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Uniqueness of finite field [duplicate]
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Finite fields are isomorphic
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Assume $L$ is the algebraic closure of $mathbbF_p$. Show there exists a unique finite field of cardinality $p^n$ containing $mathbbF_p$. The existence is easy just have to define the splitting field of $X^p^n-X$. But what about uniqueness?
abstract-algebra galois-theory finite-fields
asked Jul 28 at 12:35
Xavier Yang
440314
marked as duplicate by Dietrich Burde abstract-algebra
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Jul 28 at 12:51
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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This question already has an answer here:
Finite fields are isomorphic
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Assume $L$ is the algebraic closure of $mathbbF_p$. Show there exists a unique finite field of cardinality $p^n$ containing $mathbbF_p$. The existence is easy just have to define the splitting field of $X^p^n-X$. But what about uniqueness?
abstract-algebra galois-theory finite-fields
asked Jul 28 at 12:35
Xavier Yang
440314
marked as duplicate by Dietrich Burde abstract-algebra
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Jul 28 at 12:51
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Well, there are at most $p^n$ solutions in $overlinemathbb F_p$ to the equation $x^p^n-x=0$. On the other hand, all the elements of a field of $p^n$ elements must satisfy the equation $x^p^n-x=0$. So the elements of such a field are exactly those $p^n$ solutions.
– Saucy O'Path
Jul 28 at 12:42
@SaucyO'Path: Why the element of that field with $p^n$ must satisfy the equation? How can I deduce it without mentioning the fact that every finite field has cyclic multiplicative group.
– Xavier Yang
Jul 28 at 14:12
You don't need it to be cyclic. You just need it to be a group (which it is by definition).
– Saucy O'Path
Jul 28 at 14:14
add a comment |Â
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0
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up vote
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down vote
favorite
This question already has an answer here:
Finite fields are isomorphic
2 answers
Assume $L$ is the algebraic closure of $mathbbF_p$. Show there exists a unique finite field of cardinality $p^n$ containing $mathbbF_p$. The existence is easy just have to define the splitting field of $X^p^n-X$. But what about uniqueness?
abstract-algebra galois-theory finite-fields
asked Jul 28 at 12:35
Xavier Yang
440314
This question already has an answer here:
Finite fields are isomorphic
2 answers
Assume $L$ is the algebraic closure of $mathbbF_p$. Show there exists a unique finite field of cardinality $p^n$ containing $mathbbF_p$. The existence is easy just have to define the splitting field of $X^p^n-X$. But what about uniqueness?
This question already has an answer here:
Finite fields are isomorphic
2 answers
abstract-algebra galois-theory finite-fields
asked Jul 28 at 12:35
Xavier Yang
440314
asked Jul 28 at 12:35
Xavier Yang
440314
asked Jul 28 at 12:35
Xavier Yang
440314
asked Jul 28 at 12:35
Xavier Yang
440314
440314
marked as duplicate by Dietrich Burde abstract-algebra
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Jul 28 at 12:51
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Dietrich Burde abstract-algebra
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Jul 28 at 12:51
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Well, there are at most $p^n$ solutions in $overlinemathbb F_p$ to the equation $x^p^n-x=0$. On the other hand, all the elements of a field of $p^n$ elements must satisfy the equation $x^p^n-x=0$. So the elements of such a field are exactly those $p^n$ solutions.
– Saucy O'Path
Jul 28 at 12:42
@SaucyO'Path: Why the element of that field with $p^n$ must satisfy the equation? How can I deduce it without mentioning the fact that every finite field has cyclic multiplicative group.
– Xavier Yang
Jul 28 at 14:12
You don't need it to be cyclic. You just need it to be a group (which it is by definition).
– Saucy O'Path
Jul 28 at 14:14
add a comment |Â
Well, there are at most $p^n$ solutions in $overlinemathbb F_p$ to the equation $x^p^n-x=0$. On the other hand, all the elements of a field of $p^n$ elements must satisfy the equation $x^p^n-x=0$. So the elements of such a field are exactly those $p^n$ solutions.
– Saucy O'Path
Jul 28 at 12:42
@SaucyO'Path: Why the element of that field with $p^n$ must satisfy the equation? How can I deduce it without mentioning the fact that every finite field has cyclic multiplicative group.
– Xavier Yang
Jul 28 at 14:12
You don't need it to be cyclic. You just need it to be a group (which it is by definition).
– Saucy O'Path
Jul 28 at 14:14
Well, there are at most $p^n$ solutions in $overlinemathbb F_p$ to the equation $x^p^n-x=0$. On the other hand, all the elements of a field of $p^n$ elements must satisfy the equation $x^p^n-x=0$. So the elements of such a field are exactly those $p^n$ solutions.
