Show that $K$ is a field [duplicate]

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This question already has an answer here:

• 
  Prove that $mathbb F_8=mathbb F_2[X]/(X^3+X+1)$
  
  2 answers
  
  

Let $f = x^3+x+1 in mathbb{F}_2[x]$. We know that $K = mathbb{F}_2[x] / langle f rangle$ is a ring. I just need to show that is also a field.

Its actually the first time that I need to work with a polynomial in $mathbb{F}_2[x]$ and Im not sure how to show that $K$ is a field.

abstract-algebra finite-fields

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edited Jan 20 at 15:45

J. W. Tanner

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asked Jan 20 at 11:24

ArjihadArjihad

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Jan 20 at 12:14

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.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you know why it's enough to show that $f $ is irreducible over $mathbb{F}_2$ ?
  $endgroup$
  – Max
  Jan 20 at 11:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is given as the hint for this task but Im not sure why this is enough to show.
  $endgroup$
  – Arjihad
  Jan 20 at 11:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See this duplicate. Indeed, $f$ is irreducible over $Bbb F_2$.
  $endgroup$
  – Dietrich Burde
  Jan 20 at 11:58
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Arjihad, if f is irreducible, then it is a prime element. If it is a prime element, the ideal is prime. If it is prime, it is maximal.This is true because $F_2[x]$ is an PID
  $endgroup$
  – IAmNoOne
  Jan 20 at 12:06
  

  
  
  
  

add a comment | 

0

$begingroup$

This question already has an answer here:

• 
  Prove that $mathbb F_8=mathbb F_2[X]/(X^3+X+1)$
  
  2 answers
  
  

Let $f = x^3+x+1 in mathbb{F}_2[x]$. We know that $K = mathbb{F}_2[x] / langle f rangle$ is a ring. I just need to show that is also a field.

Its actually the first time that I need to work with a polynomial in $mathbb{F}_2[x]$ and Im not sure how to show that $K$ is a field.

abstract-algebra finite-fields

share|cite|improve this question

edited Jan 20 at 15:45

J. W. Tanner

2,6561217

asked Jan 20 at 11:24

ArjihadArjihad

390112

$endgroup$

marked as duplicate by Dietrich Burde abstract-algebra
Users with the  abstract-algebra badge can single-handedly close abstract-algebra questions as duplicates and reopen them as needed.

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Jan 20 at 12:14

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.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you know why it's enough to show that $f $ is irreducible over $mathbb{F}_2$ ?
  $endgroup$
  – Max
  Jan 20 at 11:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is given as the hint for this task but Im not sure why this is enough to show.
  $endgroup$
  – Arjihad
  Jan 20 at 11:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See this duplicate. Indeed, $f$ is irreducible over $Bbb F_2$.
  $endgroup$
  – Dietrich Burde
  Jan 20 at 11:58
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Arjihad, if f is irreducible, then it is a prime element. If it is a prime element, the ideal is prime. If it is prime, it is maximal.This is true because $F_2[x]$ is an PID
  $endgroup$
  – IAmNoOne
  Jan 20 at 12:06
  

  
  
  
  

add a comment | 

0

0

0

1

$begingroup$

This question already has an answer here:

• 
  Prove that $mathbb F_8=mathbb F_2[X]/(X^3+X+1)$
  
  2 answers
  
  

Let $f = x^3+x+1 in mathbb{F}_2[x]$. We know that $K = mathbb{F}_2[x] / langle f rangle$ is a ring. I just need to show that is also a field.

Its actually the first time that I need to work with a polynomial in $mathbb{F}_2[x]$ and Im not sure how to show that $K$ is a field.

abstract-algebra finite-fields

share|cite|improve this question

edited Jan 20 at 15:45

J. W. Tanner

2,6561217

asked Jan 20 at 11:24

ArjihadArjihad

390112

$endgroup$

This question already has an answer here:

• 
  Prove that $mathbb F_8=mathbb F_2[X]/(X^3+X+1)$
  
  2 answers
  
  

Let $f = x^3+x+1 in mathbb{F}_2[x]$. We know that $K = mathbb{F}_2[x] / langle f rangle$ is a ring. I just need to show that is also a field.

