Integer domains and fields
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While studying algebra I came across the following problematic: $F$ fields and $Rneemptyset$ a subring of $F$, then $R$ is an integer domain, moreover it was supposed that $1_Fin R$, I did not understand the reason for this assumption.
After having reasoned I arrived at the following conclusion: if $F$ is a field and $R$ a subring of $F$ it is beyond question that $R$ is a integer domain. Now, same author when they define a integer domain they do not insist on the fact that the ring must be unitary, then in this case if we are interested having a unity in $R$ we have to suppose that $1_Fin R$. However, if for us an integer domain is a commutative ring with unity that has no zero divisors, then necessarily $1_R=1_F$, otherwise we can prove that $1_R$ is a zero divisor in $F$, but this is impossible, because $F$ is a field.
It's correct? Thanks!
abstract-algebra
asked Nov 22 at 15:34
Jack J.
6401419
add a comment |
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1
down vote
favorite
While studying algebra I came across the following problematic: $F$ fields and $Rneemptyset$ a subring of $F$, then $R$ is an integer domain, moreover it was supposed that $1_Fin R$, I did not understand the reason for this assumption.
After having reasoned I arrived at the following conclusion: if $F$ is a field and $R$ a subring of $F$ it is beyond question that $R$ is a integer domain. Now, same author when they define a integer domain they do not insist on the fact that the ring must be unitary, then in this case if we are interested having a unity in $R$ we have to suppose that $1_Fin R$. However, if for us an integer domain is a commutative ring with unity that has no zero divisors, then necessarily $1_R=1_F$, otherwise we can prove that $1_R$ is a zero divisor in $F$, but this is impossible, because $F$ is a field.
It's correct? Thanks!
abstract-algebra
asked Nov 22 at 15:34
Jack J.
6401419
1
$1_F=1_F1_R=1_R$ in $R$ is all you need to show that $1_R=1_F$.
– John Douma
Nov 22 at 15:55
The reason I guess for the assumption $1_F in R$ is the choice of the definition of an `Integral domain' as a non zero ring with identity.
– R.C.Cowsik
Nov 25 at 13:55
@R.C.Cowsikthanks for your ansewer. But, if we consider $Rsubseteq F$, when $R$ is an integer domain and $F$ is a field, that is a commutative ring with unity without zero divisors, then necessarily $1_R=1_F$, otherwise $1_R$ it would be a divisor of the zero of $F$. Right?
– Jack J.
Nov 25 at 14:41
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
While studying algebra I came across the following problematic: $F$ fields and $Rneemptyset$ a subring of $F$, then $R$ is an integer domain, moreover it was supposed that $1_Fin R$, I did not understand the reason for this assumption.
After having reasoned I arrived at the following conclusion: if $F$ is a field and $R$ a subring of $F$ it is beyond question that $R$ is a integer domain. Now, same author when they define a integer domain they do not insist on the fact that the ring must be unitary, then in this case if we are interested having a unity in $R$ we have to suppose that $1_Fin R$. However, if for us an integer domain is a commutative ring with unity that has no zero divisors, then necessarily $1_R=1_F$, otherwise we can prove that $1_R$ is a zero divisor in $F$, but this is impossible, because $F$ is a field.
It's correct? Thanks!
abstract-algebra
asked Nov 22 at 15:34
Jack J.
6401419
While studying algebra I came across the following problematic: $F$ fields and $Rneemptyset$ a subring of $F$, then $R$ is an integer domain, moreover it was supposed that $1_Fin R$, I did not understand the reason for this assumption.
After having reasoned I arrived at the following conclusion: if $F$ is a field and $R$ a subring of $F$ it is beyond question that $R$ is a integer domain. Now, same author when they define a integer domain they do not insist on the fact that the ring must be unitary, then in this case if we are interested having a unity in $R$ we have to suppose that $1_Fin R$. However, if for us an integer domain is a commutative ring with unity that has no zero divisors, then necessarily $1_R=1_F$, otherwise we can prove that $1_R$ is a zero divisor in $F$, but this is impossible, because $F$ is a field.
