Related to ring homomorphism Question. [duplicate]
up vote
0
down vote
favorite
This question already has an answer here:
What are all the homomorphisms between the rings $mathbb{Z}_{18}$ and $mathbb{Z}_{15}$?
3 answers
find the number of ring homomorphism from $mathbb Z_{12}$ to $mathbb Z_{28}$
My attempt :we know that the number of ring homomorphism from $mathbb Z_m$ into $mathbb Z_n= 2^{[phi(n)-phi(n/gcd(m,n))]}$ , where $phi(n)$ denotes the numbers of prime divisors of positive integer n.
so by formula we get number of ring homomorphism from $mathbb Z_{12}$ to $mathbb Z_{28}$ is
$$2^{[phi(28)-phi(28/gcd(12,28))]}= 2^{[phi(28)-phi(28/4)]}=2^{[phi(4 times 7)-phi(7)]} =2^{[phi(4) times phi(7)-phi(7)times phi(1)]} =2^{2 times 6 - 6times 1}= 2^6$$
Finally i got $2^6$ answer
is its correct ?
abstract-algebra
asked Nov 20 at 14:03
jasmine
1,379416
marked as duplicate by rschwieb abstract-algebra
Users with the abstract-algebra badge can single-handedly close abstract-algebra questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Nov 20 at 14:24
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
up vote
0
down vote
favorite
This question already has an answer here:
What are all the homomorphisms between the rings $mathbb{Z}_{18}$ and $mathbb{Z}_{15}$?
3 answers
find the number of ring homomorphism from $mathbb Z_{12}$ to $mathbb Z_{28}$
My attempt :we know that the number of ring homomorphism from $mathbb Z_m$ into $mathbb Z_n= 2^{[phi(n)-phi(n/gcd(m,n))]}$ , where $phi(n)$ denotes the numbers of prime divisors of positive integer n.
so by formula we get number of ring homomorphism from $mathbb Z_{12}$ to $mathbb Z_{28}$ is
$$2^{[phi(28)-phi(28/gcd(12,28))]}= 2^{[phi(28)-phi(28/4)]}=2^{[phi(4 times 7)-phi(7)]} =2^{[phi(4) times phi(7)-phi(7)times phi(1)]} =2^{2 times 6 - 6times 1}= 2^6$$
Finally i got $2^6$ answer
is its correct ?
abstract-algebra
asked Nov 20 at 14:03
jasmine
1,379416
marked as duplicate by rschwieb abstract-algebra
Users with the abstract-algebra badge can single-handedly close abstract-algebra questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Nov 20 at 14:24
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
We have $gcd(12, 28) = 4$. But more importantly, why not just list them all and count them directly? $2^6$ is a bit too large to be honest.
– Arthur
Nov 20 at 14:13
@Arthur ..Is my answer is correct?? or something missing ..i have edit also
– jasmine
Nov 20 at 14:20
1
See the duplicate linked to the duplicate: math.stackexchange.com/q/358324/29335 . The logic in the two is the same. It is a duplicate mutatis mutandi.
– rschwieb
Nov 20 at 14:24
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This question already has an answer here:
What are all the homomorphisms between the rings $mathbb{Z}_{18}$ and $mathbb{Z}_{15}$?
3 answers
find the number of ring homomorphism from $mathbb Z_{12}$ to $mathbb Z_{28}$
My attempt :we know that the number of ring homomorphism from $mathbb Z_m$ into $mathbb Z_n= 2^{[phi(n)-phi(n/gcd(m,n))]}$ , where $phi(n)$ denotes the numbers of prime divisors of positive integer n.
so by formula we get number of ring homomorphism from $mathbb Z_{12}$ to $mathbb Z_{28}$ is
$$2^{[phi(28)-phi(28/gcd(12,28))]}= 2^{[phi(28)-phi(28/4)]}=2^{[phi(4 times 7)-phi(7)]} =2^{[phi(4) times phi(7)-phi(7)times phi(1)]} =2^{2 times 6 - 6times 1}= 2^6$$
Finally i got $2^6$ answer
is its correct ?
abstract-algebra
asked Nov 20 at 14:03
jasmine
1,379416
This question already has an answer here:
What are all the homomorphisms between the rings $mathbb{Z}_{18}$ and $mathbb{Z}_{15}$?
3 answers
find the number of ring homomorphism from $mathbb Z_{12}$ to $mathbb Z_{28}$
My attempt :we know that the number of ring homomorphism from $mathbb Z_m$ into $mathbb Z_n= 2^{[phi(n)-phi(n/gcd(m,n))]}$ , where $phi(n)$ denotes the numbers of prime divisors of positive integer n.
so by formula we get number of ring homomorphism from $mathbb Z_{12}$ to $mathbb Z_{28}$ is
$$2^{[phi(28)-phi(28/gcd(12,28))]}= 2^{[phi(28)-phi(28/4)]}=2^{[phi(4 times 7)-phi(7)]} =2^{[phi(4) times phi(7)-phi(7)times phi(1)]} =2^{2 times 6 - 6times 1}= 2^6$$
Finally i got $2^6$ answer
is its correct ?
