Prime subfield when $ch(F) = 0$; shouldn't it be prime subring isomorphic to $mathbb{Z}$?
up vote
1
down vote
favorite
Suppose $ch(F)=0$ and let $phi: mathbb{Z} to F$ with $phi:n mapsto n cdot 1_F$. Since the characteristic is $0$, we have $ker phi = {0 }$. By the First Isomorphism Theorem, we have $mathbb{Z} / { 0}cong text{Im}(phi)$. But the ring on the right is just $mathbb{Z}$, and the ring on the left is the prime "subfield" of $F$. Could someone point out the problem please?
abstract-algebra field-theory
asked Nov 21 at 2:22
Ovi
12.1k938108
add a comment |
up vote
1
down vote
favorite
Suppose $ch(F)=0$ and let $phi: mathbb{Z} to F$ with $phi:n mapsto n cdot 1_F$. Since the characteristic is $0$, we have $ker phi = {0 }$. By the First Isomorphism Theorem, we have $mathbb{Z} / { 0}cong text{Im}(phi)$. But the ring on the right is just $mathbb{Z}$, and the ring on the left is the prime "subfield" of $F$. Could someone point out the problem please?
abstract-algebra field-theory
asked Nov 21 at 2:22
Ovi
12.1k938108
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Suppose $ch(F)=0$ and let $phi: mathbb{Z} to F$ with $phi:n mapsto n cdot 1_F$. Since the characteristic is $0$, we have $ker phi = {0 }$. By the First Isomorphism Theorem, we have $mathbb{Z} / { 0}cong text{Im}(phi)$. But the ring on the right is just $mathbb{Z}$, and the ring on the left is the prime "subfield" of $F$. Could someone point out the problem please?
abstract-algebra field-theory
asked Nov 21 at 2:22
Ovi
12.1k938108
Suppose $ch(F)=0$ and let $phi: mathbb{Z} to F$ with $phi:n mapsto n cdot 1_F$. Since the characteristic is $0$, we have $ker phi = {0 }$. By the First Isomorphism Theorem, we have $mathbb{Z} / { 0}cong text{Im}(phi)$. But the ring on the right is just $mathbb{Z}$, and the ring on the left is the prime "subfield" of $F$. Could someone point out the problem please?
abstract-algebra field-theory
abstract-algebra field-theory
asked Nov 21 at 2:22
Ovi
12.1k938108
asked Nov 21 at 2:22
Ovi
12.1k938108
asked Nov 21 at 2:22
Ovi
12.1k938108
asked Nov 21 at 2:22
Ovi
12.1k938108
asked Nov 21 at 2:22
Ovi
12.1k938108
12.1k938108
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
2
down vote
Your $phi$ is the morphism of the initial object in the category of unital rings to the ring $F$. If the codomain is a field of prime characteristic, then the image of the morphism is not just a subring, but a subfield. This is due to the fact that the prime non-zero ideals of $Bbb Z$ are also maximal. If not, then it's just a subring.
answered Nov 21 at 2:28
Saucy O'Path
5,7341626
Sorry I'm having trouble understanding your answer. In $mathbb{Z} / { 0}cong text{Im}(phi)$, is the ring on the left not $mathbb{Z}$, or is the ring on the right not the prime subfield?
– Ovi
Nov 21 at 2:31
$operatorname{im}phi$ is not a subfield.
– Saucy O'Path
Nov 21 at 2:31
Maybe my definition of prime subfield is wrong; as I understand it, the prime subfield of $F$ is the set of all finite sums of $1$, or ${ n cdot 1_F| n in mathbb{Z} }$. I think that $text {Im}(phi) = { n cdot 1_F| n in mathbb{Z} }$. So why am I getting that this object is not a field, when it should be?
– Ovi
Nov 21 at 2:34
4
Your definition of prime subfield is not correct. The prime subfield of a field is its smallest subfield, which contains not only all the integer multiples of $1_F$, but in the case of a field of characteristic 0, all its rational multiples as well.
– Monstrous Moonshiner
Nov 21 at 2:43
1
@Ovi note that it says field generated by $1_F$, so it's the set of sums and products and quotients that you obtain by starting with $1_F$.
