Equivariant projective modules and skew group algebras Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Showing $mathbb C G$ modules are projectiveCondition for a ring on projective and free modules problemWhat is the relation between graded modules and finitely generated modulesFinite projective dimension and vanishing of ext on f.g modulesProving that free modules are flat (without appealing projective modules)Graded projective modules and vector bundles on projective varietiesFinitely generated projective modules over a simple algebraic ring extension of a polynomial ringProjective objects in BGG category $mathcalO$ are projective $U(mathfrakg)$-modules?projective resolution of finitely generated modulesFinitely generated projective modules over polynomial rings with integral coefficientsProjective injective modules over algebras.

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Equivariant projective modules and skew group algebras



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Showing $mathbb C G$ modules are projectiveCondition for a ring on projective and free modules problemWhat is the relation between graded modules and finitely generated modulesFinite projective dimension and vanishing of ext on f.g modulesProving that free modules are flat (without appealing projective modules)Graded projective modules and vector bundles on projective varietiesFinitely generated projective modules over a simple algebraic ring extension of a polynomial ringProjective objects in BGG category $mathcalO$ are projective $U(mathfrakg)$-modules?projective resolution of finitely generated modulesFinitely generated projective modules over polynomial rings with integral coefficientsProjective injective modules over algebras.










1












$begingroup$


This is a question related to the two dimensional McKay correspondence. Let $R = mathbbC[x,y]$, and $G$ a finite group acting on $R$. Recall that a $G$-equivariant $R$-module is an $R$-module with a $G$ action, such that the multiplication map $R otimes M to M$ is $G$-equivariant, i.e., $g *(rm) = (gcdot r) (g *m)$.



It is well known that the category of such modules, $Rtext-Mod_G$, is equivalent to the category of modules over the (noncommutative, unital) skew group algebra $R # G$, which is the algebra on the underlying vector space $R otimes_mathbbC mathbbC[G]$, and multiplication defined by $(rotimes g)(r' otimes g') = r(gcdot r')otimes gg'$.



I would like to know what the projective objects are in the latter category. I know that Quillen-Suslin implies that finitely generated projective $R$-modules are free, and so I suspect that all the finitely generated projective equivariant $R$-modules are also free. However I would like to know a proof of this in the setting of the skew group ring. I am aware that the group ring $mathbbC[G]$ is semisimple, so one would hope that $R # G$ also has a similar decomposition, but I can only see this as vector spaces, rather then algebras. Moreover I'm still not sure how I would use this to get projective objects in $R# G$-Mod.










share|cite|improve this question











$endgroup$











  • $begingroup$
    Almost immediately (sigh...) after posting this I found this question which is very related. I would still like to know why the decomposition of $R # G$ holds as algebras, and I think that the question about projectives is still unclear, as certainly $R# G$ can't be semisimple (there are $R# G$ modules which are not projective, right?)
    $endgroup$
    – DKS
    Mar 22 at 18:03















1












$begingroup$


This is a question related to the two dimensional McKay correspondence. Let $R = mathbbC[x,y]$, and $G$ a finite group acting on $R$. Recall that a $G$-equivariant $R$-module is an $R$-module with a $G$ action, such that the multiplication map $R otimes M to M$ is $G$-equivariant, i.e., $g *(rm) = (gcdot r) (g *m)$.



It is well known that the category of such modules, $Rtext-Mod_G$, is equivalent to the category of modules over the (noncommutative, unital) skew group algebra $R # G$, which is the algebra on the underlying vector space $R otimes_mathbbC mathbbC[G]$, and multiplication defined by $(rotimes g)(r' otimes g') = r(gcdot r')otimes gg'$.



I would like to know what the projective objects are in the latter category. I know that Quillen-Suslin implies that finitely generated projective $R$-modules are free, and so I suspect that all the finitely generated projective equivariant $R$-modules are also free. However I would like to know a proof of this in the setting of the skew group ring. I am aware that the group ring $mathbbC[G]$ is semisimple, so one would hope that $R # G$ also has a similar decomposition, but I can only see this as vector spaces, rather then algebras. Moreover I'm still not sure how I would use this to get projective objects in $R# G$-Mod.










share|cite|improve this question











$endgroup$











  • $begingroup$
    Almost immediately (sigh...) after posting this I found this question which is very related. I would still like to know why the decomposition of $R # G$ holds as algebras, and I think that the question about projectives is still unclear, as certainly $R# G$ can't be semisimple (there are $R# G$ modules which are not projective, right?)
    $endgroup$
    – DKS
    Mar 22 at 18:03













1












1








1


1



$begingroup$


This is a question related to the two dimensional McKay correspondence. Let $R = mathbbC[x,y]$, and $G$ a finite group acting on $R$. Recall that a $G$-equivariant $R$-module is an $R$-module with a $G$ action, such that the multiplication map $R otimes M to M$ is $G$-equivariant, i.e., $g *(rm) = (gcdot r) (g *m)$.



