Thread: Ring Theory - Homomorphisms from Z to Z/30Z

1. Ring Theory - Homomorphisms from Z to Z/30Z

I am reading Dummit and Foote on Ring Theory.

Problem 4, Section 7.3 RIng Homomorphisms and Quotient Rings states:

"Find all ring homomorphisms from $\mathbb{Z}$ to $\mathbb{Z} / 30 \mathbb{Z}$. In each case describe the kernal and the image."

Peter

2. Re: Ring Theory - Homomorphisms from Z to Z/30Z

well, first of all such a homomorphism must be a homomorphism of additive groups.

since 1 generates Z, it suffices to consider what h(1) is, for such a homomorphism h. in particular, <h(1)> must be a subgroup of Z/30Z.

so let's look at the additive subgroups of Z/30Z:

<0> = {0}
<1> = <7> = <11> = <13> = <17> = <19> = <23> = <29> = Z/30Z
<2> = <4> = <8> = <14> = <16> = <22> = <26> = <28> = {0,2,4,6,8,10,12,14,16,18,20,22,24,26,28}
<3> = <9> = <21> = <27> = {0,3,6,9,12,15,18,21,24,27}
<5> = <25> = {0,5,10,15,20,25}
<6> = <12> = <18> = <24> = {0,6,12,18,24}
<10> = <20> = {0,10,20}
<15> = {0,15}

but since h must also be a multiplicative homomorphism, we need h(k) = h(k*1) = h(k)h(1), that is: h(1) must be the (unique!) multiplicative identity of h(Z). so if h(Z) = Z/30Z, for example, we have to have h(1) = 1.

this severely cuts down on the number of possible ring homomorphisms.

for example, we can't have h(1) = 2, since we must have h(1) = h(1)h(1), but if h(1) = 2, h(1)h(1) = 2*2 = 4.

it is clear that each additive subgroup can have at most one multiplicative identity. some trial-and-error shows that 16 is such an identity for <2>. so we need to verify that:

if h(1) = 16, that h(km) = h(k)h(m).

h(km) = h(km*1) = km*h(1) = km*16 (mod 30).
h(k)h(m) = h(k*1)h(m*1) = (k*h(1))(m*h(1)) = km*256 (mod 30) = km*16 (mod 30) (since 256 = 16 (mod 30)).

so so far, we have 3 ring homomorphisms: h(1) = 0, h(1) = 1, and h(1) = 16.

let's continue: what we are looking for, for each image h(Z), is an element a with a2 = a. this simplifies the trial-and-error, all we need to do is square each element of h(Z). so let's look at <3>:

32 = 9
62 = 6 <--bingo!
92 = 21
122 = 24
152 = 15 <--bingo!
182 = 24
212 = 21 <--bingo!
242 = 6
272 = 9

so the multiplicative identity of <3> can only be 6,15, or 21. since 3*6 = 18, it's not 6. since 15*6 = 0, it's not 15. therefore it must be 21, so h(1) = 21 is our 4th homomorphism.

can you take it from here?

3. Re: Ring Theory - Homomorphisms from Z to Z/30Z

Thanks

Just home from work - will try to work through this now

Peter ... from Tasmania

4. Re: Ring Theory - Homomorphisms from Z to Z/30Z

Thanks Deveno ... but I need some more help to get a real understanding of what is going on here.

Problems I have are as follows (hope you can help!!!):

Issue 1

You assert that since 1 generates Z we have that <h(1)> must be a subgroup of Z/30Z

Sounds about right but can you show formally why this follows?

Issue 2 (Trying to get an understanding of the problem situation)

Folllowing your help we seem to be dealing with a situation where

the image of h, h(Z) equals either {0} , or Z/30Z, or {0,2,4, ... , 28} , or {0, 3, 6, 9, ... 27} or {0, 5, ..., 25} or {0, 6, ... 24} , {0, 10, 20} or {0,15}

(I did not use notation <1>, <2> etc as these seem to be infinite sets)

Is this correct?

