Properties of Hom and tenzor product

• 1.     Objects are middle linear maps on A x B , i.e { f : A x B ^ }

• 2.     Morphisms between f : A x B ^ C and g : A x B ^ D is a group

homomorphism h : C ^ D s.t the diagram is commutative: g = h°

An initial object in M(A,B), if it exists, has the universal property: V(f,C)gOb(M(A,B)) there exists a:init.ob^(f,C) s.t the diagram commutes. This means that if we denote the initial object by (i, A ® B) then we have A x B-^A ® B ДC and f = a . A g ModR,B g RMod, F be a free i                        ∀f abelian group on the set A×B. That is A×B→F and A×B→∀C ∈ Ab , then 3h: F ^ C an abelian group homomorphism s.t f = ho .

The elements of F are ( a , b ),( a , b ) + ( a ₂, b ₂),...

Definition: Let K be a subgroup of F generated by all elements of the following form:        (1) ( a + a ‘ , b ) - ( a , b ) - ( a ', b )                          (2)

( a , b + b ') — ( a , b ) — ( a , b')                                 (3) ( a . r , b ) — ( a , r . b )

The quotient group F / K is called the tensor product of A and B and is denoted A ® B . The coset  ( a , b ) + K of the element ( a , b ) g F is denoted by

R

• a ⊗ b . Note that in A ⊗ B we have

• (1)    ( a + a ‘ ) ® b = a ® b + a ‘® b     (2) a ® ( b + b ') = a ® b + a ® b ‘

• (3) a . r ® b = a ® r.b       (*)

  Note that    ( a , b ) н

  because of (*) . We call this

  
  

K = a ® b is a middle linear map, this is i : A x B ^ A ® B the canonical middle linear map. R

M ( A , B )

A is abelian  group, prove that

( c )□           n mm

• □       - homomorphism of abelian

fm )   f (1)    f (1) = ma and we

m

• = ma , so we can define

Now ( i , A ® B ) will be the initial object in

Property :   (a)

Hom(        Am ] : = {a g A | ma = 0}

• (b) Homp □    □,

gcd( m , n )

Proof: (a) □ m groups. Let f (1) = a ^ f (1_2_+2

m know that

• 1    • 1    • 1 о ^ 0 = f (0) = f (ml

m

Ф : Hom (□         A [ m ]

• s.t      f h^     , it is easy to check being homomorphism:

  /■>           ■>        " = f (1) _ g (1) .

Now we define ^: A[m] ^ Hom(□      s.t a i-^      ) = a (□ has generator 1, so it is enough to define f at 1), ma = 0 ^ f g Hom(□      .

Let k = k ' ^ f ( k ) = f ( 1 • 1     • 1 ⁴    'a = ( k '_ mn ) a = k'a = f ( k ') ^

k well-defined.

f ( k _ n ) = ( k _ n ) a = ka _ na = f ( k ) _ f ( n )

and

a _ a '^       ) = a _ a'

a i—>        ) = a           and            a ' н       = a ',     therefore

( g _ h )(1) = g (1) _ h (1) = a _ a ' = f (1) . So g _ h and f are equal at generator , thus they are exactly the same functions.

Now, we consider the composition of these homomorphisms, if they give identity, then Hom (          Am ] : ( ^°         p ( w (a )) = ^ ( a )(1) = a and

( ^ o        ^ ( f (1)) = f

• ^{( b} )    Hom^  □    □       ,

gcd( m , n )

From (a) we have Hom(□   □     □           □          ■, so it is enough to show that □      □         . Let m = m d, n = n ’ d,(m, n) = 1, we n       gcd( m, n )

have to find k gD      : (we can assume that 0 < k < n -1). In order to accomplish           this            mk n            must           be:

m = m’ dk:                  :         :                ',...,( d — 1) n' }

Now    we    can    construct

^:П          0         s.t nd

{0, n ',...,( d - 1) n' } ^D           _,..., d - 1}

d

It is making a sense that kn' + sn = (s + k) n' н       s + k

It is easy to see that ^ is bijective □

kn ' н     and by checking

.

gcd( m , n )

( c ) □           _n

m

According to (a) we have □□

□           k = 0} = 0 .

Property : Let A is abelian group. Prove that:

(a) V meD

(b )□     □     □                             (c )□   □□ m    n    gcd( a, n )

Proof:      ( a )   Initially we construct h ⁰: A xD          ' mA s.t

( a, k )i-> k = k' ^ k = k' + mt                                         and

( a , k ')i->       A = ( k — mt ) a + mA = ka — tma + mA = ka + mA . Because of

(a, k )i->       4 we get (a, k') = (a, k ). So the function is well-defined. Now we examine that it is middle linear map:

1)( a + a ', k ) H        '') + mA = ( ka + mA) + ( ka' + mA )

2)( a , k + n ) = ( a , k + n ) l-^

a + mA = ( ka + mA ) + ( na + mA )

3)( na , k ) I—»       - mA = ( kn ) a + mA

So, by universal property we get h: A ®Q        ' m which is homomorphism of groups s.t a ® k I—>      A

Now we define g: A / mA ^ A ® □   s.t a + mA I—>     . For simplicity lets define a + mA = [a], then [a] + [b] i-^       ® 1 = a ® I + b ® 1, thus g is homomorphism                       of                       groups.

[ a ] = [ a '] ^ a — a ' = ma ",[ a ] = a ® 1 = ( a ' + ma ") ® 1 =

= a® 1 + ma"® 1 = a'® 1 + a"® m 1 = a'® 1 + a"® 0 = a'® 1, thus g is well-defined.

( g₀         ) = g ([ ka ]) = ka ® 1 = a ® k

and

( h о

= h ( a ® 1) = [ a ]   ^ g°

a ®D

at generators of A ®D , so

they are equal at full group. g

_A ^ A ®D

A / aA       m

(b ) Denote k_m as class k in □  . h^ : 0    0

Now     we      check     whether     it

mA

0 d s.t ( k m , S n )^ ^— .

is       well-defined:

k_m = k'_m , s_n = s ‘ ^ ks = ( k ' + mt )( s ‘ + nl ) = k's' + k'nl + s ' mt + mnlt = k ' s '(mod d )

⇒ks =ks′ . So it is well-defined and now we show it is a middle linear map:

¹⁾⁽ k m + k' m , s n w

s_d = ks_d + k ′ s_d

²)( k m , s n + s П ) = ⁽ k m , s n + s П )^         ^d = ks d + ^ksd

3)( lk_m , s_n ) = ( lk_m , s_n )^

dd

⊗ → mn _d

s.t

h ( k , s ) = ks which is group homomorphism. Let g :□     □     0 s.t

Sd^..... _   _ _  _

• ( c )    □  □    □      h^ this is a middle linear map, because:

1)( r + r ', s ) I—>         s = rs + r^rs and 2)( r , s + s ‘)i—>         ) = rs + rs' ,

3)( nr , s ) = nrs = r ( ns ) ^3 ^ :Q □   □      h->  .               Now

• у :D D D H     ,      it      is      a      homomorphism      :

( r + s )l->       & 1 = r & 1 + s & 1

Now we will examine their compositions:

• r &  s H>          = p • - &  1 = p • — &  k • ¹ = p • — • k & ¹ = pl & ¹ = p &  - = r &  s

qk    qk   k qk   k  q k q k

• r l-^          , consequently, ^ о      ^ and у о