Let R and S be rings and ϕ:R→S be an isomorphism, the ϕ is __________________ monomorphism endomorphism Automorphism homomorphism
homo for sure
An example of an endomorphic map is ___________________ ϕ:Z→Q ϕ:R→C ϕ:Z→Z ϕ:C→R
i think c @Loser66
yup
Let R be the ring[0,1]of all conditions real valued function on the close interval [0,1] and ϕ:R→R defined by ϕ(f)=(1/2)f then ϕ is _______ homomorphism Endomorphism Isomorphism Epimorphism
@Loser66 are you sure or you think?
now, i am sure, since before you typed 12 f, it is nonsense
now it is 1/2 f, makes more sense
have a lot more please help
If G is agroup and H is a subgroup of G.then the order of G divides the order H the order of H divides the order G H1∩G2 has same order as H H1∩G2 has same order as G
@Loser66
the order of H divides the order of G
I don't know what is H1 or G2
ok.
Consider the subset I={ 0,3,6} of the ring equivalent of [0] is ------------------ [8,10,12,14,...] [...,-18,-9,0,9,18,...] [...,30,36,42,48,54,..] [...,2,4,6,8,10,...]
@Loser66
What is the ring?
z9
set of multiple of 9
You should understand what is going on
in \(\mathbb Z_9\) the congruence [0] is set of number divided by 9
for example 9 |9 =1 Remainder 0 , this remainder is what it is indicate to congruence 0
9|18 =2 Remainder 0, hence 18 is in the set you are looking for
Suppose I give you 20, then 9|20 = 2 Remainder 2, hence 20 is NOT in the set of congruence 0
Among the options, only {....., -18,-9,0,9,18.......} all terms are multiple of 9. Each term divided by 9 get 0 in remainder. Hence this set is the answer. Got me?
yes sir
Let ϕ:R→R′ be a homomorphism of a ring R into a ring R'. The one of these is not true ϕ(0)=0′ ϕ(−a)=−ϕ(a) ϕ(e)=e′ ϕ′(s′) is not a subring of R where S' is a subring
@Loser66
what is \(\phi '\)
im back in 30 minutes
ok sir. will be waiting
thanks
you there?
yes
ok are you back?
but I don't know what is phi ' from the last one, is it inverse of phi?
dont know . dats just the question
i think it is inverse
Actually, among them, the last one seems not true since the homomorphism doesn't give us that homomorphism is onto (surjective) , hence nothing guarantee that S' has pre-image in R
ok
Moreover, all the other choices are correct
ok
Let R be the ring[0,1]of all conditions real valued function on the close interval [0,1] and ϕ:R→R defined by ϕ(f)=1/2f what is Kerφ
0
let f is in kernel, then \(\phi (f) = 1/2 f =0\) if and only if f =0, Hence f =0 is kernel phi
Let G be a group and H1 , H2 normal subgroups of .one of these is a normal subgroup of G H1∩H2 H1∪H2 H1−H2 A×B
@Loser66
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