– Saucy O'Path
Jul 28 at 12:42
Well, there are at most $p^n$ solutions in $overlinemathbb F_p$ to the equation $x^p^n-x=0$. On the other hand, all the elements of a field of $p^n$ elements must satisfy the equation $x^p^n-x=0$. So the elements of such a field are exactly those $p^n$ solutions.
– Saucy O'Path
Jul 28 at 12:42
@SaucyO'Path: Why the element of that field with $p^n$ must satisfy the equation? How can I deduce it without mentioning the fact that every finite field has cyclic multiplicative group.
– Xavier Yang
Jul 28 at 14:12
@SaucyO'Path: Why the element of that field with $p^n$ must satisfy the equation? How can I deduce it without mentioning the fact that every finite field has cyclic multiplicative group.
– Xavier Yang
Jul 28 at 14:12
You don't need it to be cyclic. You just need it to be a group (which it is by definition).
– Saucy O'Path
Jul 28 at 14:14
You don't need it to be cyclic. You just need it to be a group (which it is by definition).
– Saucy O'Path
Jul 28 at 14:14
add a comment |Â
2 Answers
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HINT: Prove that every element of such extension is a root of $x^p^n - x$.
answered Jul 28 at 12:42
Stefan4024
28k52974
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up vote
1
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The multiplicative subgroup of nonzero elements of a field with $p^n$ elements is an abelian group of $p^n - 1$ elements; every element is a root of $X^p^n - 1 - 1$.
answered Jul 28 at 12:45
Hurkyl
107k9112253
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
HINT: Prove that every element of such extension is a root of $x^p^n - x$.
answered Jul 28 at 12:42
Stefan4024
28k52974
add a comment |Â
up vote
2
down vote
accepted
HINT: Prove that every element of such extension is a root of $x^p^n - x$.
answered Jul 28 at 12:42
Stefan4024
28k52974
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
HINT: Prove that every element of such extension is a root of $x^p^n - x$.
answered Jul 28 at 12:42
Stefan4024
28k52974
HINT: Prove that every element of such extension is a root of $x^p^n - x$.
answered Jul 28 at 12:42
Stefan4024
28k52974
answered Jul 28 at 12:42
Stefan4024
28k52974
answered Jul 28 at 12:42
Stefan4024
28k52974
answered Jul 28 at 12:42
Stefan4024
28k52974
28k52974
add a comment |Â
add a comment |Â
up vote
1
down vote
The multiplicative subgroup of nonzero elements of a field with $p^n$ elements is an abelian group of $p^n - 1$ elements; every element is a root of $X^p^n - 1 - 1$.
answered Jul 28 at 12:45
Hurkyl
107k9112253
add a comment |Â
up vote
1
down vote
The multiplicative subgroup of nonzero elements of a field with $p^n$ elements is an abelian group of $p^n - 1$ elements; every element is a root of $X^p^n - 1 - 1$.
answered Jul 28 at 12:45
Hurkyl
107k9112253
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The multiplicative subgroup of nonzero elements of a field with $p^n$ elements is an abelian group of $p^n - 1$ elements; every element is a root of $X^p^n - 1 - 1$.
answered Jul 28 at 12:45
Hurkyl
107k9112253
The multiplicative subgroup of nonzero elements of a field with $p^n$ elements is an abelian group of $p^n - 1$ elements; every element is a root of $X^p^n - 1 - 1$.
answered Jul 28 at 12:45
Hurkyl
107k9112253
answered Jul 28 at 12:45
Hurkyl
107k9112253
answered Jul 28 at 12:45
Hurkyl
107k9112253
answered Jul 28 at 12:45
Hurkyl
107k9112253
107k9112253
add a comment |Â
add a comment |Â
Well, there are at most $p^n$ solutions in $overlinemathbb F_p$ to the equation $x^p^n-x=0$. On the other hand, all the elements of a field of $p^n$ elements must satisfy the equation $x^p^n-x=0$. So the elements of such a field are exactly those $p^n$ solutions.
– Saucy O'Path
Jul 28 at 12:42
@SaucyO'Path: Why the element of that field with $p^n$ must satisfy the equation? How can I deduce it without mentioning the fact that every finite field has cyclic multiplicative group.
– Xavier Yang
Jul 28 at 14:12
You don't need it to be cyclic. You just need it to be a group (which it is by definition).
– Saucy O'Path
Jul 28 at 14:14