-algebra, then Spec(A/I) correspond to a fnte number of closed ponts of Spec A, {x 1,..., x n }, and there exsts an m N such that (m x1... m xn ) m I. Therefore, Ã = lm A/(m x1... m xn ) m x 1,...,xn,m = lm A/m m x 1... A/m m x n = x 1,...,xn,m x Spec max A (2) The morphsm D A s surjectve f and only f the morphsm A D (k) = D s njectve. By (1) ths morphsm s obvously njectve. Lemma Let φ : F 1 F 2 be a morphsm of vector space functors and let φ : F 1 F 2 be the nduced morphsm on the vector space scheme closure. It holds that Coker φ = Coker φ and φ( F 1 ) s the vector space scheme closure of the mage of F 1 n F 2. Proof. Obvously, φ( F 1 ) s the same as the mnmum vector space subscheme n F 2 that contans the mage of F 1. It follows mmedately from the functoral defnton of Coker and the vector space scheme closure that Coker φ = Coker φ. Notaton In the next proposton, gven F E, we wll denote by F the module scheme closure of F n E. Proposton Let I 1,..., I n A be blateral deal schemes and let E be an A -module. It holds that Proof. (1) I 1 E s a quas-coherent submodule of E. (2) I 1 I 2 E = (I 1 I 2) E. (3) {e E : I 1 e = 0} s a quas-coherent submodule of E. (4) (E I 1... I n) s an A -submodule of E and to take the module scheme closure s stable under base change,.e., gven a morphsm of rngs k B, then (E I 1... I n) B = (E B I 1 B... I n B ). Therefore, (E /(E I 1... I n) ) = (E /(E I 1... I n)). (5) ((E I 1... I r) (I r+1... I n) ) = (E I 1... I n). (1) The mage of the morphsm of A -modules I 1 k E E s a quas-coherent A -submodule and t concdes wth I 1 E. (2) It s enough to prove I 1 I 2 e = (I 1 I 2) e. Let us consder the commutatve dagram I 1 I 2 e E A e E Â x

18 18 AMELIA ÁLVAREZ, CARLOS SANCHO, AND PEDRO SANCHO The mage of the top horzontal morphsm s a vector space of fnte dmenson, then t s closed and t concdes wth I 1 I 2 e. The countermage of I 1 I 2 e by the bottom horzontal morphsm must contan (I I 2). Hence, (I I 2) e I I 2 e and the equalty follows. (3) Let us consder the exact sequence I 1 E E E /E I 1 0. Then the kernel of the morphsm E I 1 E s {e E : I 1 e = 0}. (4) The mage of the morphsm of A -modules E I 1... I n = (E I 1... I n ) E s the same as (E I 1... I n). Therefore, (E I 1... I n) s an A -module and the module scheme closure s stable under base change. (5) It follows from E I 1... I n = E I 1... I r I r+1... I n. Notaton From now on, when we are n the context of algebra schemes and vector spaces, gven a blateral deal scheme I A and a rght A -module E we wll understand by E I ts module scheme closure (E I ) n the vector space scheme E. 6. Maxmal quotent semsmple algebra scheme. Defnton 6.1. We say a k-algebra scheme A s smple f t does not contan any proper blateral deal. We say that an A -module E 0 s smple f t does not contan any proper A -submodule. We say that an A -module E s semsmple f t s a sum of smple A -modules. An A -module E s semsmple f and only f t s a drect sum of smple A - modules. By Proposton 4.7, the smple A -modules are k-vector spaces of fnte dmenson. Theorem 6.2. A s smple f and only f t s somorphc to the endomorphsm rng of a fnte-dmensonal vector space over a non-commutatve feld of fnte degree. Proof. If A s smple, by Proposton 4.12 A s a fnte k-algebra scheme. Now, ths theorem s a consequence of Wedderburn Theorem ([P, 3.5]). Theorem 6.3. Every smple A -module s an A -submodule of the regular module. Every A -module s a submodule of a drect sum of regular modules. Proof. If E s a smple left A -module, therefore of fnte dmenson, then E s a smple rght A -module. Hence, for every w E not null, E = w A. I.e., E s a quotent of A, as a rght A -module. Therefore, E s a submodule of A, as left modules. Let us suppose now that E s not smple. The morphsm of multplcaton E A E s obvously surjectve and t s of rght A -modules, where A acts on E A by the second factor (on the rght). Takng duals we have the desred njecton E A E = A. Corollary 6.4. [W, 3.5] Every smple G-module s a G-submodule of the regular G-module. Every G-module s a G-submodule of a drect sum of regular modules.

