Noncommutative Algebraic Geometry Homework





<#137#>Exercises from Chapter 1.<#137#>


<#138#>Exercise 1.1 ( Michael Van Opstall)<#138#>

<#139#> #math1#Hom~A(M,M) is a ring if A is a preadditive category. <#139#>

In a preadditive category, #math2#Hom~A(M,M) is an abelian group. Define multiplication by function composition. Then by the axioms for the category, there exists an identity morphism which is the multiplicative identity for this ring. The requirement that composition of morphisms is bilinear gives distributivity of multiplication.


<#140#>Exercise 1.2 ( Michael Van Opstall)<#140#>

<#141#> Let R be a ring with identity. Let CR be the category with a single object and morphisms given by elements of R. The category of additive functors from CR to Ab is equivalent to the category of left R-modules. <#141#>

Any additive functor F from CR to Ab assigns the single object an abelian group G. The morphisms (one for each element of R) become endomorphisms of G. The action F(r)(g) induces a left R-module structure on G: The identity element is taken to the identity automorphism, as required of a functor. The additivity of the functor guarantees that the map #math3##tex2html_wrap_inline1094# is a group homomorphism, so #math4#rs.g=F(rs)(g)=F(r)(F(s)(g))=r.(sg), so the requirements for a module are met.

A natural transformation of two functors S and T from CR to Ab assigns a homomorphism #tex2html_wrap_inline1106# from the group GS=S(*) to GT=T(*) such that #math5##tex2html_wrap_inline1112# for any element #math6##tex2html_wrap_inline1114#. But this is just the added requirement needed for a group homomorphism to be a module homomorphism.

So every additive functor yields a module and every natural transformation of such functors yields a module homomorphism. Conversely, given an R-module M, there is a corresponding functor FM which sends * to the underlying abelian group of M, and sends each element of #math7#R=Hom~(*,*) to the group homomorphism given by multiplication by r. And given a module homomorphism #math8##tex2html_wrap_inline1130#, the underlying group homomorphism from FM(*) to FN(*) is a natural transformation by the properties of module homomorphisms (commuting with scalars). By construction this correspondance is the inverse of the assignment of modules to functors, so the two categories are equivalent.

<#142#>Corollary 1.8<#142#> If f is a morphism in an additive category, then f is monic if and only if fg=0 implies g=0. The dual statement about epis follows from the dual proof.


#pf143#


<#145#>Exercise 1.5 (Alexandra Nichifor)<#145#>

See the other homework file.


<#146#>Exercise 1.6 (Pete Littig)<#146#>

<#800#>Show that in an additive category the maps

#eqnarraystar944#


arising in the definition of the product and coproduct are epic and monic, respectively. <#800#>

Let #math10##tex2html_wrap_inline1168# be a finite collection of objects in our additive category. For pairs #math11##tex2html_wrap_inline1170# of indices, define morphisms #math12##tex2html_wrap_inline1172# by


#math13# #displaymath1050#

By the universal properties of the product and coproduct, each morphism #math14##tex2html_wrap_inline1174# factors as #math15##tex2html_wrap_inline1176# and #math16##tex2html_wrap_inline1178# where #math17##tex2html_wrap_inline1180# and #math18##tex2html_wrap_inline1182#. With these morphisms in hand we are ready to proceed to the main line.

Let's show first that the projection maps are epic. Suppose for some object N that #tex2html_wrap_inline1186# and #tex2html_wrap_inline1188# are morphisms #math19##tex2html_wrap_inline1190# such that #math20##tex2html_wrap_inline1192#. Then

#eqnarraystar956#


from which it follows that each #tex2html_wrap_inline1194# is an epimorphism.

Analysis of the injection maps follows similarly. Let N be an object with morphisms #math22##tex2html_wrap_inline1198#satisfying #math23##tex2html_wrap_inline1200#. Then

#eqnarraystar972#


providing us with the desired result.

This completes the proof.


<#192#>Exercise 2.2 (Jeremy Calvert)<#192#>

<#802#>Let A be an abelian category having coproducts. Show that #math25##tex2html_wrap_inline1204# is a generator if and only if, for each #math26##tex2html_wrap_inline1206#, there exists an epimorphism #math27##tex2html_wrap_inline1208# defined on some coproduct of copies of N<#802#>

#math28##tex2html_wrap_inline1212# Suppose X is a generator of A. Then consider the two maps #math29##tex2html_wrap_inline1218#, and #math30##tex2html_wrap_inline1220#. Since X is a generator, #math31##tex2html_wrap_inline1224# such that #math32##tex2html_wrap_inline1226#, and hence g1 is non-zero.

Now consider the pair of maps #math33##tex2html_wrap_inline1230#and #math34##tex2html_wrap_inline1232#, where #tex2html_wrap_inline1234# is the standard quotient. Again, X a generator implies that #math35##tex2html_wrap_inline1238# so that #math36##tex2html_wrap_inline1240#. Hence #math37##tex2html_wrap_inline1242#.

Define a map #math38##tex2html_wrap_inline1244#. Clearly g1(X) is a proper subobject of #math39##tex2html_wrap_inline1248#.

