Mixing
If A is a finite integral domain and is a cyclic group with addition alone, then order of A is prime
$begingroup$
Here's my attempt at it:
Let $A$ be a finite integral domain and $ain A$ and suppose that $A=langle a rangle$. Hence, $|a|=|A|$. By definition of the order of an element, we have that $|a| cdot a = |A| cdot a = 0$. If $p$ is the characteristic of $A$ (which must be prime), then $p$ divides $|A|$. Also since $p$ is the characteristic of $A$ it must be that $p cdot a = 0$. Hence $|A| = |a|$ divides $p$. Thus, we have $p=|A|$.
Is this proof okay? This question is from Charles Pinter's A book of Abstract Algebra.
abstract-algebra proof-verification integral-domain
asked Jan 15 at 9:13
Ashish KAshish K
910613
$endgroup$
add a comment |
$begingroup$
Here's my attempt at it:
Let $A$ be a finite integral domain and $ain A$ and suppose that $A=langle a rangle$. Hence, $|a|=|A|$. By definition of the order of an element, we have that $|a| cdot a = |A| cdot a = 0$. If $p$ is the characteristic of $A$ (which must be prime), then $p$ divides $|A|$. Also since $p$ is the characteristic of $A$ it must be that $p cdot a = 0$. Hence $|A| = |a|$ divides $p$. Thus, we have $p=|A|$.
Is this proof okay? This question is from Charles Pinter's A book of Abstract Algebra.
abstract-algebra proof-verification integral-domain
asked Jan 15 at 9:13
Ashish KAshish K
910613
$endgroup$
1
$begingroup$
yep, good argumen! I would have gone over showing A is a field, but yours is actually faster.
$endgroup$
– Enkidu
Jan 15 at 9:21
add a comment |
$begingroup$
Here's my attempt at it:
Let $A$ be a finite integral domain and $ain A$ and suppose that $A=langle a rangle$. Hence, $|a|=|A|$. By definition of the order of an element, we have that $|a| cdot a = |A| cdot a = 0$. If $p$ is the characteristic of $A$ (which must be prime), then $p$ divides $|A|$. Also since $p$ is the characteristic of $A$ it must be that $p cdot a = 0$. Hence $|A| = |a|$ divides $p$. Thus, we have $p=|A|$.
Is this proof okay? This question is from Charles Pinter's A book of Abstract Algebra.
abstract-algebra proof-verification integral-domain
asked Jan 15 at 9:13
Ashish KAshish K
910613
$endgroup$
Here's my attempt at it:
Let $A$ be a finite integral domain and $ain A$ and suppose that $A=langle a rangle$. Hence, $|a|=|A|$. By definition of the order of an element, we have that $|a| cdot a = |A| cdot a = 0$. If $p$ is the characteristic of $A$ (which must be prime), then $p$ divides $|A|$. Also since $p$ is the characteristic of $A$ it must be that $p cdot a = 0$. Hence $|A| = |a|$ divides $p$. Thus, we have $p=|A|$.
Is this proof okay? This question is from Charles Pinter's A book of Abstract Algebra.
abstract-algebra proof-verification integral-domain
abstract-algebra proof-verification integral-domain
asked Jan 15 at 9:13
Ashish KAshish K
910613
asked Jan 15 at 9:13
Ashish KAshish K
910613
asked Jan 15 at 9:13
Ashish KAshish K
910613
asked Jan 15 at 9:13
Ashish KAshish K
910613
asked Jan 15 at 9:13
Ashish KAshish K
910613
910613
1
$begingroup$
yep, good argumen! I would have gone over showing A is a field, but yours is actually faster.
$endgroup$
– Enkidu
Jan 15 at 9:21
add a comment |
1
$begingroup$
yep, good argumen! I would have gone over showing A is a field, but yours is actually faster.
$endgroup$
– Enkidu
Jan 15 at 9:21
1
1
$begingroup$
yep, good argumen! I would have gone over showing A is a field, but yours is actually faster.
$endgroup$
– Enkidu
Jan 15 at 9:21
$begingroup$
yep, good argumen! I would have gone over showing A is a field, but yours is actually faster.
$endgroup$
– Enkidu
Jan 15 at 9:21
add a comment |
1 Answer
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yep it is!
