Mixing
$mathbb{F}_2[X]/(S) cong mathbb{F}_4 $
Let $S(X) = X^2 +X+1 in mathbb{F}_2[X]$
Prove that $mathbb{F}_2[X]/(S) cong mathbb{F}_4 $
What I did:
${1, X }$ is a basis of $mathbb{F}_2[X]/(S)$ and S is irreducible in $mathbb{F}_2$ so $ |mathbb{F}_2[X]/(S)| = 2^2 = 4$
I need help to prove that $mathbb{F}_2[X]/(S) cong mathbb{F}_4 $.
Thank you.
abstract-algebra quotient-group
asked Nov 21 '18 at 3:06
PerelMan
519211
add a comment |
Let $S(X) = X^2 +X+1 in mathbb{F}_2[X]$
Prove that $mathbb{F}_2[X]/(S) cong mathbb{F}_4 $
What I did:
${1, X }$ is a basis of $mathbb{F}_2[X]/(S)$ and S is irreducible in $mathbb{F}_2$ so $ |mathbb{F}_2[X]/(S)| = 2^2 = 4$
I need help to prove that $mathbb{F}_2[X]/(S) cong mathbb{F}_4 $.
Thank you.
abstract-algebra quotient-group
asked Nov 21 '18 at 3:06
PerelMan
519211
3
Do you mean $S(X) = X^2 + X + 1$? Otherwise your basis would be $1,X,X^2,X^3,X^4$ and the quotient would be $mathbb{F}_{2^5}$.
– Trevor Gunn
Nov 21 '18 at 3:14
Thank you, indeed I was wrong and $S(X) = X^2 + X + 1$
– PerelMan
Nov 21 '18 at 9:58
2
What is your definition of $Bbb{F}_4$? Many would define $Bbb{F}_4$ as the quotient ring $Bbb{F}_2[X]/(X^2+X+1)$ :-)
– Jyrki Lahtonen
Nov 21 '18 at 11:25
add a comment |
Let $S(X) = X^2 +X+1 in mathbb{F}_2[X]$
Prove that $mathbb{F}_2[X]/(S) cong mathbb{F}_4 $
What I did:
${1, X }$ is a basis of $mathbb{F}_2[X]/(S)$ and S is irreducible in $mathbb{F}_2$ so $ |mathbb{F}_2[X]/(S)| = 2^2 = 4$
I need help to prove that $mathbb{F}_2[X]/(S) cong mathbb{F}_4 $.
Thank you.
abstract-algebra quotient-group
asked Nov 21 '18 at 3:06
PerelMan
519211
Let $S(X) = X^2 +X+1 in mathbb{F}_2[X]$
Prove that $mathbb{F}_2[X]/(S) cong mathbb{F}_4 $
What I did:
${1, X }$ is a basis of $mathbb{F}_2[X]/(S)$ and S is irreducible in $mathbb{F}_2$ so $ |mathbb{F}_2[X]/(S)| = 2^2 = 4$
I need help to prove that $mathbb{F}_2[X]/(S) cong mathbb{F}_4 $.
Thank you.
abstract-algebra quotient-group
abstract-algebra quotient-group
asked Nov 21 '18 at 3:06
PerelMan
519211
asked Nov 21 '18 at 3:06
PerelMan
519211
edited Nov 21 '18 at 9:58
asked Nov 21 '18 at 3:06
PerelMan
519211
asked Nov 21 '18 at 3:06
PerelMan
519211
asked Nov 21 '18 at 3:06
PerelMan
519211
519211
3
Do you mean $S(X) = X^2 + X + 1$? Otherwise your basis would be $1,X,X^2,X^3,X^4$ and the quotient would be $mathbb{F}_{2^5}$.
– Trevor Gunn
Nov 21 '18 at 3:14
Thank you, indeed I was wrong and $S(X) = X^2 + X + 1$
– PerelMan
Nov 21 '18 at 9:58
2
What is your definition of $Bbb{F}_4$? Many would define $Bbb{F}_4$ as the quotient ring $Bbb{F}_2[X]/(X^2+X+1)$ :-)
– Jyrki Lahtonen
Nov 21 '18 at 11:25
add a comment |
3
Do you mean $S(X) = X^2 + X + 1$? Otherwise your basis would be $1,X,X^2,X^3,X^4$ and the quotient would be $mathbb{F}_{2^5}$.
– Trevor Gunn
Nov 21 '18 at 3:14
Thank you, indeed I was wrong and $S(X) = X^2 + X + 1$
– PerelMan
Nov 21 '18 at 9:58
2
What is your definition of $Bbb{F}_4$? Many would define $Bbb{F}_4$ as the quotient ring $Bbb{F}_2[X]/(X^2+X+1)$ :-)
– Jyrki Lahtonen
Nov 21 '18 at 11:25
3
3
Do you mean $S(X) = X^2 + X + 1$? Otherwise your basis would be $1,X,X^2,X^3,X^4$ and the quotient would be $mathbb{F}_{2^5}$.