In general, given #math40##tex2html_wrap_inline1250# we can use the generating property of X to find a nonzero map onto the cokernel of N. Now we have to show that iterating this process leads to a surjection from a coproduct to N. I don't know how to do this, except to beg the question. That is, to assume that A having coproducts means coproducts with indexing set any cardinality, so that we simply define the coproduct #math41##tex2html_wrap_inline1260#, where the elements of the indexing set I are in bijection with the cokernels of N, and the maps in components are as described above. That is, if #tex2html_wrap_inline1266# corresponds to the cokernel N/M, then in the component i the map #math42##tex2html_wrap_inline1272# is some #math43##tex2html_wrap_inline1274#such that #math44##tex2html_wrap_inline1276# where #tex2html_wrap_inline1278# is the standard quotient #math45##tex2html_wrap_inline1280#. Now suppose that the cokernel of #math46##tex2html_wrap_inline1282# is non-zero, so that G is not surjective. Call this cokernel #tex2html_wrap_inline1286#. But by definition #math47##tex2html_wrap_inline1288#s.t. #math48##tex2html_wrap_inline1290# satisfies #math49##tex2html_wrap_inline1292#non-zero, where #tex2html_wrap_inline1294# is the quotient #math50##tex2html_wrap_inline1296#, contradicting the assumption that #tex2html_wrap_inline1298# is the cokernel. Hence the cokernel of Gis 0, and so is surjective.

;SPMquot;#math51##tex2html_wrap_inline1304#;SPMquot; Suppose that for every N in A, there is a surjection #math52##tex2html_wrap_inline1310#. Then consider any pairs of maps #math53##tex2html_wrap_inline1312#. Since #math54##tex2html_wrap_inline1314# is a surjection, we must have that #math55##tex2html_wrap_inline1316#. Now appealing to the axiom of choice, there must be a factor in the coproduct which is not in the kernel of #math56##tex2html_wrap_inline1318#, and the restriction of G to this factor is a map which provides the generator property for X over N. Since this can be done for any #tex2html_wrap_inline1326#, X is a generator.


<#206#>Exercise 2.6 (Alexandra Nichifor)<#206#>

See the other homework file.


<#207#>Exercise 4.1 (Michael Van Opstall)<#207#>

<#208#>Let E be an injective object in an abelian category A. <#208#>

<#209#>(a) E is a cogenerator if and only if #math57##tex2html_wrap_inline1336# for each #tex2html_wrap_inline1338# in A. <#209#>

Since E is injective, #math58#Hom~(-,E) is exact. By lemma 2.10, #math59#Hom~(-,E) is faithful (i.e. E is a cogenerator) if and only if it sends nonzero objects to nonzero objects.

<#210#>(b) E is a cogenerator in #math60#Mod~R if and only if it contains a copy of every simple R-module. <#210#>

Suppose M is a simple R-module. An element of #math61#Hom~(M,E) is either zero or injective since M is simple. If it is nonzero, then it injects a copy of M into E.

Conversely, suppose E contains a copy of every simple module. Then since E is injective, it contains the injective envelope of every simple module. Let M be an R-module. Then take a nonzero #tex2html_wrap_inline1376#. By 4.1a it suffices to produce a map #math62##tex2html_wrap_inline1378# with #math63##tex2html_wrap_inline1380#. Let S be the simple module formed by taking a quotient of Rm by a maximal submodule and let #tex2html_wrap_inline1386# be the quotient map. Then #math64##tex2html_wrap_inline1388# (where #tex2html_wrap_inline1390# is the inclusion of S into E(S)) is a map from Rm into E(S) with #math65##tex2html_wrap_inline1400#.

We have an exact sequence


#math66# #displaymath213#

which by exactness of #math67#Hom~(-,E(S)) gives the exact sequence:


#math68# #displaymath215#

that is, f has a preimage #tex2html_wrap_inline1406#, and #math69##tex2html_wrap_inline1408#. Since E(S) is contained in E, composing #tex2html_wrap_inline1414# with the inclusion of E(S) into E is the desired map #tex2html_wrap_inline1420#.



<#217#>Exercises from Chapter 2.<#217#>


<#218#>Exercise 1.1 (Rebekah Hahn)<#218#>

<#803#>Let T= all nxn lower triangular matrices over the field k, and #math70#y=e2,1+e3,2+...+en,n-1 or, the matrix with ones on the subdiagonal and zeroes elsewhere.

(a) Show that yT=Ty= all strictly lower triangular matrices. <#803#>

We proceed by showing that both yT and Ty equal the set of all strictly lower triangular matrices, S.

If #math71#A=(ai,j), then yA is the matrix consisting of a row of zeroes, followed by the first n-1 rows of A, and Ay is the matrix consisting of the last n-1 columns of A, followed by a column of zeroes. Thus, Ty and yT are contained in S.

Now, let #tex2html_wrap_inline1456#. Then, let A be the matrix formed by moving the last n-1 rows of B up one row, and then putting a row of zeroes at the bottom, You should convince yourself that this matrix is lower triangular and that yA=B. We can now conclude that yT=S. The conclusion of the proof that Ty=S is similar.

<#849#>(b) Show that the simple module corresponding to ei,i is #math72#T/<#804#>yT+(1-ei,i)T<#804#>. <#849#>

I will begin by showing that #math73#S=T/<#805#>yT+(1-ei,i)T<#805#> is one dimensional, and therefore simple. From part (a), we know that yT is all strictly lower triangular matrices. Also, if A is in T, then ei,iA is the matrix with i-th row equal to the i-th row of A and zeroes elsewhere. Therefore, #math74#(1-ei,i)T is all lower triangular matrices with zeroes in the i-th row.