Its actually the first time that I need to work with a polynomial in $mathbb{F}_2[x]$ and Im not sure how to show that $K$ is a field.

This question already has an answer here:

• 
  Prove that $mathbb F_8=mathbb F_2[X]/(X^3+X+1)$
  
  2 answers
  
  

abstract-algebra finite-fields

abstract-algebra finite-fields

share|cite|improve this question

edited Jan 20 at 15:45

J. W. Tanner

2,6561217

asked Jan 20 at 11:24

ArjihadArjihad

390112

share|cite|improve this question

edited Jan 20 at 15:45

J. W. Tanner

2,6561217

asked Jan 20 at 11:24

ArjihadArjihad

390112

share|cite|improve this question

share|cite|improve this question

edited Jan 20 at 15:45

J. W. Tanner

2,6561217

edited Jan 20 at 15:45

J. W. Tanner

2,6561217

edited Jan 20 at 15:45

J. W. Tanner

2,6561217

2,6561217

asked Jan 20 at 11:24

ArjihadArjihad

390112

asked Jan 20 at 11:24

ArjihadArjihad

390112

asked Jan 20 at 11:24

ArjihadArjihad

390112

390112

marked as duplicate by Dietrich Burde abstract-algebra
Users with the  abstract-algebra badge can single-handedly close abstract-algebra questions as duplicates and reopen them as needed.

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Jan 20 at 12:14

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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Jan 20 at 12:14

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.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you know why it's enough to show that $f $ is irreducible over $mathbb{F}_2$ ?
  $endgroup$
  – Max
  Jan 20 at 11:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is given as the hint for this task but Im not sure why this is enough to show.
  $endgroup$
  – Arjihad
  Jan 20 at 11:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See this duplicate. Indeed, $f$ is irreducible over $Bbb F_2$.
  $endgroup$
  – Dietrich Burde
  Jan 20 at 11:58
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Arjihad, if f is irreducible, then it is a prime element. If it is a prime element, the ideal is prime. If it is prime, it is maximal.This is true because $F_2[x]$ is an PID
  $endgroup$
  – IAmNoOne
  Jan 20 at 12:06
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you know why it's enough to show that $f $ is irreducible over $mathbb{F}_2$ ?
  $endgroup$
  – Max
  Jan 20 at 11:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is given as the hint for this task but Im not sure why this is enough to show.
  $endgroup$
  – Arjihad
  Jan 20 at 11:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See this duplicate. Indeed, $f$ is irreducible over $Bbb F_2$.
  $endgroup$
  – Dietrich Burde
  Jan 20 at 11:58
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Arjihad, if f is irreducible, then it is a prime element. If it is a prime element, the ideal is prime. If it is prime, it is maximal.This is true because $F_2[x]$ is an PID
  $endgroup$
  – IAmNoOne
  Jan 20 at 12:06
  

  
  
  
  

$begingroup$
Do you know why it's enough to show that $f $ is irreducible over $mathbb{F}_2$ ?
$endgroup$
– Max
Jan 20 at 11:38

$begingroup$
Do you know why it's enough to show that $f $ is irreducible over $mathbb{F}_2$ ?
$endgroup$
– Max
Jan 20 at 11:38

$begingroup$
This is given as the hint for this task but Im not sure why this is enough to show.
$endgroup$
– Arjihad
Jan 20 at 11:49

$begingroup$
This is given as the hint for this task but Im not sure why this is enough to show.
$endgroup$
– Arjihad
Jan 20 at 11:49

$begingroup$
See this duplicate. Indeed, $f$ is irreducible over $Bbb F_2$.
$endgroup$
– Dietrich Burde
Jan 20 at 11:58

$begingroup$
See this duplicate. Indeed, $f$ is irreducible over $Bbb F_2$.
$endgroup$
– Dietrich Burde
Jan 20 at 11:58

$begingroup$
@Arjihad, if f is irreducible, then it is a prime element. If it is a prime element, the ideal is prime. If it is prime, it is maximal.This is true because $F_2[x]$ is an PID
$endgroup$
– IAmNoOne
Jan 20 at 12:06