19 ALGEBRA SCHEMES AND THEIR REPRESENTATIONS 19 Defnton 6.5. We say that a k-algebra scheme A s a semsmple k-algebra scheme f every quas-coherent A -module s semsmple. Proposton 6.6. A s a semsmple algebra scheme f and only f A s a semsmple A -module. Proof. If A s a semsmple algebra scheme then n partcular A s a semsmple A -module. Inversely, f A s a semsmple A -module, as by Proposton 6.3 every A -module E s a submodule of a drect sum of A s, that s semsmple, we have that E s semsmple. Then A s a semsmple algebra scheme. Defnton 6.7. A blateral deal scheme I A s sad to be a maxmal blateral deal scheme f A /I s smple. We shall call maxmal spectrum of A the set of ts maxmal blateral deal schemes, whch we wll denote by Spec max A. If A = A 1 A 2, then Spec max A = Spec max A 1 Spec max A 2 because every blateral deal scheme I A s I = I 1 I 2, where I s a blateral deal scheme of A. Therefore, every epmorphsm from a product of two k-algebra schemes to a smple k-algebra scheme factorzes through the projecton on one of the two factors. If A, B j are smple k-algebras and φ : A 1... A r B 1... B s s an epmorphsm, then there exst somorphsms φ j : A j B j ( j k, f j k) such that φ(a 1,..., a r ) = (φ 1 (a 1 ),..., φ s (a s )). Theorem 6.8. A s a semsmple k-algebra scheme f and only f t s a drect product of smple k-algebras. Proof. Let us suppose that A s a semsmple algebra scheme. We know that A s an nverse lmt of quotents A whch are fnte k-algebras. Obvously, the A -modules are A -modules, then A s a semsmple algebra. By the theory of semsmple rngs A s a drect product of smple fnte k-algebras, therefore A s a drect product of smple fnte k-algebras. Proposton 6.9. Every A -module E 0 contans an only maxmal semsmple A -submodule not null. Proof. The maxmal semsmple submodule s the sum of every semsmple submodule. As well there exst smple submodules, snce gven 0 e E, ths e s contaned n a fnte-dmensonal A -module, whch contans smple A -submodules. Proposton The dual of the maxmal semsmple submodule of A s the maxmal semsmple quotent algebra scheme of A,.e., any other semsmple quotent k-algebra scheme of A s a quotent of ths one. Proof. Let M A be the maxmal semsmple submodule. Let us see that t s blateral. We must prove that t s a rght A -module. Gven w A, t s clear that M w s a left A -submodule of A. Then t s a left A -submodule of A. It s also clear that t s semsmple, then M w M. Hence M s a rght A -submodule of A, then t s a rght A -submodule of A. Moreover, the count w : A k (.e., the unt of A ) s not null on the whole M: f m(w) = w(m) = 0 for every m M, then 0 = (m w )(w) = m(w w) = m(w ) for every w A and m = 0, then M = 0, a contradcton.