I would have gone over the intermediate step of proving it is a field. And since I feel bad just putting "yep it is" as answer, I just attach it:
as $A$ is integral domain, $xcdot: A to A$ is injective for all $xneq 0$, in particular since $A$ is finite, it is invertible. But that means that $R$ is a field. So it is a finite field, and hence has a primepower of elements. Now since it is cyclic this means that it is isomorphis to $mathbb{Z}/p^nmathbb{Z}$. but for this to be a field we need $n=1$ and we are done.
Respectively, if you are unhappy with the argument about $mathbb{Z}/p^nmathbb{Z}$, one can also (since one now knows precisely the structure of $A$) argue by saying that if it is not a prime number it is a vector space over a prime field and hence can't be cyclic.
answered Jan 15 at 9:23
EnkiduEnkidu
1,36719
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1 Answer
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$begingroup$
yep it is!
I would have gone over the intermediate step of proving it is a field. And since I feel bad just putting "yep it is" as answer, I just attach it:
as $A$ is integral domain, $xcdot: A to A$ is injective for all $xneq 0$, in particular since $A$ is finite, it is invertible. But that means that $R$ is a field. So it is a finite field, and hence has a primepower of elements. Now since it is cyclic this means that it is isomorphis to $mathbb{Z}/p^nmathbb{Z}$. but for this to be a field we need $n=1$ and we are done.
Respectively, if you are unhappy with the argument about $mathbb{Z}/p^nmathbb{Z}$, one can also (since one now knows precisely the structure of $A$) argue by saying that if it is not a prime number it is a vector space over a prime field and hence can't be cyclic.
answered Jan 15 at 9:23
EnkiduEnkidu
1,36719
$endgroup$
add a comment |
$begingroup$
yep it is!
I would have gone over the intermediate step of proving it is a field. And since I feel bad just putting "yep it is" as answer, I just attach it:
as $A$ is integral domain, $xcdot: A to A$ is injective for all $xneq 0$, in particular since $A$ is finite, it is invertible. But that means that $R$ is a field. So it is a finite field, and hence has a primepower of elements. Now since it is cyclic this means that it is isomorphis to $mathbb{Z}/p^nmathbb{Z}$. but for this to be a field we need $n=1$ and we are done.
Respectively, if you are unhappy with the argument about $mathbb{Z}/p^nmathbb{Z}$, one can also (since one now knows precisely the structure of $A$) argue by saying that if it is not a prime number it is a vector space over a prime field and hence can't be cyclic.
answered Jan 15 at 9:23
EnkiduEnkidu
1,36719
$endgroup$
add a comment |
$begingroup$
yep it is!
I would have gone over the intermediate step of proving it is a field. And since I feel bad just putting "yep it is" as answer, I just attach it:
as $A$ is integral domain, $xcdot: A to A$ is injective for all $xneq 0$, in particular since $A$ is finite, it is invertible. But that means that $R$ is a field. So it is a finite field, and hence has a primepower of elements. Now since it is cyclic this means that it is isomorphis to $mathbb{Z}/p^nmathbb{Z}$. but for this to be a field we need $n=1$ and we are done.
Respectively, if you are unhappy with the argument about $mathbb{Z}/p^nmathbb{Z}$, one can also (since one now knows precisely the structure of $A$) argue by saying that if it is not a prime number it is a vector space over a prime field and hence can't be cyclic.
answered Jan 15 at 9:23
EnkiduEnkidu
1,36719
$endgroup$
yep it is!
I would have gone over the intermediate step of proving it is a field. And since I feel bad just putting "yep it is" as answer, I just attach it:
as $A$ is integral domain, $xcdot: A to A$ is injective for all $xneq 0$, in particular since $A$ is finite, it is invertible. But that means that $R$ is a field. So it is a finite field, and hence has a primepower of elements. Now since it is cyclic this means that it is isomorphis to $mathbb{Z}/p^nmathbb{Z}$. but for this to be a field we need $n=1$ and we are done.
Respectively, if you are unhappy with the argument about $mathbb{Z}/p^nmathbb{Z}$, one can also (since one now knows precisely the structure of $A$) argue by saying that if it is not a prime number it is a vector space over a prime field and hence can't be cyclic.
answered Jan 15 at 9:23
EnkiduEnkidu
1,36719
answered Jan 15 at 9:23
EnkiduEnkidu
1,36719
answered Jan 15 at 9:23
EnkiduEnkidu
1,36719
answered Jan 15 at 9:23
EnkiduEnkidu
1,36719
1,36719
add a comment |
add a comment |
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$begingroup$
yep, good argumen! I would have gone over showing A is a field, but yours is actually faster.
$endgroup$
– Enkidu
Jan 15 at 9:21