Therefore, a complement of #math75#yT+(1-ei,i)T in T is just all matrices with a nonzero entry in the i,i-th position, and zeroes elsewhere. Thus, the quotient is one dimensional.

It remains to show that S is, indeed, the simple module corresponding to ei,i. That is, ei,i acts as the identity on S. So, let #tex2html_wrap_inline1496#. Let #math76#U=yT+(1-ei,i)T. Then A=B+U for some matrix #tex2html_wrap_inline1502#. We can choose B so that B is a matrix with its only nonzero entry in the i,i-th position, since U consists of sums of strictly lower triangular matrices and matrices with zeroes in the i-th row. Also, note that ei,iU is contained in U, since multiplication by ei,i preserves the qualities of being lower triangular and of having only zeroes in the i-th row.

Now, let ei,i act on A. Then, #math77#ei,iA=ei,i(B+U)=ei,iB+U=B+U=A. The second to last equality follows from the properties of matrix multiplication and the choice of B.

Therefore, S is the simple module corresponding to ei,i.


<#239#>Exercise 1.2 (Alexandra Nichifor)<#239#>

<#806#>Let T, U denote the rings of lower, respectively upper, triangular #math78##tex2html_wrap_inline1532# matrices over the field k.

(a) Show there is an algebra isomorphism #math79##tex2html_wrap_inline1534# given by the k-linear map #math80##tex2html_wrap_inline1536#. <#806#>

T and U are #math81##tex2html_wrap_inline1542# dimensional k vector spaces with bases #math82##tex2html_wrap_inline1546# and #math83##tex2html_wrap_inline1548# respectively.
Let #math84##tex2html_wrap_inline1550# be the k-linear map sending #math85##tex2html_wrap_inline1552#. To see that #tex2html_wrap_inline1554# gives an algebra isomorphism we need to check:
1) #tex2html_wrap_inline1556# is linear over k: by definition.
2) #math86##tex2html_wrap_inline1558#

#math87##tex2html_wrap_inline1560#
3) #math88##tex2html_wrap_inline1562#

Suffices to verify that #tex2html_wrap_inline1564# is multiplicative on basis elements #math89##tex2html_wrap_inline1566# (since each of A, B can be written as a linear combination of ei,j's and the general formula follows from distributivity in T, k-linearity of #tex2html_wrap_inline1574#, and its multiplicative property on basis elements). We have

#math90# #displaymath1576#

4)bijectivity:

Let #math91##tex2html_wrap_inline1578# be the k-linear map #math92##tex2html_wrap_inline1582# for #tex2html_wrap_inline1584#. Note that #tex2html_wrap_inline1586# is the inverse of #tex2html_wrap_inline1588#.

(b) Show there is an algebra anti-isomorphism #math93##tex2html_wrap_inline1590# given by the k-linear map #math94##tex2html_wrap_inline1592#.

Let #math95##tex2html_wrap_inline1594# be the k-linear map #math96##tex2html_wrap_inline1596#. Everything is similar to (a) above, except for part 3) where we need to check that #math97##tex2html_wrap_inline1598# instead. We have

#math98# #displaymath1600#



<#298#>Exercise 1.3 (Pete Littig)<#298#>

<#299#>Let R be a commutative noetherian ring. Prove that there is a bijection between the isomorphism classes of indecomposable injective R-modules and the prime ideals in R. <#299#>

Our proof consists of two parts. First we'll show that if P is prime then E(R/P) is an indecomposable injective. Second we'll see that every indecomposable injective E is isomorphic to E(R/P) for some prime P.

Let P be prime. It follows that P is irreducible (i.e., there are no two submodules K and L each properly containing P such that #math99##tex2html_wrap_inline1628#.) To see this, suppose that #math100##tex2html_wrap_inline1630#. Then #math101##tex2html_wrap_inline1632# whence either A = P or B = P. Since P is irreducible, E(R/P) is indecomposable.

Conversely. suppose that E is injective. Then E contains a submodule M that is isomorphic to R/P. Since E is indecomposable, it is an injective envelope for each of it's non-trivial submodules, that is #math102##tex2html_wrap_inline1652#. Since #math103##tex2html_wrap_inline1654#we have #math104##tex2html_wrap_inline1656# and consequently E is isomorphic to E(R/P).

The correspondence we have constructed is certainly surjective in both directions. It remains to show that this correspondence is also injective. The assignment #math105##tex2html_wrap_inline1662# is indeed injective, for if E1 and E2 are both sent to P then they are both isomorphic to E(R/P) and hence to each other. I'm not sure how to show injectivity in the opposite direction, but my suspicion is that #math106##tex2html_wrap_inline1672# implies P1 = P2.


<#300#>Exercise 1.4 (Michael Van Opstall)<#300#>

<#301#>

The natural transformations of the identity functor on #math107#Mod~R form a ring isomorphic to Z(R). <#301#>

Let R be a ring. Then a natural transformation T of the identity functor assigns to each right R-module A and endomorphism T(A) of A in such a way that


#math108# #displaymath302#

commutes for any A, B, and #math109##tex2html_wrap_inline1696#. Let the addition of natural transformations be #math110#(S+T)(A)=S(A)+T(A) and multiplication be given by function composition: #math111##tex2html_wrap_inline1700#. Since S(A) and T(A) are module homomorphisms, the distributive laws are satisfied, and the other ring axioms are clear.