This question already has an answer here:
What are all the homomorphisms between the rings $mathbb{Z}_{18}$ and $mathbb{Z}_{15}$?
3 answers
abstract-algebra
abstract-algebra
asked Nov 20 at 14:03
jasmine
1,379416
asked Nov 20 at 14:03
jasmine
1,379416
edited Nov 20 at 14:19
asked Nov 20 at 14:03
jasmine
1,379416
asked Nov 20 at 14:03
jasmine
1,379416
asked Nov 20 at 14:03
jasmine
1,379416
1,379416
marked as duplicate by rschwieb abstract-algebra
Users with the abstract-algebra badge can single-handedly close abstract-algebra questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Nov 20 at 14:24
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by rschwieb abstract-algebra
Users with the abstract-algebra badge can single-handedly close abstract-algebra questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Nov 20 at 14:24
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
We have $gcd(12, 28) = 4$. But more importantly, why not just list them all and count them directly? $2^6$ is a bit too large to be honest.
– Arthur
Nov 20 at 14:13
@Arthur ..Is my answer is correct?? or something missing ..i have edit also
– jasmine
Nov 20 at 14:20
1
See the duplicate linked to the duplicate: math.stackexchange.com/q/358324/29335 . The logic in the two is the same. It is a duplicate mutatis mutandi.
– rschwieb
Nov 20 at 14:24
add a comment |
1
We have $gcd(12, 28) = 4$. But more importantly, why not just list them all and count them directly? $2^6$ is a bit too large to be honest.
– Arthur
Nov 20 at 14:13
@Arthur ..Is my answer is correct?? or something missing ..i have edit also
– jasmine
Nov 20 at 14:20
1
See the duplicate linked to the duplicate: math.stackexchange.com/q/358324/29335 . The logic in the two is the same. It is a duplicate mutatis mutandi.
– rschwieb
Nov 20 at 14:24
1
1
We have $gcd(12, 28) = 4$. But more importantly, why not just list them all and count them directly? $2^6$ is a bit too large to be honest.
– Arthur
Nov 20 at 14:13
We have $gcd(12, 28) = 4$. But more importantly, why not just list them all and count them directly? $2^6$ is a bit too large to be honest.
– Arthur
Nov 20 at 14:13
@Arthur ..Is my answer is correct?? or something missing ..i have edit also
– jasmine
Nov 20 at 14:20
@Arthur ..Is my answer is correct?? or something missing ..i have edit also
– jasmine
Nov 20 at 14:20
1
1
See the duplicate linked to the duplicate: math.stackexchange.com/q/358324/29335 . The logic in the two is the same. It is a duplicate mutatis mutandi.
– rschwieb
Nov 20 at 14:24
See the duplicate linked to the duplicate: math.stackexchange.com/q/358324/29335 . The logic in the two is the same. It is a duplicate mutatis mutandi.
– rschwieb
Nov 20 at 14:24
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Hint: Given a homomorphism $phi:Bbb Z_{12}toBbb Z_{28}$, how many possibilities are there for $phi(1)$? For each such possibility, how many different possibilities are there for the rest of $phi$?
answered Nov 20 at 14:22
Arthur
109k7103186
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Hint: Given a homomorphism $phi:Bbb Z_{12}toBbb Z_{28}$, how many possibilities are there for $phi(1)$? For each such possibility, how many different possibilities are there for the rest of $phi$?
answered Nov 20 at 14:22
Arthur
109k7103186
add a comment |
up vote
1
down vote
accepted
Hint: Given a homomorphism $phi:Bbb Z_{12}toBbb Z_{28}$, how many possibilities are there for $phi(1)$? For each such possibility, how many different possibilities are there for the rest of $phi$?
answered Nov 20 at 14:22
Arthur
109k7103186
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Hint: Given a homomorphism $phi:Bbb Z_{12}toBbb Z_{28}$, how many possibilities are there for $phi(1)$? For each such possibility, how many different possibilities are there for the rest of $phi$?
answered Nov 20 at 14:22
Arthur
109k7103186
Hint: Given a homomorphism $phi:Bbb Z_{12}toBbb Z_{28}$, how many possibilities are there for $phi(1)$? For each such possibility, how many different possibilities are there for the rest of $phi$?
answered Nov 20 at 14:22
Arthur
109k7103186
answered Nov 20 at 14:22
Arthur
109k7103186
answered Nov 20 at 14:22
Arthur
109k7103186
answered Nov 20 at 14:22
Arthur
109k7103186
109k7103186
add a comment |
add a comment |
1
We have $gcd(12, 28) = 4$. But more importantly, why not just list them all and count them directly? $2^6$ is a bit too large to be honest.
– Arthur
Nov 20 at 14:13
@Arthur ..Is my answer is correct?? or something missing ..i have edit also
– jasmine
Nov 20 at 14:20
1
See the duplicate linked to the duplicate: math.stackexchange.com/q/358324/29335 . The logic in the two is the same. It is a duplicate mutatis mutandi.
– rschwieb
Nov 20 at 14:24