It's correct? Thanks!
abstract-algebra
abstract-algebra
asked Nov 22 at 15:34
Jack J.
6401419
asked Nov 22 at 15:34
Jack J.
6401419
asked Nov 22 at 15:34
Jack J.
6401419
asked Nov 22 at 15:34
Jack J.
6401419
asked Nov 22 at 15:34
Jack J.
6401419
6401419
1
$1_F=1_F1_R=1_R$ in $R$ is all you need to show that $1_R=1_F$.
– John Douma
Nov 22 at 15:55
The reason I guess for the assumption $1_F in R$ is the choice of the definition of an `Integral domain' as a non zero ring with identity.
– R.C.Cowsik
Nov 25 at 13:55
@R.C.Cowsikthanks for your ansewer. But, if we consider $Rsubseteq F$, when $R$ is an integer domain and $F$ is a field, that is a commutative ring with unity without zero divisors, then necessarily $1_R=1_F$, otherwise $1_R$ it would be a divisor of the zero of $F$. Right?
– Jack J.
Nov 25 at 14:41
add a comment |
1
$1_F=1_F1_R=1_R$ in $R$ is all you need to show that $1_R=1_F$.
– John Douma
Nov 22 at 15:55
The reason I guess for the assumption $1_F in R$ is the choice of the definition of an `Integral domain' as a non zero ring with identity.
– R.C.Cowsik
Nov 25 at 13:55
@R.C.Cowsikthanks for your ansewer. But, if we consider $Rsubseteq F$, when $R$ is an integer domain and $F$ is a field, that is a commutative ring with unity without zero divisors, then necessarily $1_R=1_F$, otherwise $1_R$ it would be a divisor of the zero of $F$. Right?
– Jack J.
Nov 25 at 14:41
1
1
$1_F=1_F1_R=1_R$ in $R$ is all you need to show that $1_R=1_F$.
– John Douma
Nov 22 at 15:55
$1_F=1_F1_R=1_R$ in $R$ is all you need to show that $1_R=1_F$.
– John Douma
Nov 22 at 15:55
The reason I guess for the assumption $1_F in R$ is the choice of the definition of an `Integral domain' as a non zero ring with identity.
– R.C.Cowsik
Nov 25 at 13:55
The reason I guess for the assumption $1_F in R$ is the choice of the definition of an `Integral domain' as a non zero ring with identity.
– R.C.Cowsik
Nov 25 at 13:55
@R.C.Cowsikthanks for your ansewer. But, if we consider $Rsubseteq F$, when $R$ is an integer domain and $F$ is a field, that is a commutative ring with unity without zero divisors, then necessarily $1_R=1_F$, otherwise $1_R$ it would be a divisor of the zero of $F$. Right?
– Jack J.
Nov 25 at 14:41
@R.C.Cowsikthanks for your ansewer. But, if we consider $Rsubseteq F$, when $R$ is an integer domain and $F$ is a field, that is a commutative ring with unity without zero divisors, then necessarily $1_R=1_F$, otherwise $1_R$ it would be a divisor of the zero of $F$. Right?
– Jack J.
Nov 25 at 14:41
add a comment |
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1
$1_F=1_F1_R=1_R$ in $R$ is all you need to show that $1_R=1_F$.
– John Douma
Nov 22 at 15:55
The reason I guess for the assumption $1_F in R$ is the choice of the definition of an `Integral domain' as a non zero ring with identity.
– R.C.Cowsik
Nov 25 at 13:55
@R.C.Cowsikthanks for your ansewer. But, if we consider $Rsubseteq F$, when $R$ is an integer domain and $F$ is a field, that is a commutative ring with unity without zero divisors, then necessarily $1_R=1_F$, otherwise $1_R$ it would be a divisor of the zero of $F$. Right?
– Jack J.
Nov 25 at 14:41