There exist atmost one finite field (upto isomorphism) of any given order: true or false?

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Is the following statement true or false:

There exists at most one finite field (up to isomorphism) of any given order.

I think this statement is false, because I know that $mathbbZ/pmathbbZ$ is a field if $p$ is prime, but $mathbbZ/nmathbbZ$ will not be a field if $n = 4,6,8, ldots$

Is that correct?

abstract-algebra

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edited Jul 17 at 7:35

Louis

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asked Jul 17 at 6:25

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  Note that the fact $mathbbZ / 4 mathbbZ$ is not a field does not imply there can't be some other ring which is a field with four elements. And, in fact, there is a field of four elements.
  – Hurkyl
  Jul 17 at 8:19
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Note that the statement says "at most one" i.e. zero or one. Just because there is no finite field of order 6 (for example) does not make the statement false.
  – gandalf61
  Jul 17 at 8:50
  
  

  
  
  
  

add a comment | 

up vote
-1
down vote

favorite

Is the following statement true or false:

There exists at most one finite field (up to isomorphism) of any given order.

I think this statement is false, because I know that $mathbbZ/pmathbbZ$ is a field if $p$ is prime, but $mathbbZ/nmathbbZ$ will not be a field if $n = 4,6,8, ldots$

Is that correct?

abstract-algebra

share|cite|improve this question

edited Jul 17 at 7:35

Louis

2,28011228

asked Jul 17 at 6:25

stupid

55819

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Note that the fact $mathbbZ / 4 mathbbZ$ is not a field does not imply there can't be some other ring which is a field with four elements. And, in fact, there is a field of four elements.
  – Hurkyl
  Jul 17 at 8:19
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Note that the statement says "at most one" i.e. zero or one. Just because there is no finite field of order 6 (for example) does not make the statement false.
  – gandalf61
  Jul 17 at 8:50
  
  

  
  
  
  

add a comment | 

up vote
-1
down vote

favorite

up vote
-1
down vote

favorite

Is the following statement true or false:

There exists at most one finite field (up to isomorphism) of any given order.

I think this statement is false, because I know that $mathbbZ/pmathbbZ$ is a field if $p$ is prime, but $mathbbZ/nmathbbZ$ will not be a field if $n = 4,6,8, ldots$

Is that correct?

abstract-algebra

share|cite|improve this question

edited Jul 17 at 7:35

Louis

2,28011228

asked Jul 17 at 6:25

stupid

55819

Is the following statement true or false:

There exists at most one finite field (up to isomorphism) of any given order.

I think this statement is false, because I know that $mathbbZ/pmathbbZ$ is a field if $p$ is prime, but $mathbbZ/nmathbbZ$ will not be a field if $n = 4,6,8, ldots$

Is that correct?

abstract-algebra

share|cite|improve this question

edited Jul 17 at 7:35

Louis

2,28011228

asked Jul 17 at 6:25

stupid

55819

share|cite|improve this question

share|cite|improve this question

edited Jul 17 at 7:35

Louis

2,28011228

edited Jul 17 at 7:35

Louis

2,28011228

edited Jul 17 at 7:35

Louis

2,28011228

2,28011228

asked Jul 17 at 6:25

stupid

55819

asked Jul 17 at 6:25

stupid

55819

asked Jul 17 at 6:25

stupid

55819

55819

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Note that the fact $mathbbZ / 4 mathbbZ$ is not a field does not imply there can't be some other ring which is a field with four elements. And, in fact, there is a field of four elements.
  – Hurkyl
  Jul 17 at 8:19
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Note that the statement says "at most one" i.e. zero or one. Just because there is no finite field of order 6 (for example) does not make the statement false.
  – gandalf61
  Jul 17 at 8:50
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Note that the fact $mathbbZ / 4 mathbbZ$ is not a field does not imply there can't be some other ring which is a field with four elements. And, in fact, there is a field of four elements.
  – Hurkyl
  Jul 17 at 8:19
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Note that the statement says "at most one" i.e. zero or one. Just because there is no finite field of order 6 (for example) does not make the statement false.
  – gandalf61
  Jul 17 at 8:50
  
  

  
  
  
  

1

1

Note that the fact $mathbbZ / 4 mathbbZ$ is not a field does not imply there can't be some other ring which is a field with four elements. And, in fact, there is a field of four elements.
– Hurkyl
Jul 17 at 8:19

Note that the fact $mathbbZ / 4 mathbbZ$ is not a field does not imply there can't be some other ring which is a field with four elements. And, in fact, there is a field of four elements.
– Hurkyl
Jul 17 at 8:19

2

2

Note that the statement says "at most one" i.e. zero or one. Just because there is no finite field of order 6 (for example) does not make the statement false.
– gandalf61
Jul 17 at 8:50

Note that the statement says "at most one" i.e. zero or one. Just because there is no finite field of order 6 (for example) does not make the statement false.
– gandalf61
Jul 17 at 8:50

add a comment | 

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The statement is true. You can visit Wikipedia. Notice that if $p$ and $q$ are two different primes, $mathbb Z_p$ and $mathbb Z_q$ have different offer, so the statement doesn't apply here. Your statement says that there exists only one (up to isomorphism) finite field of order $p$, and that this result can be applied to every $p$. I guess you are thinking that there is only one finite field, which is obviously false as you have pointed out.

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answered Jul 17 at 6:42

Dog_69

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accepted

The statement is true. You can visit Wikipedia. Notice that if $p$ and $q$ are two different primes, $mathbb Z_p$ and $mathbb Z_q$ have different offer, so the statement doesn't apply here. Your statement says that there exists only one (up to isomorphism) finite field of order $p$, and that this result can be applied to every $p$. I guess you are thinking that there is only one finite field, which is obviously false as you have pointed out.

share|cite|improve this answer

answered Jul 17 at 6:42

Dog_69

3551320

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up vote
1
down vote



accepted

The statement is true. You can visit Wikipedia. Notice that if $p$ and $q$ are two different primes, $mathbb Z_p$ and $mathbb Z_q$ have different offer, so the statement doesn't apply here. Your statement says that there exists only one (up to isomorphism) finite field of order $p$, and that this result can be applied to every $p$. I guess you are thinking that there is only one finite field, which is obviously false as you have pointed out.

share|cite|improve this answer

answered Jul 17 at 6:42

Dog_69

3551320

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

The statement is true. You can visit Wikipedia. Notice that if $p$ and $q$ are two different primes, $mathbb Z_p$ and $mathbb Z_q$ have different offer, so the statement doesn't apply here. Your statement says that there exists only one (up to isomorphism) finite field of order $p$, and that this result can be applied to every $p$. I guess you are thinking that there is only one finite field, which is obviously false as you have pointed out.

share|cite|improve this answer

answered Jul 17 at 6:42

Dog_69

3551320

The statement is true. You can visit Wikipedia. Notice that if $p$ and $q$ are two different primes, $mathbb Z_p$ and $mathbb Z_q$ have different offer, so the statement doesn't apply here. Your statement says that there exists only one (up to isomorphism) finite field of order $p$, and that this result can be applied to every $p$. I guess you are thinking that there is only one finite field, which is obviously false as you have pointed out.

share|cite|improve this answer

answered Jul 17 at 6:42

Dog_69

3551320

share|cite|improve this answer

share|cite|improve this answer

answered Jul 17 at 6:42

Dog_69

3551320

answered Jul 17 at 6:42

Dog_69

3551320

answered Jul 17 at 6:42

Dog_69

3551320

3551320

add a comment | 

add a comment | 

 

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