Fall 2019 Algebra Qualifying Exam Solutions

1. Let V be a 5-dimensional vector space over a field F .

(a) Let T : V Ñ V be a linear transformation with characteristic poly-

nomial px ´ 1q3px ´ 2q2 and minimal polynomial px ´ 1q2px ´ 2q.i. Write down a matrix which represents T in Jordan normal form.

ii. Write down the matrix which represents T in rational normal

form.

(b) Instead, let T : V Ñ V be a nilpotent linear transformation which

has exactly one 2-dimensional invariant subspace.

i. How many similarity classes of such linear maps T are there?

ii. Assuming finally that F is the finite field Fq with q elements,

find an explicit formula for the number of such linear maps T .

3 points

7- point

3 points

7- points

in. in

.

ii. This b- the FED - module F%⇒⑤" " ¥ '

toFE "% ⇒

to Helga,

¥ F " % me ⇒

to" "

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K'

-3×+2I 2814575×-2

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O Invariant factors

soil: :i¥

(bi?

one -it is angie Jordan block

I log? )

ii. Orbit sie = index of centralizer ii GL .sc/Fe )

y pinvertible matrices

adgadega )order

→o aiolis. Hetmanate ,

order ( q- I ) q

"

i. Its qb ( q'

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2. Let A be a finite-dimensional algebra over a field F equipped with a non-

degenerate symmetric bilinear form p¨, ¨q : A ˆ A Ñ F . Let x1, . . . , xn be

a basis for A and y1, . . . , yn be the dual basis with respect to the given

form, i.e., pxi, yjq “ �i,j for all i, j “ 1, . . . , n.

(a) Show that the element

z :“nÿ

i“1

xiyi

is well defined independent of the initial choice of the basis x1, . . . , xn.

(b) Assume for the remainder of the question that the form p¨, ¨q is in-variant, which means that pab, cq “ pa, bcq for all a, b, c P A. Show

that pra, bs, cq “ pa, rb, csq where r¨, .¨s is the commutator.

(c) Let a P A be any element and suppose that ra, xis “ ∞nj“1 �ijxj and

ra, yis “ ∞nj“1 µijyj for scalars �ij , µij P F . Show that �ij ` µji “ 0.

Deduce that z lies in the center of the algebra A. (You may find the

identity ra, xys “ ra, xsy ` xra, ys helpful here.)

6 points

4 points

5+5 points

(a) Let Ki ,- . . in be another basis with dual basis yi ,

. . .

,yin .

Say Kj = § aijxi , y .

= E bij y'

.

c j j

⇒ ( xj , y ;) = aij-

- bij

Then Z = ? xiyi

= §.

bij Ki yj'

I -

- jxjgj =Eai ; ,.

,.

I the sane !

iij

Cb) ( Laid , c) = ( ab - ba,

c) = Cab,c) - Caba )

= Ca ,b c) - Ccb ,

a ) = ( a ,be - cb )

= Ca ,Eb

, d)¢

Cc )

Xij = ( [ aiki )

, yj ) =- ( C Kisa ]

, gj ) = - Gci ,Easy ;] )

Mj ; -

. ( Ki ,last )

- taji-

- of

To show 2- central ,show [ a

, 2-1=0 Ha .

[ a ,zI= § La , Kisi ] = ? @aiciTyitxiEa.gi] )

= Xij Kj 's i t § " ilnijyj

= taji ) xjyi

= Of

3. In this question, R is a ring and e P R is an idempotent, so that eRe is

another ring with identity element e.

(a) What does it mean to say that R is semisimple? State the Artin-

Wedderburn Theorem.

(b) If V is a completely reducible R-module of finite length, show that

the algebra EndRpV q is semisimple. Deduce for a semisimple ring R

that eRe is semisimple too.

(c) Assuming that R is left Artinian, show that JpeReq “ eJpRqe, whereJ denotes Jacobson radical.

4 point

41-4 points

8 points

(a) R is Semi agile if RR is completely reducible module .

AW : R is semiapie ⇒ R I Mn,

CD , ) x --

- x Mnr C Dr )

for r 70 ,ni

, - Mr 71 ,

division algebras Di ,. . ,

Dr .