Issue 3

I am trying to get a sense of the various functions h involved - in terms of explicit functional rules

For example when h(Z) = Z/30Z what is the actual homomorphism - h(z) = ??? as a functional rule

Similarly if h(Z) = {0, 2, 4, ... 28} what is the actual function h - ie h(z) = ???

I will get a better picture of the situation if someone can clarify

Peter

5. Re: Ring Theory - Homomorphisms from Z to Z/30Z

to make it clear what is going on, i will use [k], for an element of Z/30Z and k for an element of Z.

since h must be an additive homomorphism:

$h(k) = h(\sum_{i = 1}^k 1) = \sum_{i = 1}^k h(1)$.

that is: for any integer k, h(k) = h(k*1) = [k*h(1)].

or, written another way:

h(k) = h(1+1+...+1) (k times)

= h(1)+h(1)...+h(1) (k times, where the sum is mod 30).

for any group G and any element g of G <g> is a subgroup of G. since 1 generates Z (additively), h(1) generates h(Z), because of the additive homomorphism property of h.

to avoid confusion, i should have written:

<[0]>, <[1]>, etc. (or used an overline, to indicate i mean the elements of Z/30Z. this is a common abuse of notation, though, and when one is working mod n, one often omits the brackets:

something like 14 + 16 = 0 doesn't even make sense in "the regular integers").

specifically, when h(Z) = {[0],[2],[4],...,[28]} = <[2]>, the actual homomorphism is:

h(k) = [16k] (or h(k) = 16k (mod 30)).

let's just work out the image of h and the kernel of h given only that h(1) = [16] (naturally we have to have h(0) = [0]).

h(2) = [16] + [16] = [32] = [2]
h(3) = [16] + [16] + [16] = [48] = [18]
h(4) = [64] = [4]
h(5) = [80] = [20]
h(6) = [96] = [6]
h(7) = [112] = [22]
h(8) = [128] = [8]
h(9) = [144] = [24]
h(10) = [160] = [10]
h(11) = [176] = [26]
h(12) = [192] = [12]
h(13) = [208] = [28]
h(14) = [224] = [14]
h(15) = [240] = [0]
h(16) = [256] = [16]...and we repeat: h(17) = [2], h(18) = [18], etc.

that is:

h(k+15n) = h(k) + h(15n) = [16k] + [16*15n] = [16k] + [0] = [16k] = h(k) (since 15*16 is a multiple of 30, so certainly 15*16n is, as well).

this makes it clear that ker(h) = {k in Z: k = 15n} = 15Z.

as an additive group, <[16]> is thus isomorphic to Z/15Z, which has 15 elements. and indeed <[16]> = <[2]> has 30/2 = 15 elements (the "even" ones). explicitly:

[16] <--> [1] (where [1] on the right-hand side is (mod 15), and [16] on the left-hand side is (mod 30)).
[2] <--> [2]
[18] <--> [3]
[4] <--> [4]
[20] <--> [5]
[6] <--> [6]
[22] <--> [7]
[8] <--> [8]
[24] <--> [9]
[10] <--> [10]
[26] <--> [11]
[12] <--> [12]
[28] <--> [13]
[14] <--> [14]
[15] <--> [0]

note that all we are doing is sending k (mod 30) to k (mod 15). so even though it looks like we are "jumbling" the order, all the additive and multiplicative properties of the ring Z/15Z are preserved:

[8]*[24] = [192] = [12] (mod 30) and:

[8]*[9] = [72] = [12] (mod 15), and:

[8] + [24] = [32] = [2] (mod 30)

[8] + [9] = [17] = [2] (mod 15).

it's more transparent if we use coset notation:

2k + 30Z is mapped to 2k + 15Z, since 15 is ODD, 16 +30Z (the first positive even integer > 15) gets sent to 16 + 15Z = (1 + 15Z) + (15 + 15Z) = 1 + 15Z + 15Z

(since 15 is IN the coset 15Z...translating the multiples of 15 by 15, gives you back the same infinite set) = (1 + 15Z) + (0 + 15Z) = (1+0) + 15Z = 1 + 15Z.