It is well known that the category of such modules, $Rtext-Mod_G$, is equivalent to the category of modules over the (noncommutative, unital) skew group algebra $R # G$, which is the algebra on the underlying vector space $R otimes_mathbbC mathbbC[G]$, and multiplication defined by $(rotimes g)(r' otimes g') = r(gcdot r')otimes gg'$.



I would like to know what the projective objects are in the latter category. I know that Quillen-Suslin implies that finitely generated projective $R$-modules are free, and so I suspect that all the finitely generated projective equivariant $R$-modules are also free. However I would like to know a proof of this in the setting of the skew group ring. I am aware that the group ring $mathbbC[G]$ is semisimple, so one would hope that $R # G$ also has a similar decomposition, but I can only see this as vector spaces, rather then algebras. Moreover I'm still not sure how I would use this to get projective objects in $R# G$-Mod.










share|cite|improve this question











$endgroup$




This is a question related to the two dimensional McKay correspondence. Let $R = mathbbC[x,y]$, and $G$ a finite group acting on $R$. Recall that a $G$-equivariant $R$-module is an $R$-module with a $G$ action, such that the multiplication map $R otimes M to M$ is $G$-equivariant, i.e., $g *(rm) = (gcdot r) (g *m)$.



It is well known that the category of such modules, $Rtext-Mod_G$, is equivalent to the category of modules over the (noncommutative, unital) skew group algebra $R # G$, which is the algebra on the underlying vector space $R otimes_mathbbC mathbbC[G]$, and multiplication defined by $(rotimes g)(r' otimes g') = r(gcdot r')otimes gg'$.



I would like to know what the projective objects are in the latter category. I know that Quillen-Suslin implies that finitely generated projective $R$-modules are free, and so I suspect that all the finitely generated projective equivariant $R$-modules are also free. However I would like to know a proof of this in the setting of the skew group ring. I am aware that the group ring $mathbbC[G]$ is semisimple, so one would hope that $R # G$ also has a similar decomposition, but I can only see this as vector spaces, rather then algebras. Moreover I'm still not sure how I would use this to get projective objects in $R# G$-Mod.







abstract-algebra algebraic-geometry representation-theory homological-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 22 at 18:08







DKS

















asked Mar 22 at 17:53









DKSDKS

633414




633414











  • $begingroup$
    Almost immediately (sigh...) after posting this I found this question which is very related. I would still like to know why the decomposition of $R # G$ holds as algebras, and I think that the question about projectives is still unclear, as certainly $R# G$ can't be semisimple (there are $R# G$ modules which are not projective, right?)
    $endgroup$
    – DKS
    Mar 22 at 18:03
















  • $begingroup$
    Almost immediately (sigh...) after posting this I found this question which is very related. I would still like to know why the decomposition of $R # G$ holds as algebras, and I think that the question about projectives is still unclear, as certainly $R# G$ can't be semisimple (there are $R# G$ modules which are not projective, right?)
    $endgroup$
    – DKS
    Mar 22 at 18:03















$begingroup$
Almost immediately (sigh...) after posting this I found this question which is very related. I would still like to know why the decomposition of $R # G$ holds as algebras, and I think that the question about projectives is still unclear, as certainly $R# G$ can't be semisimple (there are $R# G$ modules which are not projective, right?)
$endgroup$
– DKS
Mar 22 at 18:03




$begingroup$
Almost immediately (sigh...) after posting this I found this question which is very related. I would still like to know why the decomposition of $R # G$ holds as algebras, and I think that the question about projectives is still unclear, as certainly $R# G$ can't be semisimple (there are $R# G$ modules which are not projective, right?)
$endgroup$
– DKS
Mar 22 at 18:03










1 Answer
1






active

oldest

votes


















4





+50







$begingroup$

A first observation is that the centre $Z=Z(R#G)$ is isomorphic to the fixed point ring $R^G$. Next, we can define a ring homomorphism
$$ R# G to mathrmEnd_Z(R), quad rotimes g mapsto (tmapsto rg(t)). $$
Auslander showed that this is actually an isomorphism.