( could write to instead of X, okay too ) .

(b) V E LY '

④ - - to Lir . hi ,.

.

,Lr pairwise ¥ inner .

Let Di = Enda ( Li ) a division algebra by Schur 's Lemma .

As Home Chi, Lj 7=0 for it's ,

Endear ) E Enda (hi ' ) to - -- to Enda ( Lr

" )

E Mn,

CD, ) ④ - - to Mnr C Dr ) which is semi suplex

for ere,

note e Re E Enda C Re )°P.

As R is seminole,

Re ear is or . of finite length ,so

Erdal Re )"

is semi suite by previous part.

You car remove the op-

R is senisyile ⇒ R"

is semis gig

G) In on Artinian rig ,

JCR ) is nepotist 2 - said ideal

Such that RIJCR )is seuiapie Characterisation )

To use this for question ,

need to check that R Artinian

⇒ ere Arhiiar.

1- Let J,

713 - - -

be achari of C left ) ideals u

'

ere .

Then RJ ,3 RJ z

) - - -

is onein R

,so stabling

RJn=RJn+ ,

= -- -

Now multiply by e :

eRJn=eRJn+ ,

= . ..

But e R Jn = e Re Jn = Jn, so this does

the job : Jn = Jat ,= - - -

f )

As JCR ) is mlpotiit 2- ailed ideal of R,

EJ CR) e is so in. ere

.

Remains to showe Rey

e JCR ) e

is seuinpie .

x

E ( Rtm ) e- where E -

- et JCR ) E PYJCR)

in -

scmisiiple

This is titled semisimple by (b) ¢

4. Let G be a finite group. Adopt the usual notation for the character table

of G. In particular, C1 “ t1u, C2, . . . , Cn are the conjugacy classes and

�1 “ 1,�2, . . . ,�n are the irreducible characters.

(a) Let ⇢ : G Ñ GLnpCq be a finite-dimensional representation with

associated character �. Prove that ker ⇢ “ tg P G | �pgq “ �p1qu.(b) Use the row and column orthogonality relations to work out the val-

ues of ↵,�, � and � in the following character table:

C1 C2 C3 C4 C5 C6 C7 C8 C9

�1 1 1 1 1 1 1 1 1 1

�2 2 ´2 ´1 � ´� ´↵ 1 ↵ �

�3 2 ´2 ´1 � ´↵ ´� 1 � ↵

�4 3 3 0 ´1 � ↵ 0 ↵ �

�5 3 3 0 ´1 ↵ � 0 � ↵

�6 4 ´4 1 � ´1 ´1 ´1 1 1

�7 4 4 1 � ´1 ´1 1 ´1 ´1

�8 5 5 ´1 1 0 0 ´1 0 0

�9 � ´� 0 � 1 1 0 ´1 ´1

(c) Let G be a group with the character table computed in (b). Work

out the character table of the group H “ G{ZpGq, explaining your

steps. What group is H?

6 points

- #

# I I 20 30 12 12 20 12 12

→

6 points→

→

→

→

6+2 points

(a) Take gfG .As g

'-

-

I some N,

the mu'

- poly . of 51g )

divides xn - I ,which has district huai factors .

Here,

51g) is draginainable , say to ("

o

-

-

. con ),

each Ci is aroot of unity .

¥kit - - tale kilt - - Henk n

,

So Xlg ) : C,

t -- - t Cn -

neg equality Cf. - in

Now Xlg ) -

-Xfl ) ⇐ Cit - - ten ' n

ya , .iq .

⇐ C,

= - -.

-

- Cn = I

⇐ f (g) = I

⇐ g e Ker Sp

(b) Note

dipare

!by franc 't

( Xaxi , X,) = Haik ) -

-I

i. Xixi - X,

is a character of degree 3 with

no trivial constituent,so could only be

Xc,

or Xs ( using Stl ,

'

else what 's Xixq ? )

Also heated,

so L, p are real as values of Xytxs

Now dot columns together using this to see :

O = C,

.cz ⇒ S 2=36 ⇒ 8=6

O -

- Cy . Cz ⇒ 27=0 ⇒zO = Cy . Cs ⇒ Lt B -

- I

} .