– Trevor Gunn
Nov 21 '18 at 3:14
Do you mean $S(X) = X^2 + X + 1$? Otherwise your basis would be $1,X,X^2,X^3,X^4$ and the quotient would be $mathbb{F}_{2^5}$.
– Trevor Gunn
Nov 21 '18 at 3:14
Thank you, indeed I was wrong and $S(X) = X^2 + X + 1$
– PerelMan
Nov 21 '18 at 9:58
Thank you, indeed I was wrong and $S(X) = X^2 + X + 1$
– PerelMan
Nov 21 '18 at 9:58
2
2
What is your definition of $Bbb{F}_4$? Many would define $Bbb{F}_4$ as the quotient ring $Bbb{F}_2[X]/(X^2+X+1)$ :-)
– Jyrki Lahtonen
Nov 21 '18 at 11:25
What is your definition of $Bbb{F}_4$? Many would define $Bbb{F}_4$ as the quotient ring $Bbb{F}_2[X]/(X^2+X+1)$ :-)
– Jyrki Lahtonen
Nov 21 '18 at 11:25
add a comment |
2 Answers
2
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What Trevor Gunn said is right. The statement you want to prove is not correct. First you need to show that $S(X) = X^5 + X^2 + 1$ is irreducible over $mathbb{F}_2$. Then it will follow that $F[X]/langle S rangle$ is a field spanned by the elements $1, X, X^2, X^3, X^4$ and is thus isomorphic to $mathbb{F}_{32}$ (remember that there is exactly one field with 32 elements).
Edit: OP has now changed the polynomial in question to $S(X) = X^2 + X + 1$. In this case the quotient ring $F[X]/langle S(X) rangle$ is a field spanned by the elements $1, X$ and thus has cardinality $2^2 = 4$, as mentioned in the question. The identification with $mathbb{F}_4$ then comes from the fact that, up to isomorphism, there is only one field of cardinality $4$.
answered Nov 21 '18 at 3:32
Monstrous Moonshiner
2,25011337
I fixed my question thank you. Now $F[X]/langle S rangle$ is a field spawned by $1,X $
– PerelMan
Nov 21 '18 at 10:00
add a comment |
It is easy to see that $x^2+x+1$ is irreducible since it has degree $2$ and it has no root. Hence the quotient ring is necessarily a field. Since the quotient has $4$ elements and a finite field is uniquely determined by the number of its elements, you get $mathbb{F}_2[x]/(x^2+x+1)congmathbb{F}_4$.
answered Nov 21 '18 at 10:28
Levent
2,690925
add a comment |
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2 Answers
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2 Answers
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What Trevor Gunn said is right. The statement you want to prove is not correct. First you need to show that $S(X) = X^5 + X^2 + 1$ is irreducible over $mathbb{F}_2$. Then it will follow that $F[X]/langle S rangle$ is a field spanned by the elements $1, X, X^2, X^3, X^4$ and is thus isomorphic to $mathbb{F}_{32}$ (remember that there is exactly one field with 32 elements).
Edit: OP has now changed the polynomial in question to $S(X) = X^2 + X + 1$. In this case the quotient ring $F[X]/langle S(X) rangle$ is a field spanned by the elements $1, X$ and thus has cardinality $2^2 = 4$, as mentioned in the question. The identification with $mathbb{F}_4$ then comes from the fact that, up to isomorphism, there is only one field of cardinality $4$.
answered Nov 21 '18 at 3:32
Monstrous Moonshiner
2,25011337
I fixed my question thank you. Now $F[X]/langle S rangle$ is a field spawned by $1,X $
– PerelMan
Nov 21 '18 at 10:00
add a comment |
What Trevor Gunn said is right. The statement you want to prove is not correct. First you need to show that $S(X) = X^5 + X^2 + 1$ is irreducible over $mathbb{F}_2$. Then it will follow that $F[X]/langle S rangle$ is a field spanned by the elements $1, X, X^2, X^3, X^4$ and is thus isomorphic to $mathbb{F}_{32}$ (remember that there is exactly one field with 32 elements).
Edit: OP has now changed the polynomial in question to $S(X) = X^2 + X + 1$. In this case the quotient ring $F[X]/langle S(X) rangle$ is a field spanned by the elements $1, X$ and thus has cardinality $2^2 = 4$, as mentioned in the question. The identification with $mathbb{F}_4$ then comes from the fact that, up to isomorphism, there is only one field of cardinality $4$.
answered Nov 21 '18 at 3:32
Monstrous Moonshiner
2,25011337
I fixed my question thank you. Now $F[X]/langle S rangle$ is a field spawned by $1,X $
– PerelMan
Nov 21 '18 at 10:00
add a comment |
What Trevor Gunn said is right. The statement you want to prove is not correct. First you need to show that $S(X) = X^5 + X^2 + 1$ is irreducible over $mathbb{F}_2$. Then it will follow that $F[X]/langle S rangle$ is a field spanned by the elements $1, X, X^2, X^3, X^4$ and is thus isomorphic to $mathbb{F}_{32}$ (remember that there is exactly one field with 32 elements).