Every element r in the center of R gives a natural transformation by assigning to each module the endomorphism given by right multiplication by r. Composition of these morphisms corresponds to multiplication in the ring, so this map is a homomorphism.

This map is obviously injective. It remains to show that it is surjective. Let T be any natural transformation of the identity functor. Then in particular, T assigns to R an element of #math112#Hom~(R,R) which is in the center of #math113#Hom~(R,R), i.e.


#math114# #displaymath310#

commutes for all f.

Since our ring has 1, #math115#Hom~(R,R) is isomorphic to the opposite ring of R, and hence T(R) is given by right multiplication by a central element of R.

Now this determines the T(M) for all free modules M (as right multiplication by some central element of R). Any other module is a quotient of a free module, so we have, for a module A and some free module M,


#math116# #displaymath316#

Since the quotient map f is surjective, there is a unique T(A) which completes the diagram, and multiplication by the same central element of Ris such a map. So every natural transformation of the identity functor is given by assigning to each module A the endomorphism given by right multiplication by a central element of R.


<#322#>Exercise 1.5 (Adam Nyman)<#322#>

<#808#>If R is commutative and noetherian, the map from Spec R to the indecomposable injectives of Mod R sending #tex2html_wrap_inline1758# to #math117##tex2html_wrap_inline1760# is a bijection. <#808#>

We will show that the map #math118##tex2html_wrap_inline1762#implements the bijection.

First, the map is well-defined. Since #math119##tex2html_wrap_inline1764# contains no non-trivial direct sums, and since it is an essential submodule of its injective hull, #math120##tex2html_wrap_inline1766# is indecomposable.

Now we show the map is surjective. Let E be an indecomposable injective. Since R is noetherian, we can choose an #tex2html_wrap_inline1772# such that its annihilator is as large as possible. This condition ensures that the annihilator is a prime ideal, say #tex2html_wrap_inline1774#. Thus, #math121##tex2html_wrap_inline1776# is isomorphic to a submodule of E. But #math122##tex2html_wrap_inline1780# is also a submodule of its injective hull, #math123##tex2html_wrap_inline1782#. Since E and #math124##tex2html_wrap_inline1786# are injective, there are maps #math125##tex2html_wrap_inline1788# and #math126##tex2html_wrap_inline1790# which are mutually inverse (use the universal property of injectives).

Next we show that the map is injective. Thus, suppose #tex2html_wrap_inline1792# and #tex2html_wrap_inline1794# are distinct elements in Spec R. Let x be in #tex2html_wrap_inline1800# but not in #tex2html_wrap_inline1802#. Then multiplication by x kills #math127##tex2html_wrap_inline1806# but acts injectively on #math128##tex2html_wrap_inline1808# (since #tex2html_wrap_inline1810# is prime). We claim x acts injectively on #math129##tex2html_wrap_inline1814#. For, let #math130##tex2html_wrap_inline1816#. Then since #math131##tex2html_wrap_inline1818# and #math132##tex2html_wrap_inline1820# is essential in its hull, M=0. Thus #math133##tex2html_wrap_inline1824# is not isomorphic to #math134##tex2html_wrap_inline1826#.


<#349#>Exercise 1.6 (Rebekah Hahn)<#349#>

<#350#>Let #math135##tex2html_wrap_inline1828#. Show that the center, say Z, of R is #math136#k[x2,y2] and that this is isomorphic to the polynomial ring in two variables. By finding a k-vector space basis for R, show R is free of rank four as a Z-module. Find all simple R-modules when k is algebraically closed. <#350#>



First, notice that R is essentially the polynomial ring k[x,y], except that x and y are anticommutative (xy=-yx). So, elements of R can be written as polynomials in the two variables x and y. In fact, the set #math137##tex2html_wrap_inline1864# is a basis for R as a k vector space.

To find the center, Z, of R, it suffices to find everything that commutes with x and y. We have the relations

#math138# #displaymath1878#


#math139# #displaymath1880#

Furthermore, we can write #math140##tex2html_wrap_inline1882# as #math141##tex2html_wrap_inline1884#.

Now, #math142##tex2html_wrap_inline1886# if and only if #math143#xf(x,y)=f(x,y)x and #math144#yf(x,y)=f(x,y)y if and only if
#math145##tex2html_wrap_inline1892# and
#math146##tex2html_wrap_inline1894# if and only if
#math147##tex2html_wrap_inline1896# and
#math148##tex2html_wrap_inline1898# if and only if

#math149#ai,j = 0 whenever either i is odd or j is odd if and only if #math150##tex2html_wrap_inline1906#. Thus, #math151#Z=k[x2,y2]

To see that this is isomorphic to the polynomial ring k[u,v] (the map would be send x2 to u, y2 to v), recall that R is essentially a polynomial ring already, except for the anticommutativity of x and y. So, since Z is commutative and #math152#k[x2,y2] would be isomorphic to k[u,v] if #math153#k[x2,y2] were the usual subring of the polynomial ring, Z is isomorphic to k[u,v].