We therefore have an equivalence between the category $mathrmproj(R# G)$ of finitely generated projective $R# G$-modules, and the category $mathrmadd_Z(R)$ of $Z$-module direct summands of $R^n$ for $ngeq1$.



We also have an equivalence between $mathrmproj(R# G)$ and the semisimple category $mathrmrep_mathbb C,G=mathrmmod,mathbb C[G]$.



This sends a $G$-representation $V$ to $Rotimes_mathbb CV$, with its natural $R#G$-module structure. This will be projective since there exists some $W$ such that $Voplus W$ is a free $mathbb C[G]$-module. Alternatively it is extension of scalars along the inclusion $mathbb C[G]subset R#G$.



Conversely it sends a projective $R#G$-module $P$ to the quotient $P/(xP+yP)$. Alternatively this is extension of scalars along the quotient map $R#Gtomathbb C[G]$ which sends $x,y$ to zero.



This latter equivalence shows that the isomorphism classes of indecomposable projective $R#G$-modules are in bijection with the isomorphism classes of irreducible $G$-representations.



Update: The paper by Auslander is 'On the purity of the branch locus' Amer J Math 84 (1962), and note that he takes $R=[[x,y]]$ with $G$ acting linearly.



A good reference is chapter 5 of 'Cohen-Macaulay Representations' by Leuschke and Wiegand.



Note that if we pass to the quotient fields, then this is just Galois theory. Let $L=mathbb C(x,y)$ and $K=L^G$. Then $L/K$ is a Galois extension with Galois group $G$, and so $L#Gcong mathrmEnd_K(L)$.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    Thank you for your excellent answer. Do you happen to have a reference for the result of Auslander that you quoted?
    $endgroup$
    – DKS
    Apr 2 at 12:00












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1 Answer
1






active

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votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4





+50







$begingroup$

A first observation is that the centre $Z=Z(R#G)$ is isomorphic to the fixed point ring $R^G$. Next, we can define a ring homomorphism
$$ R# G to mathrmEnd_Z(R), quad rotimes g mapsto (tmapsto rg(t)). $$
Auslander showed that this is actually an isomorphism.



We therefore have an equivalence between the category $mathrmproj(R# G)$ of finitely generated projective $R# G$-modules, and the category $mathrmadd_Z(R)$ of $Z$-module direct summands of $R^n$ for $ngeq1$.



We also have an equivalence between $mathrmproj(R# G)$ and the semisimple category $mathrmrep_mathbb C,G=mathrmmod,mathbb C[G]$.



This sends a $G$-representation $V$ to $Rotimes_mathbb CV$, with its natural $R#G$-module structure. This will be projective since there exists some $W$ such that $Voplus W$ is a free $mathbb C[G]$-module. Alternatively it is extension of scalars along the inclusion $mathbb C[G]subset R#G$.



Conversely it sends a projective $R#G$-module $P$ to the quotient $P/(xP+yP)$. Alternatively this is extension of scalars along the quotient map $R#Gtomathbb C[G]$ which sends $x,y$ to zero.



This latter equivalence shows that the isomorphism classes of indecomposable projective $R#G$-modules are in bijection with the isomorphism classes of irreducible $G$-representations.



Update: The paper by Auslander is 'On the purity of the branch locus' Amer J Math 84 (1962), and note that he takes $R=[[x,y]]$ with $G$ acting linearly.



A good reference is chapter 5 of 'Cohen-Macaulay Representations' by Leuschke and Wiegand.



Note that if we pass to the quotient fields, then this is just Galois theory. Let $L=mathbb C(x,y)$ and $K=L^G$. Then $L/K$ is a Galois extension with Galois group $G$, and so $L#Gcong mathrmEnd_K(L)$.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    Thank you for your excellent answer. Do you happen to have a reference for the result of Auslander that you quoted?
    $endgroup$
    – DKS
    Apr 2 at 12:00
















4





+50







$begingroup$

A first observation is that the centre $Z=Z(R#G)$ is isomorphic to the fixed point ring $R^G$. Next, we can define a ring homomorphism
$$ R# G to mathrmEnd_Z(R), quad rotimes g mapsto (tmapsto rg(t)). $$
Auslander showed that this is actually an isomorphism.