$begingroup$
@Arjihad, if f is irreducible, then it is a prime element. If it is a prime element, the ideal is prime. If it is prime, it is maximal.This is true because $F_2[x]$ is an PID
$endgroup$
– IAmNoOne
Jan 20 at 12:06

add a comment | 

2 Answers
2

active

oldest

votes

1

$begingroup$

One more hint:
The polynomial $f=x^3+x+1$ is irreducible because it is a cubic polynomial which has no root in $mathbf F_2$. As $mathbf F_2[x]$ is a P.I.D., this implies the ideal $(f)$ is prime, so that $mathbf F_2[x]/(f)$ is an integral domain.

share|cite|improve this answer

answered Jan 20 at 11:57

BernardBernard

122k740116

$endgroup$

add a comment | 

0

$begingroup$

List the conditions for a ring to be a field.

• Has multiplicative identity (namely $1$)


• Every element except $0$ has a multiplicative inverse.


• Multiplication commutes (obvious)

So the core problem is:

$$f text{ is irreducible over } mathbb F Leftrightarrow text{All elements }pinmathbb F/langle frangle text{ except $0$ have inverses}.$$

See how this resembles proving that $mathbb Z/pmathbb Z$ forms a field ($p$ prime)? When we prove that every non-zero element in $mathbb Z/pmathbb Z$ has an inverse, we use Bezout's identity:

Suppose $(a,b)=1$. There exist $m,ninmathbb Z$ that $$ma+nb=1.$$

And the proof of that relies on the generalized Euclidean algorithm, which, in turn, requires division with remainder. Now we know that, in $mathbb F_2[x]$ we can perform polynomial division analogous to division with remainder in integers. So you can generalize Bezout's identity to obtain the result.

share|cite|improve this answer

answered Jan 20 at 12:04

TreborTrebor

85415

$endgroup$

add a comment | 

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

1

$begingroup$

One more hint:
The polynomial $f=x^3+x+1$ is irreducible because it is a cubic polynomial which has no root in $mathbf F_2$. As $mathbf F_2[x]$ is a P.I.D., this implies the ideal $(f)$ is prime, so that $mathbf F_2[x]/(f)$ is an integral domain.

share|cite|improve this answer

answered Jan 20 at 11:57

BernardBernard

122k740116

$endgroup$

add a comment | 

1

$begingroup$

One more hint:
The polynomial $f=x^3+x+1$ is irreducible because it is a cubic polynomial which has no root in $mathbf F_2$. As $mathbf F_2[x]$ is a P.I.D., this implies the ideal $(f)$ is prime, so that $mathbf F_2[x]/(f)$ is an integral domain.

share|cite|improve this answer

answered Jan 20 at 11:57

BernardBernard

122k740116

$endgroup$

add a comment | 

1

1

1

$begingroup$

One more hint:
The polynomial $f=x^3+x+1$ is irreducible because it is a cubic polynomial which has no root in $mathbf F_2$. As $mathbf F_2[x]$ is a P.I.D., this implies the ideal $(f)$ is prime, so that $mathbf F_2[x]/(f)$ is an integral domain.

share|cite|improve this answer

answered Jan 20 at 11:57

BernardBernard

122k740116

$endgroup$

One more hint:
The polynomial $f=x^3+x+1$ is irreducible because it is a cubic polynomial which has no root in $mathbf F_2$. As $mathbf F_2[x]$ is a P.I.D., this implies the ideal $(f)$ is prime, so that $mathbf F_2[x]/(f)$ is an integral domain.

share|cite|improve this answer

answered Jan 20 at 11:57

BernardBernard

122k740116

share|cite|improve this answer

share|cite|improve this answer

answered Jan 20 at 11:57

BernardBernard

122k740116

answered Jan 20 at 11:57

BernardBernard

122k740116

answered Jan 20 at 11:57

BernardBernard

122k740116

122k740116

add a comment | 

add a comment | 

0

$begingroup$

List the conditions for a ring to be a field.

• Has multiplicative identity (namely $1$)


• Every element except $0$ has a multiplicative inverse.