Recall that #math154##tex2html_wrap_inline1938# is a k-vector space basis for R. Consider the set #math155##tex2html_wrap_inline1944#. This is a generating set for R as a Z module since any element of the k-basis, #math156##tex2html_wrap_inline1952#, can be obtained by multiplying an element of #math157##tex2html_wrap_inline1954# by and element of Z and k is contained in Z. And, since #math158##tex2html_wrap_inline1962#is k linearly independent, it must be Z linearly independent, since Z is just k with some higher degree terms thrown in.

Finally, we can address the issue of simple modules of R. From a theorem in class, we know that all of the simple modules of R are finite dimensional over k, and that a given simple module Mis annihilated by a maximal ideal of the center, Z, as well as by a maximal ideal of R itself. Since Z is just a polynomial ring in 2 variables, we know all of its maximal ideals. They are of the form #math159##tex2html_wrap_inline1986# where #math160##tex2html_wrap_inline1988#.

At this point, we need to address cases. If #tex2html_wrap_inline1990#, then the maximal ideals of R containing #math161##tex2html_wrap_inline1994# are the ideals #math162##tex2html_wrap_inline1996#and #math163##tex2html_wrap_inline1998#. So the simple ideals corresponding to this maximal ideal are #math164##tex2html_wrap_inline2000# and #math165##tex2html_wrap_inline2002#. Similarly, if #tex2html_wrap_inline2004#, we get the ideals #math166##tex2html_wrap_inline2006# and #math167##tex2html_wrap_inline2008#. But, if neither #tex2html_wrap_inline2010# or #tex2html_wrap_inline2012# is zero, then it turns out that the ideal #math168##tex2html_wrap_inline2014# is itself maximal in R, so there is just one simple module associated with this ideal, namely, #math169##tex2html_wrap_inline2018#.


<#395#>Exercise 1.7 (Patrick Perkins)<#395#>

<#396#> Let k be an algebraically closed field and let #math170##tex2html_wrap_inline2022# be the quotient of the two-variable free algebra by the relation xy+yx=0. Describe the points in #math171#X=Mod~A, the noncommutative affine space with coordinate ring A.<#396#>

The points in the space X are given by subcategories of #math172#Mod~A of the form #math173#Mod~A/I where A/I is a simple artinian quotient of A. The subalgebra #math174#R=k[x2,y2] is a commutative ring and we may identify Spec(R) with #math175##tex2html_wrap_inline2044#. We can write

#math176# A=R+Rx+Ry+Rxy

as a finite R-module. The sum is direct and R is precisely Z(A), the center of A. The inclusion #math177##tex2html_wrap_inline2054# determines a morphism #math178##tex2html_wrap_inline2056#. We can thus picture X as a covering space of the affine plane.


<#397#>Lemma<#397#> <#398#>If A/I is simple artinian then #tex2html_wrap_inline2062# is a maximal ideal of R. <#398#>


#pf399#

This proof applies whenever A is a finite module over its center R, and R is noetherian. By the Nullstellesatz, there exist #math179##tex2html_wrap_inline2116# such that #math180##tex2html_wrap_inline2118#. Let J be the ideal of A generated by the central elements #math181##tex2html_wrap_inline2124# and #tex2html_wrap_inline2126#. There is a surjection #math182##tex2html_wrap_inline2128#. Using the R-module decomposition of A given above, we can write

#math183# A/J = k+kx+ky+kxy

as a 4-dimensional vector space. We have relations #math184##tex2html_wrap_inline2134#, #tex2html_wrap_inline2136# and xy=-yx. This shows that A/J is a generalized quaternion algebra, see [#P##1###, 1.6]. Thus if both #tex2html_wrap_inline2142# and #tex2html_wrap_inline2144# are non-zero, then A/J is a simple algebra isomorphic to M2(k) and so I=J. (It is interesting to note that if k were not algebraically closed, then A/J could be a division algebra). Suppose, for example, that #tex2html_wrap_inline2156#. Let T be the simple module associated with R/I. Then y2 kills T. Since y is a normal element of A, Ty is a submodule of T. If Ty=T then #math185#T=Ty=(Ty)y=Ty2=0, a contradiction. Thus Ty=0 and #tex2html_wrap_inline2180#. It follows that A/I is a simple quotient of #math186##tex2html_wrap_inline2184#. Thus #math187##tex2html_wrap_inline2186# where #math188##tex2html_wrap_inline2188#. Thus, viewing X = Mod R as a covering of #math189##tex2html_wrap_inline2194# we get the following picture.

X = Mod #math190#k;SPMlt;x,y;SPMgt;/(xy+yx)
units ;SPMlt;.18in, .18in;SPMgt; (<#413#> .<#413#>) x from -10 to 10, y from -10 to 10 (<#414#> .<#414#>) <#415#>-10.3<#415#><#416#>10<#416#><#417#>-10.3<#417#><#418#>10<#418#> units ;SPMlt;.18in, .18in;SPMgt; point at .3 .3 x from -10 to 10, y from -10 to 10 <#419#>-10<#419#><#420#>10.3<#420#><#421#>-10<#421#><#422#>10.3<#422#> <#423#>#tex2html_wrap_inline2200#<#423#> at .15 .15 span ;SPMlt;4pt;SPMgt; -9.8 -9.8 9.8 9.8 /


There is a unique point of X over each point in #math191##tex2html_wrap_inline2204# away from the axes. Away from the origin, every point on an axis splits into two points in X. There is a unique point in X over the origin.