20 20 AMELIA ÁLVAREZ, CARLOS SANCHO, AND PEDRO SANCHO Therefore, M s a k-algebra scheme. M as an M -module s semsmple because as an A -module t s semsmple. Hence, by Proposton 6.6 M s a semsmple k-algebra scheme. If B s a semsmple quotent of A then B s a B -semsmple module, then t s a semsmple A -submodule of A. Therefore B M and B s a quotent of M. Notaton We wll denote by M the maxmal semsmple quotent algebra scheme of A. If E s a smple A -module then the mage of the natural morphsm A End k (E) s a smple k-algebra scheme, then t s a quotent of M. Then E s an M -module. If E s a semsmple A -module then t s a semsmple M -module. Obvously, Spec max A = Spec max M = {Set of somorphsm classes of smple A -modules} Defnton We call the radcal (deal) of a k-algebra scheme the kernel of the quotent morphsm from the algebra scheme to ts maxmal semsmple quotent algebra scheme. Let E be an A -module and let I be the radcal of A. E s semsmple f and only f t s an M -module,.e., f t s cancelled by I. If 0 E 1 E s the maxmal semsmple A -submodule of E, then E 1 = {e E : I e = 0} or equvalently, E 1 = {e E : I (k) e = 0}. Proposton Let E be an A -module and let I be the radcal of A. Let E 1 be the maxmal semsmple submodule of E, then E 1 = (E A M ) = (E /E I ) Proof. By base change t s enough to prove that E 1 = Hom k (E /E I, k). However, Hom k (E /E I, k) dentfes wth the vectors e Hom k (E, k) = E such that e(e I ) = 0. As e(w ) = w( e) for every w E and I, t follows that e E holds that e(e I ) = 0 f and only f e E 1. The functor F (E) := E 1 from the category of A -modules to the category of M -modules s a left exact functor represented by M, because F (E) = E 1 = Hom A (M, E) Let us consder the quotent E = E/E 1 and E 1 the maxmal semsmple A - submodule of E. Let E 2 := π 1 (E 1), where π : E E s the quotent morphsm. Then E 1 E 2 and E 2 /E 1 = E 1. So on we construct a canoncal chan E 1 E 2 E 3..., such that every quotent E /E +1 s a semsmple A -module and E /E +1 = {ē E/E 1 : I ē = 0}. Inductvely we deduce that E = {e E : I e = 0} Agan, as n Proposton 6.13, we obtan that E = (E A A /I ) = (E /E I ) Notaton Gven an A -module E, we wll denote by E 1 E 2... the canoncal chan of A -submodules we have just constructed. We wll denote G(E) := =1 E /E 1, where E 0 = 0 and G I E := (E I 1 /E I ). =1

21 ALGEBRA SCHEMES AND THEIR REPRESENTATIONS 21 Proposton Let E be an A -module. Then (G I E ) = G(E) In case of the regular A -module A, the canoncal chan of semsmple factors s A 1 A 2... A where A = (A /I ). Lemma Let E be a fntely-generated A -module and let I be the radcal of A. There exsts an n >> 0 such that I n E = 0. Proof. In the natural chan E 1 E 2... of E, an ncluson E n E n+1 s an equalty when E n = E by Proposton 6.9. Because E s of fnte dmenson the equalty E = E n must be true for some n N. Therefore I n E = 0. Theorem Let E be an A -module and let I be the radcal of A. It holds that Proof. (1) E = lm E (2) E = lm n E /E I n. In partcular, n=0 E I n = 0. (1) Every e E s ncluded n a fnte-dmensonal A -submodule E of E. Therefore, there exsts a n N such that I n e = 0. Then E = lm E. (2) As E = lm E, takng duals and rememberng that E = (E /E I ), t follows that E = lm n E /E I. Proposton (Nakayama) Let E, E be A -modules and let M be the