• Multiplication commutes (obvious)

So the core problem is:

$$f text{ is irreducible over } mathbb F Leftrightarrow text{All elements }pinmathbb F/langle frangle text{ except $0$ have inverses}.$$

See how this resembles proving that $mathbb Z/pmathbb Z$ forms a field ($p$ prime)? When we prove that every non-zero element in $mathbb Z/pmathbb Z$ has an inverse, we use Bezout's identity:

Suppose $(a,b)=1$. There exist $m,ninmathbb Z$ that $$ma+nb=1.$$

And the proof of that relies on the generalized Euclidean algorithm, which, in turn, requires division with remainder. Now we know that, in $mathbb F_2[x]$ we can perform polynomial division analogous to division with remainder in integers. So you can generalize Bezout's identity to obtain the result.

share|cite|improve this answer

answered Jan 20 at 12:04

TreborTrebor

85415

$endgroup$

add a comment | 

0

$begingroup$

List the conditions for a ring to be a field.

• Has multiplicative identity (namely $1$)


• Every element except $0$ has a multiplicative inverse.


• Multiplication commutes (obvious)

So the core problem is:

$$f text{ is irreducible over } mathbb F Leftrightarrow text{All elements }pinmathbb F/langle frangle text{ except $0$ have inverses}.$$

See how this resembles proving that $mathbb Z/pmathbb Z$ forms a field ($p$ prime)? When we prove that every non-zero element in $mathbb Z/pmathbb Z$ has an inverse, we use Bezout's identity:

Suppose $(a,b)=1$. There exist $m,ninmathbb Z$ that $$ma+nb=1.$$

And the proof of that relies on the generalized Euclidean algorithm, which, in turn, requires division with remainder. Now we know that, in $mathbb F_2[x]$ we can perform polynomial division analogous to division with remainder in integers. So you can generalize Bezout's identity to obtain the result.

share|cite|improve this answer

answered Jan 20 at 12:04

TreborTrebor

85415

$endgroup$

add a comment | 

0

0

0

$begingroup$

List the conditions for a ring to be a field.

• Has multiplicative identity (namely $1$)


• Every element except $0$ has a multiplicative inverse.


• Multiplication commutes (obvious)

So the core problem is:

$$f text{ is irreducible over } mathbb F Leftrightarrow text{All elements }pinmathbb F/langle frangle text{ except $0$ have inverses}.$$

See how this resembles proving that $mathbb Z/pmathbb Z$ forms a field ($p$ prime)? When we prove that every non-zero element in $mathbb Z/pmathbb Z$ has an inverse, we use Bezout's identity:

Suppose $(a,b)=1$. There exist $m,ninmathbb Z$ that $$ma+nb=1.$$

And the proof of that relies on the generalized Euclidean algorithm, which, in turn, requires division with remainder. Now we know that, in $mathbb F_2[x]$ we can perform polynomial division analogous to division with remainder in integers. So you can generalize Bezout's identity to obtain the result.

share|cite|improve this answer

answered Jan 20 at 12:04

TreborTrebor

85415

$endgroup$

List the conditions for a ring to be a field.

• Has multiplicative identity (namely $1$)


• Every element except $0$ has a multiplicative inverse.


• Multiplication commutes (obvious)

So the core problem is:

$$f text{ is irreducible over } mathbb F Leftrightarrow text{All elements }pinmathbb F/langle frangle text{ except $0$ have inverses}.$$

See how this resembles proving that $mathbb Z/pmathbb Z$ forms a field ($p$ prime)? When we prove that every non-zero element in $mathbb Z/pmathbb Z$ has an inverse, we use Bezout's identity:

Suppose $(a,b)=1$. There exist $m,ninmathbb Z$ that $$ma+nb=1.$$

And the proof of that relies on the generalized Euclidean algorithm, which, in turn, requires division with remainder. Now we know that, in $mathbb F_2[x]$ we can perform polynomial division analogous to division with remainder in integers. So you can generalize Bezout's identity to obtain the result.

share|cite|improve this answer

answered Jan 20 at 12:04

TreborTrebor

85415

share|cite|improve this answer

share|cite|improve this answer

answered Jan 20 at 12:04

TreborTrebor

85415

answered Jan 20 at 12:04

TreborTrebor

85415

answered Jan 20 at 12:04

TreborTrebor

85415

85415

add a comment | 

add a comment | 

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