Edit: OP has now changed the polynomial in question to $S(X) = X^2 + X + 1$. In this case the quotient ring $F[X]/langle S(X) rangle$ is a field spanned by the elements $1, X$ and thus has cardinality $2^2 = 4$, as mentioned in the question. The identification with $mathbb{F}_4$ then comes from the fact that, up to isomorphism, there is only one field of cardinality $4$.
answered Nov 21 '18 at 3:32
Monstrous Moonshiner
2,25011337
What Trevor Gunn said is right. The statement you want to prove is not correct. First you need to show that $S(X) = X^5 + X^2 + 1$ is irreducible over $mathbb{F}_2$. Then it will follow that $F[X]/langle S rangle$ is a field spanned by the elements $1, X, X^2, X^3, X^4$ and is thus isomorphic to $mathbb{F}_{32}$ (remember that there is exactly one field with 32 elements).
Edit: OP has now changed the polynomial in question to $S(X) = X^2 + X + 1$. In this case the quotient ring $F[X]/langle S(X) rangle$ is a field spanned by the elements $1, X$ and thus has cardinality $2^2 = 4$, as mentioned in the question. The identification with $mathbb{F}_4$ then comes from the fact that, up to isomorphism, there is only one field of cardinality $4$.
answered Nov 21 '18 at 3:32
Monstrous Moonshiner
2,25011337
edited Nov 21 '18 at 17:36
answered Nov 21 '18 at 3:32
Monstrous Moonshiner
2,25011337
answered Nov 21 '18 at 3:32
Monstrous Moonshiner
2,25011337
answered Nov 21 '18 at 3:32
Monstrous Moonshiner
2,25011337
2,25011337
I fixed my question thank you. Now $F[X]/langle S rangle$ is a field spawned by $1,X $
– PerelMan
Nov 21 '18 at 10:00
add a comment |
I fixed my question thank you. Now $F[X]/langle S rangle$ is a field spawned by $1,X $
– PerelMan
Nov 21 '18 at 10:00
I fixed my question thank you. Now $F[X]/langle S rangle$ is a field spawned by $1,X $
– PerelMan
Nov 21 '18 at 10:00
I fixed my question thank you. Now $F[X]/langle S rangle$ is a field spawned by $1,X $
– PerelMan
Nov 21 '18 at 10:00
add a comment |
It is easy to see that $x^2+x+1$ is irreducible since it has degree $2$ and it has no root. Hence the quotient ring is necessarily a field. Since the quotient has $4$ elements and a finite field is uniquely determined by the number of its elements, you get $mathbb{F}_2[x]/(x^2+x+1)congmathbb{F}_4$.
answered Nov 21 '18 at 10:28
Levent
2,690925
add a comment |
It is easy to see that $x^2+x+1$ is irreducible since it has degree $2$ and it has no root. Hence the quotient ring is necessarily a field. Since the quotient has $4$ elements and a finite field is uniquely determined by the number of its elements, you get $mathbb{F}_2[x]/(x^2+x+1)congmathbb{F}_4$.
answered Nov 21 '18 at 10:28
Levent
2,690925
add a comment |
It is easy to see that $x^2+x+1$ is irreducible since it has degree $2$ and it has no root. Hence the quotient ring is necessarily a field. Since the quotient has $4$ elements and a finite field is uniquely determined by the number of its elements, you get $mathbb{F}_2[x]/(x^2+x+1)congmathbb{F}_4$.
answered Nov 21 '18 at 10:28
Levent
2,690925
It is easy to see that $x^2+x+1$ is irreducible since it has degree $2$ and it has no root. Hence the quotient ring is necessarily a field. Since the quotient has $4$ elements and a finite field is uniquely determined by the number of its elements, you get $mathbb{F}_2[x]/(x^2+x+1)congmathbb{F}_4$.
answered Nov 21 '18 at 10:28
Levent
2,690925
answered Nov 21 '18 at 10:28
Levent
2,690925
answered Nov 21 '18 at 10:28
Levent
2,690925
answered Nov 21 '18 at 10:28
Levent
2,690925
2,690925
add a comment |
add a comment |
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3
Do you mean $S(X) = X^2 + X + 1$? Otherwise your basis would be $1,X,X^2,X^3,X^4$ and the quotient would be $mathbb{F}_{2^5}$.
– Trevor Gunn
Nov 21 '18 at 3:14
Thank you, indeed I was wrong and $S(X) = X^2 + X + 1$
– PerelMan
Nov 21 '18 at 9:58
2
What is your definition of $Bbb{F}_4$? Many would define $Bbb{F}_4$ as the quotient ring $Bbb{F}_2[X]/(X^2+X+1)$ :-)
– Jyrki Lahtonen
Nov 21 '18 at 11:25