– jgon
Nov 21 at 2:48
|
show 2 more comments
up vote
2
down vote
The image of $varphi$ is not the prime subfield of $F$, which is isomorphic to $mathbb{Q}$. The problem is that $phi$ is a morphism of rings, and not a morphism of fields. Therefore, you could think of $operatorname{im}(varphi)$ as being a sort of "prime subring" associated to $F$, but it is not a field. In general, for any field of characteristic 0, there is a unique field homomorphism $psi:mathbb{Q} to F$ whose image is the prime subfield associated to $F$, and which is isomorphic to $mathbb{Q}$.
answered Nov 21 at 2:41
Monstrous Moonshiner
2,25511337
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Your $phi$ is the morphism of the initial object in the category of unital rings to the ring $F$. If the codomain is a field of prime characteristic, then the image of the morphism is not just a subring, but a subfield. This is due to the fact that the prime non-zero ideals of $Bbb Z$ are also maximal. If not, then it's just a subring.
answered Nov 21 at 2:28
Saucy O'Path
5,7341626
Sorry I'm having trouble understanding your answer. In $mathbb{Z} / { 0}cong text{Im}(phi)$, is the ring on the left not $mathbb{Z}$, or is the ring on the right not the prime subfield?
– Ovi
Nov 21 at 2:31
$operatorname{im}phi$ is not a subfield.
– Saucy O'Path
Nov 21 at 2:31
Maybe my definition of prime subfield is wrong; as I understand it, the prime subfield of $F$ is the set of all finite sums of $1$, or ${ n cdot 1_F| n in mathbb{Z} }$. I think that $text {Im}(phi) = { n cdot 1_F| n in mathbb{Z} }$. So why am I getting that this object is not a field, when it should be?
– Ovi
Nov 21 at 2:34
4
Your definition of prime subfield is not correct. The prime subfield of a field is its smallest subfield, which contains not only all the integer multiples of $1_F$, but in the case of a field of characteristic 0, all its rational multiples as well.
– Monstrous Moonshiner
Nov 21 at 2:43
1
@Ovi note that it says field generated by $1_F$, so it's the set of sums and products and quotients that you obtain by starting with $1_F$.
– jgon
Nov 21 at 2:48
|
show 2 more comments
up vote
2
down vote
Your $phi$ is the morphism of the initial object in the category of unital rings to the ring $F$. If the codomain is a field of prime characteristic, then the image of the morphism is not just a subring, but a subfield. This is due to the fact that the prime non-zero ideals of $Bbb Z$ are also maximal. If not, then it's just a subring.
answered Nov 21 at 2:28
Saucy O'Path
5,7341626
Sorry I'm having trouble understanding your answer. In $mathbb{Z} / { 0}cong text{Im}(phi)$, is the ring on the left not $mathbb{Z}$, or is the ring on the right not the prime subfield?
– Ovi
Nov 21 at 2:31
$operatorname{im}phi$ is not a subfield.
– Saucy O'Path
Nov 21 at 2:31
Maybe my definition of prime subfield is wrong; as I understand it, the prime subfield of $F$ is the set of all finite sums of $1$, or ${ n cdot 1_F| n in mathbb{Z} }$. I think that $text {Im}(phi) = { n cdot 1_F| n in mathbb{Z} }$. So why am I getting that this object is not a field, when it should be?
– Ovi
Nov 21 at 2:34
4
Your definition of prime subfield is not correct. The prime subfield of a field is its smallest subfield, which contains not only all the integer multiples of $1_F$, but in the case of a field of characteristic 0, all its rational multiples as well.
– Monstrous Moonshiner
Nov 21 at 2:43
1
@Ovi note that it says field generated by $1_F$, so it's the set of sums and products and quotients that you obtain by starting with $1_F$.