<#424#>Exercise 2.1 (Pete Littig)<#424#>

<#810#>Suppose that we adopt the opposite convention for the concatenation of paths- that is, we write #math192##tex2html_wrap_inline2210# to mean first traverse #tex2html_wrap_inline2212#, then traverse #tex2html_wrap_inline2214#. Write k * Q for the path algebra obtained in this way. Show that #math193##tex2html_wrap_inline2218#. <#810#>

The result is trivial, but we give the details. For #tex2html_wrap_inline2220# and #tex2html_wrap_inline2222# in (kQ)op let us denote their product as #math194##tex2html_wrap_inline2226#. Define #math195##tex2html_wrap_inline2228# by #math196##tex2html_wrap_inline2230# (and extend k-additively.) Since k * Q and (kQ)op share the same additive group structure, #tex2html_wrap_inline2238# is clearly an isomorphism of groups. Multiplicatively we have

#eqnarraystar434#


so that #tex2html_wrap_inline2240# is multiplicative. It follows that #tex2html_wrap_inline2242# is an algebra isomorphism.


<#436#>Exercise 2.4 (Alexandra Nichifor)<#436#>

<#811#>If Q is a quiver, let Qop denote the ;SPMquot;same;SPMquot; quiver with the arrows reversed. Show that #math198##tex2html_wrap_inline2248#. <#811#>

Let #math199##tex2html_wrap_inline2250# be the arrows in Q and let #math200##tex2html_wrap_inline2252# be the arrows in Qop, where #math201##tex2html_wrap_inline2256# denotes the arrow #tex2html_wrap_inline2258# reversed. Then kQ is the path algebra over k with basis all paths in Q, #math202##tex2html_wrap_inline2266#, including the nullpaths #math203##tex2html_wrap_inline2268#, and with multiplication given by path concatenation in Q. Similarly, kQop is the path algebra over k with basis #math204##tex2html_wrap_inline2276# and multiplication given by path concatenation in Qop.

By definition, the opposite of a ring #math205##tex2html_wrap_inline2280# is #math206##tex2html_wrap_inline2282#, where #math207##tex2html_wrap_inline2284#.

Let #math208##tex2html_wrap_inline2286# be the k-linear transformation #math209##tex2html_wrap_inline2288# and #math210##tex2html_wrap_inline2290#. Then, since #math211#kQ=(kQ)op as vector spaces over k, #tex2html_wrap_inline2296# can be viewed as #math212##tex2html_wrap_inline2298# and provides a k-vector space isomorphism (since it is a one to one correspondence between the bases of the two spaces).

Since #math213##tex2html_wrap_inline2300#, to see that #tex2html_wrap_inline2302# is an algebra isomorpism it only remains to check that it respects the multiplication structure of the two rings. It suffices to check on the basis elements: #math214##tex2html_wrap_inline2304# = #math215##tex2html_wrap_inline2306#= #math216##tex2html_wrap_inline2308#=
#math217##tex2html_wrap_inline2310#. Clearly this works if any of the paths is a null path.


<#476#>Exercise 2.5 (Suren Fernando)<#476#>

<#477#>(a) Show that v is a sink if and only if Sv is projective. <#477#>

Let #tex2html_wrap_inline2316# be the set of arrows that begin at v. For each #math218##tex2html_wrap_inline2320#, let #tex2html_wrap_inline2322# be the end vertex of #tex2html_wrap_inline2324#. Since (1) v being a sink is equivalent to #tex2html_wrap_inline2328# being empty and

#math219# #displaymath2330#

is a minimal projective resolution of Sv (this is shown in the notes), we see that v is a sink if and only if Sv is projective.


<#482#>

(b) Show that v is a source if and only if Sv is injective. <#482#>

We recall from the notes that if u and v are vertices of Q, then

#math220# #displaymath2348#

If Sv is injective, then #math221#Ext~kQ(Su,Sv)=0 for each vertex u that is distinct from v, and so there are no arrows ending at v. In other words, if Sv is injective then v is a source. Suppose on the other hand that v is a source and Sv is not injective. Let Ev be the injective envelope of Sv, and let x belong to #math222##tex2html_wrap_inline2374#. Since Ev is an essential extension of Sv, there is an #tex2html_wrap_inline2380# such that #math223##tex2html_wrap_inline2382#. It follows that there is a path in Q that ends at v, contrary to the assumption that v is a source. Therefore Sv is injective if v is a source.


<#486#>Exercise 2.6 (Suren Fernando)<#486#>

<#487#> Show that #math224#Hom~Q(Pu,Sv) is zero if #tex2html_wrap_inline2396#, and is isomorphic to k is u=v. <#487#>

Problem 2.6 is stated and proved on pp 11-12 (I was unaware of this when I picked the problem - I would put a smiley face here if I knew my emoticons better ...). The solution to (a) is essentially the same as that in the notes. The solution to (b) is slightly different.

We note first that every #math225##tex2html_wrap_inline2402# is determined by its value #math226##tex2html_wrap_inline2404#, because Pu is a cyclic right kQ-module with generator #math227##tex2html_wrap_inline2410#.

