If $A$ and $B$ are integral domains and $A subseteq B$ then $A$ has characteristic $p$ iff $B$ has...
$begingroup$
Here's my attempt :
The unity of $A$ and the unity of $B$ are both same, so, we'll call it $1$ unambiguously.
$( Leftarrow )$ Suppose $A$ has characteristic $p$ then $p cdot 1 = 0$. Since $1$ is also in $B$, it must be that characteristic of $B$, say, $p_{B}$ divides $p$. Also, $p_{B} ne 1$ thus $p_{B}=p$.
$( Rightarrow )$ Suppose $B$ has characteristic $p$ then $p cdot 1 = 0$. Since $1$ is also in $A$, it must be that characteristic of $A$, say, $p_{A}$ divides $p$. Also, $p_{A} ne 1$ thus $p_{A}=p$.
Is the proof correct? This problem is in Pinter's A book of Abstract Algebra.
abstract-algebra integral-domain
asked Jan 15 at 15:16
Ashish KAshish K
856613
$endgroup$
|
show 3 more comments
$begingroup$
Here's my attempt :
The unity of $A$ and the unity of $B$ are both same, so, we'll call it $1$ unambiguously.
$( Leftarrow )$ Suppose $A$ has characteristic $p$ then $p cdot 1 = 0$. Since $1$ is also in $B$, it must be that characteristic of $B$, say, $p_{B}$ divides $p$. Also, $p_{B} ne 1$ thus $p_{B}=p$.
$( Rightarrow )$ Suppose $B$ has characteristic $p$ then $p cdot 1 = 0$. Since $1$ is also in $A$, it must be that characteristic of $A$, say, $p_{A}$ divides $p$. Also, $p_{A} ne 1$ thus $p_{A}=p$.
Is the proof correct? This problem is in Pinter's A book of Abstract Algebra.
abstract-algebra integral-domain
asked Jan 15 at 15:16
Ashish KAshish K
856613
$endgroup$
2
$begingroup$
Yes, it is correct.
$endgroup$
– jgon
Jan 15 at 15:20
$begingroup$
@астонвіллаолофмэллбэрг The assumption is that A and B are both integral domains. So they must contain unity, isn't it?
$endgroup$
– Ashish K
Jan 15 at 15:28
$begingroup$
Does Pinter's definition of a ring include the fact that there exists an element $1$ in every ring? There are other definitions where this condition is not necessary.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 15 at 15:29
1
$begingroup$
@AshishK Maybe even more to the point: in a ring with nonzero identity, the characteristic is exactly the additive order of $1$. This links the two things concisely.
$endgroup$
– rschwieb
Jan 15 at 15:30
1
$begingroup$
@AshishK Ok , that was my only clarification. Your answer is fine.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 15 at 15:58
|
show 3 more comments
$begingroup$
Here's my attempt :
The unity of $A$ and the unity of $B$ are both same, so, we'll call it $1$ unambiguously.
$( Leftarrow )$ Suppose $A$ has characteristic $p$ then $p cdot 1 = 0$. Since $1$ is also in $B$, it must be that characteristic of $B$, say, $p_{B}$ divides $p$. Also, $p_{B} ne 1$ thus $p_{B}=p$.
$( Rightarrow )$ Suppose $B$ has characteristic $p$ then $p cdot 1 = 0$. Since $1$ is also in $A$, it must be that characteristic of $A$, say, $p_{A}$ divides $p$. Also, $p_{A} ne 1$ thus $p_{A}=p$.
Is the proof correct? This problem is in Pinter's A book of Abstract Algebra.
abstract-algebra integral-domain
asked Jan 15 at 15:16
Ashish KAshish K
856613
$endgroup$
Here's my attempt :
The unity of $A$ and the unity of $B$ are both same, so, we'll call it $1$ unambiguously.
$( Leftarrow )$ Suppose $A$ has characteristic $p$ then $p cdot 1 = 0$. Since $1$ is also in $B$, it must be that characteristic of $B$, say, $p_{B}$ divides $p$. Also, $p_{B} ne 1$ thus $p_{B}=p$.
$( Rightarrow )$ Suppose $B$ has characteristic $p$ then $p cdot 1 = 0$. Since $1$ is also in $A$, it must be that characteristic of $A$, say, $p_{A}$ divides $p$. Also, $p_{A} ne 1$ thus $p_{A}=p$.