We therefore have an equivalence between the category $mathrmproj(R# G)$ of finitely generated projective $R# G$-modules, and the category $mathrmadd_Z(R)$ of $Z$-module direct summands of $R^n$ for $ngeq1$.



We also have an equivalence between $mathrmproj(R# G)$ and the semisimple category $mathrmrep_mathbb C,G=mathrmmod,mathbb C[G]$.



This sends a $G$-representation $V$ to $Rotimes_mathbb CV$, with its natural $R#G$-module structure. This will be projective since there exists some $W$ such that $Voplus W$ is a free $mathbb C[G]$-module. Alternatively it is extension of scalars along the inclusion $mathbb C[G]subset R#G$.



Conversely it sends a projective $R#G$-module $P$ to the quotient $P/(xP+yP)$. Alternatively this is extension of scalars along the quotient map $R#Gtomathbb C[G]$ which sends $x,y$ to zero.



This latter equivalence shows that the isomorphism classes of indecomposable projective $R#G$-modules are in bijection with the isomorphism classes of irreducible $G$-representations.



Update: The paper by Auslander is 'On the purity of the branch locus' Amer J Math 84 (1962), and note that he takes $R=[[x,y]]$ with $G$ acting linearly.



A good reference is chapter 5 of 'Cohen-Macaulay Representations' by Leuschke and Wiegand.



Note that if we pass to the quotient fields, then this is just Galois theory. Let $L=mathbb C(x,y)$ and $K=L^G$. Then $L/K$ is a Galois extension with Galois group $G$, and so $L#Gcong mathrmEnd_K(L)$.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    Thank you for your excellent answer. Do you happen to have a reference for the result of Auslander that you quoted?
    $endgroup$
    – DKS
    Apr 2 at 12:00














4





+50







4





+50



4




+50



$begingroup$

A first observation is that the centre $Z=Z(R#G)$ is isomorphic to the fixed point ring $R^G$. Next, we can define a ring homomorphism
$$ R# G to mathrmEnd_Z(R), quad rotimes g mapsto (tmapsto rg(t)). $$
Auslander showed that this is actually an isomorphism.



We therefore have an equivalence between the category $mathrmproj(R# G)$ of finitely generated projective $R# G$-modules, and the category $mathrmadd_Z(R)$ of $Z$-module direct summands of $R^n$ for $ngeq1$.



We also have an equivalence between $mathrmproj(R# G)$ and the semisimple category $mathrmrep_mathbb C,G=mathrmmod,mathbb C[G]$.



This sends a $G$-representation $V$ to $Rotimes_mathbb CV$, with its natural $R#G$-module structure. This will be projective since there exists some $W$ such that $Voplus W$ is a free $mathbb C[G]$-module. Alternatively it is extension of scalars along the inclusion $mathbb C[G]subset R#G$.



Conversely it sends a projective $R#G$-module $P$ to the quotient $P/(xP+yP)$. Alternatively this is extension of scalars along the quotient map $R#Gtomathbb C[G]$ which sends $x,y$ to zero.



This latter equivalence shows that the isomorphism classes of indecomposable projective $R#G$-modules are in bijection with the isomorphism classes of irreducible $G$-representations.



Update: The paper by Auslander is 'On the purity of the branch locus' Amer J Math 84 (1962), and note that he takes $R=[[x,y]]$ with $G$ acting linearly.



A good reference is chapter 5 of 'Cohen-Macaulay Representations' by Leuschke and Wiegand.



Note that if we pass to the quotient fields, then this is just Galois theory. Let $L=mathbb C(x,y)$ and $K=L^G$. Then $L/K$ is a Galois extension with Galois group $G$, and so $L#Gcong mathrmEnd_K(L)$.






share|cite|improve this answer











$endgroup$



A first observation is that the centre $Z=Z(R#G)$ is isomorphic to the fixed point ring $R^G$. Next, we can define a ring homomorphism
$$ R# G to mathrmEnd_Z(R), quad rotimes g mapsto (tmapsto rg(t)). $$
Auslander showed that this is actually an isomorphism.