Now suppose #tex2html_wrap_inline2412#. Then,

#math228# #displaymath2414#

since #math229##tex2html_wrap_inline2416# are orthogonal idempotents. Consequently #math230#Hom~Q(Pu,Sv)=0 if #tex2html_wrap_inline2420#.

Next suppose u=v. Then, since each #math231##tex2html_wrap_inline2424# is uniquely determined by the choice of an arbitrary element of Sv for #math232##tex2html_wrap_inline2428#, the rule #math233##tex2html_wrap_inline2430#is a bijection between #math234#Hom~Q(Pu,Sv) and Sv. The inverse map is given by the rule #math235##tex2html_wrap_inline2436#, where #math236##tex2html_wrap_inline2438# is deter mined by the condition #math237##tex2html_wrap_inline2440#. It is clear that this sets up a k-linear isomorphism between Sv and #math238#Hom~Q(Pu,Sv).


<#488#>Exercise 2.9 (Michael Van Opstall)<#488#>

<#814#>Let Q be a quiver with vertices #math239##tex2html_wrap_inline2450#. Let M be a matrix with mij equal to the number of arrows from ui to uj. Then the ij entry in Mn is the number of paths of length n from ui to uj. <#814#>

The statement is obviously true if n=1. Suppose it is true of Mm. Write the elements of Mm as #tex2html_wrap_inline2476#. Then the ij entry of Mm+1 is #math240##tex2html_wrap_inline2482#. But #tex2html_wrap_inline2484# is the number of paths from ui to uk of length m and mkj is a path from uk to uj of length 1. Since every path from ui to uj can be decomposed in this way (trivially), this is all the paths from ui to uj.


<#496#>Exercise 5.1 (Michael Van Opstall)<#496#>

<#497#>The inclusion of lower triangular 2#tex2html_wrap_inline2506#2 matrices into the ring of #tex2html_wrap_inline2508# matrices is epic. <#497#>

We must show that if f and g are two morphisms which agree on lower triangular matrices, they agree on all matrices.

Since f and g are morphisms and agree on e11, e21, and e22,


#eqnarraystar501#


which gives


#math242# f(e11)(f(e12)-g(e12))=f(e12)f(e21)(f(e12)-g(e12))=0

So


#math243# 0=f(e11+e22)(f(e12)-g(e12))=1(f(e12)-g(e12))=(f(e12)-g(e12))

and f and g agree on the strictly upper triangular portion as well.


<#538#>Exercise 5.2 (Michael Van Opstall)<#538#>

<#815#>

The inclusion

#math244# #displaymath539#

is epi. <#815#>

Call the left hand ring A. Let eij be the matrix with 1 in entry ij and 0 elsewhere. Let #math245##tex2html_wrap_inline2534# be the matrix with x in entry ij and 0 elsewhere. Let #math246##tex2html_wrap_inline2540# be the matrix with 1/x in entry ij and 0 elsewhere. We must show that a homomorphism from M2(k(x)) is determined by the values it takes on A. Let f and g be two homomorphisms which agree on A.

From exercise 4.1, we know that the image of e12 is also determined by A. We have


#eqnarraystar551#


so


#math248# #displaymath561#

but #math249#e12+e21 is a unit, so #math250##tex2html_wrap_inline2562#. Right multiplication and the same argument lead to #math251##tex2html_wrap_inline2564#. So also, #math252##tex2html_wrap_inline2566# for all n. And the inverses are also determined: #math253##tex2html_wrap_inline2570#. So we have that f and g agree on #math254##tex2html_wrap_inline2576#.

This generates the whole ring, though: #math255##tex2html_wrap_inline2578# is the element with xn in the upper right and zeros elsewhere. Similar for the 1/xn and the lower left corner.


<#582#>Exercise 5.4 (Rebekah Hahn)<#582#>

<#583#>Let R be a commutative PID. Show that an R-module is injective if and only if it is divisible. <#583#>

Assume that D is injective. Let #tex2html_wrap_inline2590# and #math256##tex2html_wrap_inline2592#be given. Consider the R-module homomorphism,f, from the ideal (r) into D which is induced by sending r to d, and the inclusion of (r) into R. The injectivity of D guarantees the existence of a map #tex2html_wrap_inline2612# such that g(x)=f(x) for all #tex2html_wrap_inline2616#. Now, #math257##tex2html_wrap_inline2618# and #tex2html_wrap_inline2620# so, D is divisible.

For the other direction, assume that D is divisible. Then, for every ideal J and R-module homomorphism #tex2html_wrap_inline2630#, we need to show that there exists an R-module homomorphism #tex2html_wrap_inline2634# such that h=k|J.

Since R is a PID, J=(r) for some #math258##tex2html_wrap_inline2642#. Since D is divisible, there exists #tex2html_wrap_inline2646# so that h(r)=rc. Define #math259##tex2html_wrap_inline2650# to be the R-module homomorphism which sends 1 to c. Then, #math260#h(r)=rc=rk(1)=k(r), so h=k|J, whence D is injective.