****************

i suspect that what is actually giving you trouble here, is that you aren't fully grasping how quotient groups work. the multiplicative structure of Z is built upon the additive structure of Z (it's the analogue of "powers" in a "generic" group). the multiplicative structure of Z/nZ is inherited from that of Z. in particular, the finer structure of the subgroups of Z/nZ has to do with the set of DIVISORS of n. Z/nZ is a cyclic group. the way an element [k] of Z/nZ behaves has everything to do with gcd(k,n). it's easy to see that if k divides n, then:

[k] has order n/k in Z/nZ.

what is perhaps more difficult to see, is that if gcd(k,n) = d, that [k] acts pretty much the same as [d] (insofar as the ADDITIVE structure of Z/nZ is concerned).

6. Re: Ring Theory - Homomorphisms from Z to Z/30Z

Will now work through your post in detail

Peter

7. Re: Ring Theory - Homomorphisms from Z to Z/30Z

Still checking various theorems and results in cyclic groups and doing arithmetic to make sure I fully understand your posts.

Regarding your first post you write:

it is clear that each additive subgroup can have at most one multiplicative identity. some trial-and-error shows that 16 is such an identity for <2>. so we need to verify that:

if h(1) = 16, that h(km) = h(k)h(m).

h(km) = h(km*1) = km*h(1) = km*16 (mod 30).
h(k)h(m) = h(k*1)h(m*1) = (k*h(1))(m*h(1)) = km*256 (mod 30) = km*16 (mod 30) (since 256 = 16 (mod 30)).

In this we are assuming that h is an additive group homomorphism and need to prove that h is also a multiplicative homomorphism and hence a ring homomorphism

so to interpret h(km) = h(km*1) = km*h(1) in the above, we have that the term km is k and m multiplied (ring multiplication operation) but the operation * means km additions of 1 ie 1+1 +1 + .... km times and hence your notational differentiation of the operations writing km not k*m for the ring multiplication of k and m and km*1 for km additions of 1

and hence what is happening when we write h(km*1) = km*h(1) we are smply using the relation or equation from the additive group homomorphism h(k) = h(k*1) = [k*h(1)

Am i making sense? Can you confirm that my reasoning is OK?

Another problem that bothered me when I was checking some arithmetic.

You write in your first post:

h(1) must be the (unique!) multiplicative identity of h(Z)

and later you write:

some trial-and-error shows that 16 is such an identity for <2>.

Some of my checking went as follows:

For h(Z) = <2> we claim h(1) = 16 so check:

[16][16] = [256] = 2 --> fine OK!

[16][2] = [32] = [2] ---> fine, OK

[16][3] = [48] = [18] ---> ??? shouldn't this be 3?

Can you help clarify this?

Peter

8. Re: Ring Theory - Homomorphisms from Z to Z/30Z

[16] is not a multiplicative identity for all of Z/30Z, only for the sub-ring <[2]> = {[0],[2],[6],[8],...,[26],[28]}.

recall that a (multiplicative) identity is an element e such that a*e = a, for all non-zero a.

[16][16] = [256] = [16] (mod 30).....256 = 240 + 16 = 8*30 + 16
[16][2] = [32] = [2] (mod 30).
[16][4] = [64] = [4] (mod 30)
[16][6] = [96] = [6] (mod 30)

in fact, if 2k is an even integer between 0 and 28:

[16][2k] = [32k] = [30k] + [2k] = [30][k] + [2k] = [0][k] + [2k] = [0k] + [2k] = [0] + [2k] = [2k].

this highlights the strange fact that a multiplicative identity of a sub-ring S of a ring R need not be the same as the identity of R (but for this to be true, the identity of R needs NOT to be in S).

9. Re: Ring Theory - Homomorphisms from Z to Z/30Z

Thanks for that ... yes, obvious now

Peter