Is the proof correct? This problem is in Pinter's A book of Abstract Algebra.
abstract-algebra integral-domain
abstract-algebra integral-domain
asked Jan 15 at 15:16
Ashish KAshish K
856613
asked Jan 15 at 15:16
Ashish KAshish K
856613
asked Jan 15 at 15:16
Ashish KAshish K
856613
asked Jan 15 at 15:16
Ashish KAshish K
856613
asked Jan 15 at 15:16
Ashish KAshish K
856613
856613
2
$begingroup$
Yes, it is correct.
$endgroup$
– jgon
Jan 15 at 15:20
$begingroup$
@астонвіллаолофмэллбэрг The assumption is that A and B are both integral domains. So they must contain unity, isn't it?
$endgroup$
– Ashish K
Jan 15 at 15:28
$begingroup$
Does Pinter's definition of a ring include the fact that there exists an element $1$ in every ring? There are other definitions where this condition is not necessary.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 15 at 15:29
1
$begingroup$
@AshishK Maybe even more to the point: in a ring with nonzero identity, the characteristic is exactly the additive order of $1$. This links the two things concisely.
$endgroup$
– rschwieb
Jan 15 at 15:30
1
$begingroup$
@AshishK Ok , that was my only clarification. Your answer is fine.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 15 at 15:58
|
show 3 more comments
2
$begingroup$
Yes, it is correct.
$endgroup$
– jgon
Jan 15 at 15:20
$begingroup$
@астонвіллаолофмэллбэрг The assumption is that A and B are both integral domains. So they must contain unity, isn't it?
$endgroup$
– Ashish K
Jan 15 at 15:28
$begingroup$
Does Pinter's definition of a ring include the fact that there exists an element $1$ in every ring? There are other definitions where this condition is not necessary.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 15 at 15:29
1
$begingroup$
@AshishK Maybe even more to the point: in a ring with nonzero identity, the characteristic is exactly the additive order of $1$. This links the two things concisely.
$endgroup$
– rschwieb
Jan 15 at 15:30
1
$begingroup$
@AshishK Ok , that was my only clarification. Your answer is fine.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 15 at 15:58
2
2
$begingroup$
Yes, it is correct.
$endgroup$
– jgon
Jan 15 at 15:20
$begingroup$
Yes, it is correct.
$endgroup$
– jgon
Jan 15 at 15:20
$begingroup$
@астонвіллаолофмэллбэрг The assumption is that A and B are both integral domains. So they must contain unity, isn't it?
$endgroup$
– Ashish K
Jan 15 at 15:28
$begingroup$
@астонвіллаолофмэллбэрг The assumption is that A and B are both integral domains. So they must contain unity, isn't it?
$endgroup$
– Ashish K
Jan 15 at 15:28
$begingroup$
Does Pinter's definition of a ring include the fact that there exists an element $1$ in every ring? There are other definitions where this condition is not necessary.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 15 at 15:29
$begingroup$
Does Pinter's definition of a ring include the fact that there exists an element $1$ in every ring? There are other definitions where this condition is not necessary.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 15 at 15:29
1
1
$begingroup$
@AshishK Maybe even more to the point: in a ring with nonzero identity, the characteristic is exactly the additive order of $1$. This links the two things concisely.
$endgroup$
– rschwieb
Jan 15 at 15:30
$begingroup$
@AshishK Maybe even more to the point: in a ring with nonzero identity, the characteristic is exactly the additive order of $1$. This links the two things concisely.
$endgroup$
– rschwieb
Jan 15 at 15:30
1
1
$begingroup$
@AshishK Ok , that was my only clarification. Your answer is fine.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 15 at 15:58
$begingroup$
@AshishK Ok , that was my only clarification. Your answer is fine.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 15 at 15:58
|
show 3 more comments
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2
$begingroup$
Yes, it is correct.
$endgroup$
– jgon
Jan 15 at 15:20
$begingroup$
@астонвіллаолофмэллбэрг The assumption is that A and B are both integral domains. So they must contain unity, isn't it?
$endgroup$
– Ashish K
Jan 15 at 15:28
$begingroup$
Does Pinter's definition of a ring include the fact that there exists an element $1$ in every ring? There are other definitions where this condition is not necessary.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 15 at 15:29
1
$begingroup$
@AshishK Maybe even more to the point: in a ring with nonzero identity, the characteristic is exactly the additive order of $1$. This links the two things concisely.
$endgroup$
– rschwieb
Jan 15 at 15:30
1
$begingroup$
@AshishK Ok , that was my only clarification. Your answer is fine.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 15 at 15:58