– jgon
Nov 21 at 2:48
|
show 2 more comments
up vote
2
down vote
up vote
2
down vote
Your $phi$ is the morphism of the initial object in the category of unital rings to the ring $F$. If the codomain is a field of prime characteristic, then the image of the morphism is not just a subring, but a subfield. This is due to the fact that the prime non-zero ideals of $Bbb Z$ are also maximal. If not, then it's just a subring.
answered Nov 21 at 2:28
Saucy O'Path
5,7341626
Your $phi$ is the morphism of the initial object in the category of unital rings to the ring $F$. If the codomain is a field of prime characteristic, then the image of the morphism is not just a subring, but a subfield. This is due to the fact that the prime non-zero ideals of $Bbb Z$ are also maximal. If not, then it's just a subring.
answered Nov 21 at 2:28
Saucy O'Path
5,7341626
edited Nov 21 at 2:32
answered Nov 21 at 2:28
Saucy O'Path
5,7341626
answered Nov 21 at 2:28
Saucy O'Path
5,7341626
answered Nov 21 at 2:28
Saucy O'Path
5,7341626
5,7341626
Sorry I'm having trouble understanding your answer. In $mathbb{Z} / { 0}cong text{Im}(phi)$, is the ring on the left not $mathbb{Z}$, or is the ring on the right not the prime subfield?
– Ovi
Nov 21 at 2:31
$operatorname{im}phi$ is not a subfield.
– Saucy O'Path
Nov 21 at 2:31
Maybe my definition of prime subfield is wrong; as I understand it, the prime subfield of $F$ is the set of all finite sums of $1$, or ${ n cdot 1_F| n in mathbb{Z} }$. I think that $text {Im}(phi) = { n cdot 1_F| n in mathbb{Z} }$. So why am I getting that this object is not a field, when it should be?
– Ovi
Nov 21 at 2:34
4
Your definition of prime subfield is not correct. The prime subfield of a field is its smallest subfield, which contains not only all the integer multiples of $1_F$, but in the case of a field of characteristic 0, all its rational multiples as well.
– Monstrous Moonshiner
Nov 21 at 2:43
1
@Ovi note that it says field generated by $1_F$, so it's the set of sums and products and quotients that you obtain by starting with $1_F$.
– jgon
Nov 21 at 2:48
|
show 2 more comments
Sorry I'm having trouble understanding your answer. In $mathbb{Z} / { 0}cong text{Im}(phi)$, is the ring on the left not $mathbb{Z}$, or is the ring on the right not the prime subfield?
– Ovi
Nov 21 at 2:31
$operatorname{im}phi$ is not a subfield.
– Saucy O'Path
Nov 21 at 2:31
Maybe my definition of prime subfield is wrong; as I understand it, the prime subfield of $F$ is the set of all finite sums of $1$, or ${ n cdot 1_F| n in mathbb{Z} }$. I think that $text {Im}(phi) = { n cdot 1_F| n in mathbb{Z} }$. So why am I getting that this object is not a field, when it should be?
– Ovi
Nov 21 at 2:34
4
Your definition of prime subfield is not correct. The prime subfield of a field is its smallest subfield, which contains not only all the integer multiples of $1_F$, but in the case of a field of characteristic 0, all its rational multiples as well.
– Monstrous Moonshiner
Nov 21 at 2:43
1
@Ovi note that it says field generated by $1_F$, so it's the set of sums and products and quotients that you obtain by starting with $1_F$.
– jgon
Nov 21 at 2:48
Sorry I'm having trouble understanding your answer. In $mathbb{Z} / { 0}cong text{Im}(phi)$, is the ring on the left not $mathbb{Z}$, or is the ring on the right not the prime subfield?
– Ovi
Nov 21 at 2:31
Sorry I'm having trouble understanding your answer. In $mathbb{Z} / { 0}cong text{Im}(phi)$, is the ring on the left not $mathbb{Z}$, or is the ring on the right not the prime subfield?
– Ovi
Nov 21 at 2:31
$operatorname{im}phi$ is not a subfield.
– Saucy O'Path
Nov 21 at 2:31
$operatorname{im}phi$ is not a subfield.
– Saucy O'Path
Nov 21 at 2:31
Maybe my definition of prime subfield is wrong; as I understand it, the prime subfield of $F$ is the set of all finite sums of $1$, or ${ n cdot 1_F| n in mathbb{Z} }$. I think that $text {Im}(phi) = { n cdot 1_F| n in mathbb{Z} }$. So why am I getting that this object is not a field, when it should be?