<#584#>Exercises from Appendix 1.<#584#>


<#585#>Exercise 2.7 (Pete Littig)<#585#>

<#586#>Show that the inclusion #math261##tex2html_wrap_inline2664# is not an epimorphism in the category of fields. <#586#>

Let #math262##tex2html_wrap_inline2666# denote complex conjugation. Observe that #tex2html_wrap_inline2668# and #math263##tex2html_wrap_inline2670# agree on #math264##tex2html_wrap_inline2672#, but clearly do not agree on the whole of #math265##tex2html_wrap_inline2674#. It follows that i is not an epimorphism.


<#589#>Some Extra Exercises.<#589#>


<#590#>Exercise A (Pat Perkins)<#590#>

Let k be an algebraically closed field. Let R be the subring #math266##tex2html_wrap_inline2682# of M2(k[x]). Then Z(R)=k[x] and R is a finite module over its center. Thus, using the lemma in my solution to Homework 1.7, if R/I is a simple artinian quotient of R then #math267##tex2html_wrap_inline2694# for some #math268##tex2html_wrap_inline2696#. Note that #tex2html_wrap_inline2698# is central in R for any #tex2html_wrap_inline2702#. If #math269##tex2html_wrap_inline2704# then clearly #math270##tex2html_wrap_inline2706# so #math271##tex2html_wrap_inline2708# and there is a unique corresponding 2 dimensional simple module. If #tex2html_wrap_inline2710# then #math272##tex2html_wrap_inline2712# and there are two non-isomorphic 1 dimensional simple modules, (k,0) and (0,k). Observe that #math273##tex2html_wrap_inline2718# is a 2-sided ideal of R and that #math274##tex2html_wrap_inline2722#. This makes the vector space (k,k) into a right R-module which is a non-split extension of (k,0) by (0,k). Similarly, the set #math275##tex2html_wrap_inline2732# is a 2-sided ideal of R and #math276##tex2html_wrap_inline2736#. This makes (k,k) into a right R-module which is a non-split extension of (0,k) by (k,0). Thus the picture of the space Mod R is precisely the same as the picture of Mod S where #math277##tex2html_wrap_inline2750#.


<#612#>Exercise B. (Rebekah Hahn)<#612#>

<#613#>Let #math278##tex2html_wrap_inline2752#, where q is not a root of unity.

(a) Show that A/(xy-1)A is an infinite dimensional simple module.

(b) Show that #math279##tex2html_wrap_inline2758# is not simple. <#613#>

<#614#>(a)<#614#> Note, first of all, that proving this proves that any module of the form #math280##tex2html_wrap_inline2760# where #math281##tex2html_wrap_inline2762# is an infinite dimensional simple module, since the module #math282##tex2html_wrap_inline2764# can be mapped into the module A/(xy-1)A via the isomorphism induced by taking x to x and y to #tex2html_wrap_inline2774#. Also, since we have already shown that all finite dimensional simples take the form #math283##tex2html_wrap_inline2776# or #math284##tex2html_wrap_inline2778#, it suffices to show that A/(xy-1)A is simple.

We begin by noting that any element of #math285#M=A/(xy-1)A can be written as the sum of a polynomial in x and a polynomial in y since we can replace any terms like xiyj by yj-i if j;SPMgt;ior xi-j if #tex2html_wrap_inline2796#.

Now, let N be a nonzero submodule of M, with #tex2html_wrap_inline2802#. Then we can write #math286#n=f(x)+g(y), f,g polynomials. Suppose that #math287##tex2html_wrap_inline2808#, and consider the element #math288##tex2html_wrap_inline2810# of N.

#math289# #displaymath2814#

which is a polynomial in y. From which we can conclude that there are elements in N which contain only powers of y.

Let #tex2html_wrap_inline2822# be such an element with least y degree and suppose degm=t. If t=0, then #tex2html_wrap_inline2830# and #math290##tex2html_wrap_inline2832# so M is simple. Otherwise, #math291##tex2html_wrap_inline2836#, with #math292##tex2html_wrap_inline2838#. Now, notice that

#math293# #displaymath2840#

so that

#math294# #displaymath2842#

But, as long as #math295##tex2html_wrap_inline2844#, then it has lower y degree than n, which would be a contradiction. Therefore, #math296#n(1-qtxy)=0. But, since q is not a root of unity, this holds if and only if #math297##tex2html_wrap_inline2854# for all i;SPMlt;t, in which case n=yt. Yet, notice that #math298##tex2html_wrap_inline2860# which also has lower y degree than n, again a contradiction. Hence, n must have had degree equal to 0, which, as has already been shown, implies that N=M and M is simple.

<#646#>(b)<#646#> Now, let #math299##tex2html_wrap_inline2872#. If #math300##tex2html_wrap_inline2874#, then M=A, which is clearly not simple. Otherwise, one of #math301##tex2html_wrap_inline2878# is nonzero, and, we can assume, without loss of generality, that #math302##tex2html_wrap_inline2880#.

CLAIM: My is a nontrivial submodule of M.

First, since y is normal in A, #math303#MyA=MAy=My so that My is, indeed, a submodule of M. Further, #tex2html_wrap_inline2896#, so that #tex2html_wrap_inline2898#. Finally, notice that #math304##tex2html_wrap_inline2900# where #math305##tex2html_wrap_inline2902#. That is, #math306##tex2html_wrap_inline2904#. But, if N denotes the submodule of M/yM corresponding to My then N=0. Hence, #tex2html_wrap_inline2914# and it is, indeed, a nontrivial submodule of M.