maxmal semsmple quotent algebra scheme of A. Proof. (1) E = 0 E A M = 0. (2) A morphsm of A -modules E E s surjectve the morphsm E A M E A M s surjectve. (3) The morphsm E E s an somorphsm the morphsm G I E G I E s an somorphsm. (1) If E A M = 0 then E 1 = (E A M ) = 0, then E = 0 and E = 0. (2) (1) must be appled to the cokernels. (3) If the morphsm G I E G I E s an somorphsm, then so s the morphsm between the completons, whch concde wth the vector space schemes themselves by Theorem 6.17 (2). If E s an A -module of fnte dmenson we defne the character assocated to E, χ E : A k, to be χ E (w) := tr h w where h w s the homotethy on E of factor w A and tr h w s the trace of such lnear endomorphsm. But the trace of h w : E E s the same as the trace of

22 22 AMELIA ÁLVAREZ, CARLOS SANCHO, AND PEDRO SANCHO the nduced endomorphsm h w : G(E) := E /E 1 E /E 1 =: G(E). So we have the commutatve dagram A χ E k χ E /E 1 M Proposton Let us suppose that car k = 0. Then χ E = χ E f and only f G(E) and G(E ) are somorphc M -modules. Proposton Let k be an algebracally closed feld. The characters assocated to the smple A -modules are lnearly ndependent. Notaton Let E be an R-module scheme (respectvely of R-algebras) and let R S be an extenson of commutatve rngs. We wll denote by E S the S-module scheme (respectvely of S-algebras) E S = (E R S) assocated to the S-module E R S. We wll say that E s E S under the base change R S. Proposton Let A be an algebra scheme, let M be ts maxmal semsmple quotent algebra scheme and let k K be an extenson of commutatve felds. The maxmal semsmple quotent K-algebra scheme of A K s a quotent of M K. If k s algebracally closed, then the maxmal semsmple quotent K-algebra scheme of A K s the same as M K. Proof. Let 0 A 1 A 2... be the canoncal fltraton of A -modules of A. Let us consder the fltraton A 1 k K A 2 k K... of A k K. If E s a smple A K -module, then t njects nto A k K and for some there exsts an njecton E (A k K)/(A 1 k K) of A K-modules. However, (A k K)/(A 1 k K) s an M K -module, then E s an M K-module. In concluson, every morphsm from A K to a smple algebra factorzes through M K. Therefore, the maxmal semsmple quotent K-algebra scheme of A K s a quotent of M K. If k s algebracally closed then M = M n (k), then M K = M n (K) s semsmple and the maxmal semsmple quotent K-algebra scheme of A K s somorphc to M K. Proposton Let M be the maxmal semsmple quotent algebra scheme of A and let N be the maxmal semsmple quotent algebra scheme of B. Then the maxmal semsmple quotent algebra scheme of A B s a quotent of M N. If k s algebracally closed, then M N s the maxmal semsmple quotent algebra scheme of A B and an A B -module s smple f and only f t s a tensoral product of a smple A -module and a smple B -module. Proof. Let E be a smple A B -module. In partcular dm k E <. Let us consder the canoncal chan E 1 E 2... E n = E of E as an A -module. As B commutes wth A n A B, then t has to leave stable the chan. Snce E s smple, then E = E 1,.e., E s a semsmple A -module. Lkewse, E s a semsmple B -module. In concluson, E s an M N -module, then t s an M N -module. Therefore, every morphsm from A B to a smple algebra scheme factorzes through M N, then the maxmal semsmple quotent algebra scheme of A B s a quotent of M N.

