CC BY
2
Samatboyeva M. T.
student
National University of Uzbekistan named after Mirzo Ulugbek PROPERTIES OF HOM AND TENZOR PRODUCT
Abstract: The article presents the notion of Horn and tenzor product as
well as several intriguing properties of them that are related to D , ^ --• and ^ .
Keywords: Hom, tenzor product, free group, quotient group, universal property, middle linear, initial object.
Firstly, let's recall the notion of tenzor product. Let A e ModR, B eRMod ^ we will construct A 0 B - an abelian group. Let
R
C e Ab.
Definition: A middle linear map f: A x B ^ C is defined by
(1)f(a + a2, b) = f (a, b)+f (a2, b) and
(2)f O, bi + b2 ) = f (a, b1) + f (a, b2)
(3)f (a.r, b) = f (a,r .b)
Fix ^ and B as above. Let's define category M(A,B):
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) e Ob(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 0 B) then we
a
have AxB^A<S)B^C and f = a<= . A eModR,B e RMod, F be a free
V/
abelian group on the set AxB. That is AxB^Fand AxB^-\/Ce Ab, then 3h: F —» C an abelian group homomorphism s.t f = h° .
The elements of F are (a, b),(a, b)+ (a, b ),..
Definition: Let K be a subgroup of F generated by all elements of the following form: (1) (a + ab) - (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 0 B. The coset (a, b) + K of the element (a, b) e F is denoted by
a 0 b. Note that in A 0 B we have
(1) (a + a) 0 b = a 0 b + a 0 b (2) a 0 (b + b') = a 0 b + a 0 b'
(3) a.r 0 b = a 0 r.b (*)
i
Note that K = a®b is a middle linear map, this is
because of (*). We call this i: A x B ^ A 0B the canonical middle linear map.
R
Now (i, A0B) will be the initial object in M(A,B)
Property : (a) V/wg□ 'A is abelian group, prove that
Hom( A[m\ := {a e ma = 0}
(b)Hom(□ □ □ , (c) □ n n n
..........m U
Proof: (a) □ □ - homomorphism of abelian
m m
groups. Let f (1) = a ^ f (1 ■ 1 ■ ■ f (1) ^ f (1) ^ ^ f(1) = ma and we
V-"-' v-V-'
m
know that
1.1. .1 r\
^ 0 = f (0) = f (ml) = ma, so we can define
m
(p : Hom( □ A[m]
s.t f b , it is easy to check being homomorphism:
/i-> " ^ " =/(T) + g(T). _
Now we define ^ : A[m] ^ Hom(U s.t a^ ) = a (□ has
generator 1, so it is enough to define f at 1 ), ma = 0 ^ f e Hom(U .
Let k=k'^ f(k ) = /C""; " ~ 1 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'
ah^ = a and a' = 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(U A[m] : (< <p(^(a)) = ^(a)(\) = a and
y(/(T )) = /
(è) □ □
.....
From (a) we have Hom(D □ □ □ , so it is
enough to show that □ □ . Let m = m'd,n = n'd,(m',ri) = 1, we
have to find k e □ : (we can assume that 0 < k < n -1). In order to
accomplish this mk\ must be:
m = m'dk\ : : ',..., (<i — 1)«'}
Now
we
can
construct
(p\U
□
s.t
,..., d -1}
It is
making a sense S + k
that kn i-> and by checking
kn' + sri = (s + k)ri h^
It is easy to see that <7? is bijective □
□
{C) □
□
According to {a) we have □ □ □
Property : Let A is abelian group. Prove that:
{a) VMS□ □ _
£ = 0} = 0.
(b) □
□ □
Proof:
(a, k )h-> A
k = k' ^ k = k + mt
(c) □ □ □ 0
(a) Initially we construct h :Ax□
mA s.t and
(a, k')
A = (k- mt)a + mA = ka- tma + mA = ka + mA. Because of
(a, k )h-> A 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 ) r ') + mA = (ka + mA) + (ka' + mA)
2)(a, k + n ) = (a, k + n) \-> a + mA = (ka + mA) + (na + mA)
3)(na, k ) \—> - mA = (kn)a + mA
So, by universal property we get h\A® □ 'm which is homomorphism of groups s.t a 0 k 1—» A
Now we define g: A/ mA ^ A0□ s.t a + mA^> . For simplicity
lets define a + mA = [a], then [a] + [b] 0 T = a 0 1 + b 0 1, thus g is
homomorphism of groups. [a] = [a'] ^ a - a' = ma",[a] = a 01 = (a' + ma") 01 =
= a' 01 + md' 01 = d 0 1 + a" 0 m1 = d 0 1 + a" 0 0 = d 0 1, thus g is well-defined. (go ) = g ([ka]) = ka 0 1 = a 0 k and
(ho = h(a0 1) = [a] ^ go at generators of A0C , so
they are equal at full group, g o
=> A®U
ru
mA
n n
mA
s.t (k , s )h->
V m' n J
(b)Denote kmas class £ in □ .
Now we check whether it is well-defined:
km = k', sh = s" ^ ks = (k' + mt)(s ' + nl) = k's' + k'nl + s'mt + mnlt = k's'(modd)
u
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www.iupr.ru
258
ksd = ks'd . So it is well-defined and now we show it is a middle linear
map:
1 )(k +k' ,s ) is, = ks, + k's,
/V m m? n / a a a
2 + <) = (*„, A + P^ _X = ^ + K
3)(lkm , Sn ) = (lkm , Sn )h-> ^ ^ □ m □ .. □ s.t
h(k, s) = ks which is group homomorphism. Let g :□ □ □ s.t
UL
" _ _ _ _ And their compositions give identity maps: (ho h(sd 01n) = sd
and
(go s ) = g(ks,) = ks 01 = sk 01 = k 0s
vo ... o\ m n m n m n
(c) □ □ □ i-^ this is a middle linear map, because: l)(r + r',s) s = rs + r's and 2)(r,s + s') ) = rs + rs',
3)(??r,,s) = nrs = r(ns) □ □ I—> . Now
(//:□□□ i-B , it is a homomorphism : (r + s) i—^ 8>l = r<8>l + 5<8>l
Now we will examine their compositions:
q k q k k q k k q k q k
r k> k> , consequently, <p o and y/ o .
References:
1. Thomas W. Hungerford, Algebra. 2000, USA.
2. Evan Chen, An Infinitely Large Napkin. August 31, 2020
3. ttps://artofproblemsolving.com/