We therefore have an equivalence between the category $mathrmproj(R# G)$ of finitely generated projective $R# G$-modules, and the category $mathrmadd_Z(R)$ of $Z$-module direct summands of $R^n$ for $ngeq1$.



We also have an equivalence between $mathrmproj(R# G)$ and the semisimple category $mathrmrep_mathbb C,G=mathrmmod,mathbb C[G]$.



This sends a $G$-representation $V$ to $Rotimes_mathbb CV$, with its natural $R#G$-module structure. This will be projective since there exists some $W$ such that $Voplus W$ is a free $mathbb C[G]$-module. Alternatively it is extension of scalars along the inclusion $mathbb C[G]subset R#G$.



Conversely it sends a projective $R#G$-module $P$ to the quotient $P/(xP+yP)$. Alternatively this is extension of scalars along the quotient map $R#Gtomathbb C[G]$ which sends $x,y$ to zero.



This latter equivalence shows that the isomorphism classes of indecomposable projective $R#G$-modules are in bijection with the isomorphism classes of irreducible $G$-representations.



Update: The paper by Auslander is 'On the purity of the branch locus' Amer J Math 84 (1962), and note that he takes $R=[[x,y]]$ with $G$ acting linearly.



A good reference is chapter 5 of 'Cohen-Macaulay Representations' by Leuschke and Wiegand.



Note that if we pass to the quotient fields, then this is just Galois theory. Let $L=mathbb C(x,y)$ and $K=L^G$. Then $L/K$ is a Galois extension with Galois group $G$, and so $L#Gcong mathrmEnd_K(L)$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Apr 2 at 14:34

























answered Apr 2 at 8:23









Andrew HuberyAndrew Hubery

58625




58625











  • $begingroup$
    Thank you for your excellent answer. Do you happen to have a reference for the result of Auslander that you quoted?
    $endgroup$
    – DKS
    Apr 2 at 12:00

















  • $begingroup$
    Thank you for your excellent answer. Do you happen to have a reference for the result of Auslander that you quoted?
    $endgroup$
    – DKS
    Apr 2 at 12:00
















$begingroup$
Thank you for your excellent answer. Do you happen to have a reference for the result of Auslander that you quoted?
$endgroup$
– DKS
Apr 2 at 12:00





$begingroup$
Thank you for your excellent answer. Do you happen to have a reference for the result of Auslander that you quoted?
$endgroup$
– DKS
Apr 2 at 12:00


