23 ALGEBRA SCHEMES AND THEIR REPRESENTATIONS 23 Let k be an algebracally closed feld. As End k (V ) k End k (V ) = End k (V k V ) (dm k V, V < ), M = End k (V ) and N = End k (V j ), then j M N =,j End k (V V j) Corollary Let k be an algebracally closed feld. k-algebra schemes f and only f A B s semsmple. A, B are semsmple Let us gve some fnal examples of the applcaton of the representaton theory of algebra schemes to the representaton theory of algebrac groups. Let G = Spec A be an algebrac group. It s easy to prove that G s unpotent f and only f A s local (.e., t only contans one blateral deal scheme) and that G s trangulable f and only f A s basc (.e., M = k). It s also easy to prove that subschemes and quotents of a basc algebra scheme (respectvely local and basc) are basc (respectvely local and basc). Corollary Subgroups, quotents, drect products of trangulable (respectvely unpotent) groups are trangulable (respectvely unpotent) groups. We shall say that X A s a dense subset n A f the mnmum vector space subscheme of A that contans X s A. Dually, X s dense n A f the only a A such that a(x) = 0 for all x X s a = 0. Proposton Let k be an algebracally closed feld, let χ : A k be a morphsm of functors of k-algebras and let X A be a dense subset n A. A s local f and only f anyone of the followng condtons holds: Proof. (1) For every x X, x χ(x) belongs to the radcal of A. (2) For every x X and every morphsm of k-algebra functors φ : A End k (E), where dm k E <, φ(x χ(x)) s nlpotent. (1) If A s local then x χ(x) belongs to the kernel of χ, whch s the radcal of A. Let us see the nverse. Let E be a smple A -module and let φ : A End k (E) be the natural epmorphsm. Gven an x X, as x χ(x) belongs to the radcal of A t holds that φ(x χ(x)) s nlpotent, then χ E (x χ(x)) = 0. Therefore χ E (x) = nχ(x), where n = dm k E. Then χ E = nχ because X s dense n A. From here t follows that E = k and χ E = χ. In concluson, A s local. (2) The proof above works here. Let χ : A k be a morphsm of k-algebra schemes. An A -module E s called χ-unpotent f there exsts a fltraton of A -modules 0 E 1 E 2 E such that A operates on E /E 1 va χ for all (and E = E ). Then A s local f and only f A s χ-unpotent. If E E s an epmorphsm of A -modules and E s χ-unpotent then E s χ-unpotent. If E s χ-unpotent then E n k... k E s χ-unpotent.

24 24 AMELIA ÁLVAREZ, CARLOS SANCHO, AND PEDRO SANCHO Corollary Let k be an algebracally closed feld and let G Gl n be an ntegral algebrac group. G s unpotent f and only f every closed pont g G s a unpotent matrx,.e, g d s nlpotent. 7. Separable algebra schemes. See for nstance [P, 10] for a study of separable algebras. Defnton 7.1. We call a k-algebra scheme A separable f and only f under every base change k K, A K := (A k K) s a semsmple K-algebra scheme. Defnton 7.2. Let F be a k-algebra functor. We wll call centre of F the k-algebra subfunctor of F, that we denote by Z(F ), defned by Z(F )(C) := {a F (C) a = a} where a : F C F C, b a b and a : F C F C, b b a. w = w It holds that Z(A )(C) = {w A (C) A C A C } concdes wth the centre of the C-algebra (A k C) = A (C). Z(A ) s a k-algebra scheme: Z(A ) s the kernel of the morphsm φ : A End k (A), w w w, and End k (A) s ncluded n the k-vector space scheme Hom k (A, Ā). It holds that Z(A B ) = Z(A ) Z(B ): B s a k-vector space scheme somorphc to k, then A B s a rght and left A -module somorphc to A. Now t s easly seen that Z(A B ) Z(A ) B. Hence, Z(A B ) Z(A ) Z(B ). Notaton 7.3. Gven a rng (R, +, ) we denote by (R, +, ) the rng that s the same as R as a set, wth the same addton and whose product s defned by a b := b a. Theorem 7.4. Let A be a k-algebra scheme. The next condtons are equvalent: (1) A s separable. (2) A k s a drect product of algebras of matrces, where k s an algebracally closed feld. (3) A s semsmple and ts centre s a separable algebra scheme. (4) A k A s semsmple. Proof. 1) 2) It s obvous. 2) 1) If A k k s a drect product of algebras of matrces then by base change t s also a drect product of algebras of matrces, whose radcal s null. If A, or any base change of t, has radcal not null, then by base change t would not be null ether. 1), 2) 3) Z(A ) k = Z(A k) = k, then Z(A ) s separable. Obvously A s semsmple. 3) 2) A s a drect product of smple algebras. As the centre of a drect product s the drect product of the centres, we can assume that A s smple, that s a fnte k-algebra. We can wrte A = A. In ths case Z(A ) s a feld, because A = End K (E) and Z(A ) = Z(K). Therefore, A s an Azumaya Z(A )-algebra and A k k = A Z(A ) Z(A ) k k = A Z(A ) k whch s a drect product of algebras of matrces.