– Ovi
Nov 21 at 2:34
Maybe my definition of prime subfield is wrong; as I understand it, the prime subfield of $F$ is the set of all finite sums of $1$, or ${ n cdot 1_F| n in mathbb{Z} }$. I think that $text {Im}(phi) = { n cdot 1_F| n in mathbb{Z} }$. So why am I getting that this object is not a field, when it should be?
– Ovi
Nov 21 at 2:34
4
4
Your definition of prime subfield is not correct. The prime subfield of a field is its smallest subfield, which contains not only all the integer multiples of $1_F$, but in the case of a field of characteristic 0, all its rational multiples as well.
– Monstrous Moonshiner
Nov 21 at 2:43
Your definition of prime subfield is not correct. The prime subfield of a field is its smallest subfield, which contains not only all the integer multiples of $1_F$, but in the case of a field of characteristic 0, all its rational multiples as well.
– Monstrous Moonshiner
Nov 21 at 2:43
1
1
@Ovi note that it says field generated by $1_F$, so it's the set of sums and products and quotients that you obtain by starting with $1_F$.
– jgon
Nov 21 at 2:48
@Ovi note that it says field generated by $1_F$, so it's the set of sums and products and quotients that you obtain by starting with $1_F$.
– jgon
Nov 21 at 2:48
|
show 2 more comments
up vote
2
down vote
The image of $varphi$ is not the prime subfield of $F$, which is isomorphic to $mathbb{Q}$. The problem is that $phi$ is a morphism of rings, and not a morphism of fields. Therefore, you could think of $operatorname{im}(varphi)$ as being a sort of "prime subring" associated to $F$, but it is not a field. In general, for any field of characteristic 0, there is a unique field homomorphism $psi:mathbb{Q} to F$ whose image is the prime subfield associated to $F$, and which is isomorphic to $mathbb{Q}$.
answered Nov 21 at 2:41
Monstrous Moonshiner
2,25511337
add a comment |
up vote
2
down vote
The image of $varphi$ is not the prime subfield of $F$, which is isomorphic to $mathbb{Q}$. The problem is that $phi$ is a morphism of rings, and not a morphism of fields. Therefore, you could think of $operatorname{im}(varphi)$ as being a sort of "prime subring" associated to $F$, but it is not a field. In general, for any field of characteristic 0, there is a unique field homomorphism $psi:mathbb{Q} to F$ whose image is the prime subfield associated to $F$, and which is isomorphic to $mathbb{Q}$.
answered Nov 21 at 2:41
Monstrous Moonshiner
2,25511337
add a comment |
up vote
2
down vote
up vote
2
down vote
The image of $varphi$ is not the prime subfield of $F$, which is isomorphic to $mathbb{Q}$. The problem is that $phi$ is a morphism of rings, and not a morphism of fields. Therefore, you could think of $operatorname{im}(varphi)$ as being a sort of "prime subring" associated to $F$, but it is not a field. In general, for any field of characteristic 0, there is a unique field homomorphism $psi:mathbb{Q} to F$ whose image is the prime subfield associated to $F$, and which is isomorphic to $mathbb{Q}$.
answered Nov 21 at 2:41
Monstrous Moonshiner
2,25511337
The image of $varphi$ is not the prime subfield of $F$, which is isomorphic to $mathbb{Q}$. The problem is that $phi$ is a morphism of rings, and not a morphism of fields. Therefore, you could think of $operatorname{im}(varphi)$ as being a sort of "prime subring" associated to $F$, but it is not a field. In general, for any field of characteristic 0, there is a unique field homomorphism $psi:mathbb{Q} to F$ whose image is the prime subfield associated to $F$, and which is isomorphic to $mathbb{Q}$.
answered Nov 21 at 2:41
Monstrous Moonshiner
2,25511337
answered Nov 21 at 2:41
Monstrous Moonshiner
2,25511337
answered Nov 21 at 2:41
Monstrous Moonshiner
2,25511337
answered Nov 21 at 2:41
Monstrous Moonshiner
2,25511337
2,25511337
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3007167%2fprime-subfield-when-chf-0-shouldnt-it-be-prime-subring-isomorphic-to-m%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