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Serbia Índice Etimología Historia Geografía Entorno natural División administrativa Política Demografía Economía Cultura Deportes Véase también Notas Referencias Bibliografía Enlaces externos Menú de navegación44°49′00″N 20°28′00″E / 44.816666666667, 20.46666666666744°49′00″N 20°28′00″E / 44.816666666667, 20.466666666667U.S. Department of Commerce (2015)«Informe sobre Desarrollo Humano 2018»Kosovo-Metohija.Neutralna Srbija u NATO okruzenju.The SerbsTheories on the Origin of the Serbs.Serbia.Earls: Webster's Quotations, Facts and Phrases.Egeo y Balcanes.Kalemegdan.Southern Pannonia during the age of the Great Migrations.Culture in Serbia.History.The Serbian Origin of the Montenegrins.Nemanjics' period (1186-1353).Stefan Uros (1355-1371).Serbian medieval history.Habsburg–Ottoman Wars (1525–1718).The Ottoman Empire, 1700-1922.The First Serbian Uprising.Miloš, prince of Serbia.3. 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Josip Broz.El nuevo orden y la resistencia.La conquista del poder.Algunos aspectos de la economía yugoslava a mediados de 1962.Albania-Kosovo crisis.De Kosovo a Kosova: una visión demográfica.La crisis de la economía yugoslava y la política de "estabilización".Milosevic: el poder de un absolutista."Serbia under Milošević: politics in the 1990s"Milosevic cavó en Kosovo la tumba de la antigua Yugoslavia.La ONU exculpa a Serbia de genocidio en la guerra de Bosnia.Slobodan Milosevic, el burócrata que supo usar el odio.Es la fuerza contra el sufrimiento de muchos inocentes.Matanza de civiles al bombardear la OTAN un puente mientras pasaba un tren.Las consecuencias negativas de los bombardeos de Yugoslavia se sentirán aún durante largo tiempo.Kostunica advierte que la misión de Europa en Kosovo es ilegal.Las 24 horas más largas en la vida de Slobodan Milosevic.Serbia declara la guerra a la mafia por matar a Djindjic.Tadic presentará "quizás en diciembre" la solicitud de entrada en la UE.Montenegro declara su independencia de Serbia.Serbia se declara estado soberano tras separación de Montenegro.«Accordance with International Law of the Unilateral Declaration of Independence by the Provisional Institutions of Self-Government of Kosovo (Request for Advisory Opinion)»Mladic pasa por el médico antes de la audiencia para extraditarloDatos de Serbia y Kosovo.The Carpathian Mountains.Position, Relief, Climate.Transport.Finding birds in Serbia.U Srbiji do 2010. godine 10% teritorije nacionalni parkovi.Geography.Serbia: Climate.Variability of Climate In Serbia In The Second Half of The 20thc Entury.BASIC CLIMATE CHARACTERISTICS FOR THE TERRITORY OF SERBIA.Fauna y flora: Serbia.Serbia and Montenegro.Información general sobre Serbia.Republic of Serbia Environmental Protection Agency (SEPA).Serbia recycling 15% of waste.Reform process of the Serbian energy sector.20-MW Wind Project Being Developed in Serbia.Las Naciones Unidas. Paz para Kosovo.Aniversario sin fiesta.Population by national or ethnic groups by Census 2002.Article 7. Coat of arms, flag and national anthem.Serbia, flag of.Historia.«Serbia and Montenegro in Pictures»Serbia.Serbia aprueba su nueva Constitución con un apoyo de más del 50%.Serbia. Population.«El nacionalista Nikolic gana las elecciones presidenciales en Serbia»El europeísta Borís Tadic gana la segunda vuelta de las presidenciales serbias.Aleksandar Vucic, de ultranacionalista serbio a fervoroso europeístaKostunica condena la declaración del "falso estado" de Kosovo.Comienza el debate sobre la independencia de Kosovo en el TIJ.La Corte Internacional de Justicia dice que Kosovo no violó el derecho internacional al declarar su independenciaKosovo: Enviado de la ONU advierte tensiones y fragilidad.«Bruselas recomienda negociar la adhesión de Serbia tras el acuerdo sobre Kosovo»Monografía de Serbia.Bez smanjivanja Vojske Srbije.Military statistics Serbia and Montenegro.Šutanovac: Vojni budžet za 2009. godinu 70 milijardi dinara.Serbia-Montenegro shortens obligatory military service to six months.No hay justicia para las víctimas de los bombardeos de la OTAN.Zapatero reitera la negativa de España a reconocer la independencia de Kosovo.Anniversary of the signing of the Stabilisation and Association Agreement.Detenido en Serbia Radovan Karadzic, el criminal de guerra más buscado de Europa."Serbia presentará su candidatura de acceso a la UE antes de fin de año".Serbia solicita la adhesión a la UE.Detenido el exgeneral serbobosnio Ratko Mladic, principal acusado del genocidio en los Balcanes«Lista de todos los Estados Miembros de las Naciones Unidas que son parte o signatarios en los diversos instrumentos de derechos humanos de las Naciones Unidas»versión pdfProtocolo Facultativo de la Convención sobre la Eliminación de todas las Formas de Discriminación contra la MujerConvención contra la tortura y otros tratos o penas crueles, inhumanos o degradantesversión pdfProtocolo Facultativo de la Convención sobre los Derechos de las Personas con DiscapacidadEl ACNUR recibe con beneplácito el envío de tropas de la OTAN a Kosovo y se prepara ante una posible llegada de refugiados a Serbia.Kosovo.