'

. Lips are root ofsi

- x - I

0 = Cg .Cq ⇒ Lp =- I sol5L

Note now we can also fu 'd the conjugacy class sizes .

fee table . .

(c) Z (G) =classes of size I

.

'

.2- (G) = C

,u Cz

,size 2

,while 161=120

So I Gza 1=60 .

If G= { 2-3,

2-2=1,

so it's either tl or-

I on each cnn.gs .

The inner 's ofGhica ) are same as ones of G on which z is +1

Classes in Gaal are either wage's ofI or 2 classes u

'

G . . .

Go Cz Czucz C4 CSU Cg 808

- Observe its

X ,I I I l I daagile group

Xy 3 0 - I B &

of #Xs 3

0 -

I L B

I0 - I - I So it must

Xz 4

Xp 5 - II O o

be Asf

5. The dihedral group G “ xa, b | a3 “ b2 “ 1, bab “ a

´1y acts on the

C-algebra S “ Crx, ys by algebra automorphisms so that

a ¨ x “ !x, a ¨ y “ !´1

y, b ¨ x “ y,

where ! “ e2⇡i{3

. Let R :“ SG

be the invariant subalgebra.

(a) Show that R “ Crx3 ` y3, xys.

(b) Show that R Ñ S is an integral extension.

(c) Find an explicit monic polynomial fptq P Rrts such that fpxq “ 0.

10 points

5 points5 points

(a) Connie 2- = E cijxiyj e S

a . z = E a wit'

si'

gj

So need Cij

= O unless i =j ( mod 3)

b. z =E cjixiyj

So need cij

= g- i t ij .

Shows R is spanned by x3ty3 and xy

( you can get anysin ty

"as monomials ri these ! )

.

'

.R -

.

I [ x 'ty

'

, xy ]¢(b) This is a general fact about invariants of finite groups .

Gwen'

ZE S,

consider f It ) = GITA( t -

g.z )

Its monic in RIH with z as aroot ⇒ integral .

a) Use the recipe from (b) !

Htt -

. getCt - g . x )

= ( t - a ) Ct - a x ) C t - ok ) Ct - y ) C t - og ) C t - aig )

= ( thx' I ( thy 3)

= t -

Cx3ty3 ) t'

+ ×3y3=

6. Work over an algebraically closed field F of characteristic zero.

(a) Let X be an a�ne variety with coordinate algebra F rXs. State the

Nullstellensatz. Then use it to show that a subset S Ñ X is dense

(in the Zariski topology) if and only if the following property holds

for all f P F rXs:

pfpsq “ 0 for all s P Sq ñ f “ 0.

(b) Given a�ne varieties X and Y and dense subsets S Ñ X and T Ñ Y ,

prove that S ˆ T is dense in X ˆ Y .

(c) Show that the integer lattice Znis dense in F

n.

51-5 points

6 points

4 points

(a) Nss :

{ 797¥73to { do.e.gqy.es }

c-

I

V ( J ) -

-{ see X I f IN t ft J }

ICS ) = { fe FIX ] I f- IN -

- O toes }

These maps are mutually inverse,

inclusion reversing bijection -

for second part , we need to show SIX ⇒ ICS ) = Co ) .

Claim J -

-

V CICS ) ) .

PI . SEV CIC :Dclose

,dhence JEVCI ( s )) .

If SEV C J ) for J -

-D-

,V C Ifs )) E V ( ICV ( JI )) -

-V (5)

This shows V I Ifs ) ) E Jf

Hence, 5 = X ⇐ VCICSD -

-

X

⇐ ICS ) = IN ) = Co )I

(b) There areseveral proofs .

All should note somewhere

That F [ XX YI =FIX ] ④ FLY ) by def . of product .

Say O E FEXXY ] is zero onSXT .

RTP 0=0 .

Write O = § fi ④ gi E FIX ] ④ FLY ],

fi 's hi . rid .

for anyTET

, § fi gift ) f FIX ) is zero onS

,

here zero as S is dense is X.

As fi 's are bring udgseidut,

this mistier gift 1=0 Htt,

So go-

= O as T is dense in Y

.

'

. O = Of

G) This follows from (b) by nitration on n as

27 is dense in F C being infinite ! ) .