- El jefe de la Minuk denuncia que los serbios boicotearon las legislativas por 'presiones'.Bosnia and Herzegovina. Population.Datos básicos de Montenegro, historia y evolución política.Serbia y Montenegro. Indicador: Tasa global de fecundidad (por 1000 habitantes).Serbia y Montenegro. Indicador: Tasa bruta de mortalidad (por 1000 habitantes).Population.Falleció el patriarca de la Iglesia Ortodoxa serbia.Atacan en Kosovo autobuses con peregrinos tras la investidura del patriarca serbio IrinejSerbian in Hungary.Tasas de cambio."Kosovo es de todos sus ciudadanos".Report for Serbia.Country groups by income.GROSS DOMESTIC PRODUCT (GDP) OF THE REPUBLIC OF SERBIA 1997–2007.Economic Trends in the Republic of Serbia 2006.National Accounts Statitics.Саопштења за јавност.GDP per inhabitant varied by one to six across the EU27 Member States.Un pacto de estabilidad para Serbia.Unemployment rate rises in Serbia.Serbia, Belarus agree free trade to woo investors.Serbia, Turkey call investors to Serbia.Success Stories.U.S. Private Investment in Serbia and Montenegro.Positive trend.Banks in Serbia.La Cámara de Comercio acompaña a empresas madrileñas a Serbia y Croacia.Serbia Industries.Energy and mining.Agriculture.Late crops, fruit and grapes output, 2008.Rebranding Serbia: A Hobby Shortly to Become a Full-Time Job.Final data on livestock statistics, 2008.Serbian cell-phone users.U Srbiji sve više računara.Телекомуникације.U Srbiji 27 odsto gradjana koristi Internet.Serbia and Montenegro.Тренд гледаности програма РТС-а у 2008. и 2009.години.Serbian railways.General Terms.El mercado del transporte aéreo en Serbia.Statistics.Vehículos de motor registrados.Planes ambiciosos para el transporte fluvial.Turismo.Turistički promet u Republici Srbiji u periodu januar-novembar 2007. godine.Your Guide to Culture.Novi Sad - city of culture.Nis - european crossroads.Serbia. Properties inscribed on the World Heritage List .Stari Ras and Sopoćani.Studenica Monastery.Medieval Monuments in Kosovo.Gamzigrad-Romuliana, Palace of Galerius.Skiing and snowboarding in Kopaonik.Tara.New7Wonders of Nature Finalists.Pilgrimage of Saint Sava.Exit Festival: Best european festival.Banje u Srbiji.«The Encyclopedia of world history»Culture.Centenario del arte serbio.«Djordje Andrejevic Kun: el único pintor de los brigadistas yugoslavos de la guerra civil española»About the museum.The collections.Miroslav Gospel – Manuscript from 1180.Historicity in the Serbo-Croatian Heroic Epic.Culture and Sport.Conversación con el rector del Seminario San Sava.'Reina Margot' funde drama, historia y gesto con música de Goran Bregovic.Serbia gana Eurovisión y España decepciona de nuevo con un vigésimo puesto.Home.Story.Emir Kusturica.Tercer oro para Paskaljevic.Nikola Tesla Year.Home.Tesla, un genio tomado por loco.Aniversario de la muerte de Nikola Tesla.El Museo Nikola Tesla en Belgrado.El inventor del mundo actual.República de Serbia.University of Belgrade official statistics.University of Novi Sad.University of Kragujevac.University of Nis.Comida. Cocina serbia.Cooking.Montenegro se convertirá en el miembro 204 del movimiento olímpico.España, campeona de Europa de baloncesto.El Partizan de Belgrado se corona campeón por octava vez consecutiva.Serbia se clasifica para el Mundial de 2010 de Sudáfrica.Serbia Name Squad For Northern Ireland And South Korea Tests.Fútbol.- El Partizán de Belgrado se proclama campeón de la Liga serbia.Clasificacion final Mundial de balonmano Croacia 2009.Serbia vence a España y se consagra campeón mundial de waterpolo.Novak Djokovic no convence pero gana en Australia.Gana Ana Ivanovic el Roland Garros.Serena Williams gana el US Open por tercera vez.Biography.Bradt Travel Guide SerbiaThe Encyclopedia of World War IGobierno de SerbiaPortal del Gobierno de SerbiaPresidencia de SerbiaAsamblea Nacional SerbiaMinisterio de Asuntos exteriores de SerbiaBanco Nacional de SerbiaAgencia Serbia para la Promoción de la Inversión y la ExportaciónOficina de Estadísticas de SerbiaCIA. Factbook 2008Organización nacional de turismo de SerbiaDiscover SerbiaConoce SerbiaNoticias de SerbiaSerbiaWorldCat1512028760000 0000 9526 67094054598-2n8519591900570825ge1309191004530741010url17413117006669D055771Serbia