Isomorphism between two versions of $GF(2^3)$ [duplicate]
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This question already has an answer here:
Finding an isomorphism between these two finite fields
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I have $GF(2^3)$ generated by $Pi_1(alpha)=x^3+x+1$ and $GF(2^3)$ generated by $Pi_2(alpha)=x^3+x^2+1$.
$bullet$ $Pi_1(alpha)=x^3+x+1$ $000=0, 100=1,010=alpha,001=alpha^2,110=alpha^3, 011=alpha^4,111=alpha^5,101=alpha^6$
$bullet$ $Pi_2(alpha)=x^3+x^2+1$ $000=0, 100=1,010=lambda,001=lambda^2,101=lambda^3, 111=lambda^4,110=lambda^5,011=lambda^6$
The exampla say that $alpha$ and $lambda^3$ both have minimal polynomial $Pi_1$ and thus $alpha iff lambda^3$ form an isomorphism between the two version.
How can I see in a practical way this fact? Someone can explain me this concept?
abstract-algebra finite-fields cyclotomic-fields
asked Dec 19 '18 at 8:53
AlessarAlessar
313115
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marked as duplicate by Jyrki Lahtonen abstract-algebra
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Dec 19 '18 at 10:06
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.
add a comment |
$begingroup$
This question already has an answer here:
Finding an isomorphism between these two finite fields
2 answers
I have $GF(2^3)$ generated by $Pi_1(alpha)=x^3+x+1$ and $GF(2^3)$ generated by $Pi_2(alpha)=x^3+x^2+1$.
$bullet$ $Pi_1(alpha)=x^3+x+1$ $000=0, 100=1,010=alpha,001=alpha^2,110=alpha^3, 011=alpha^4,111=alpha^5,101=alpha^6$
$bullet$ $Pi_2(alpha)=x^3+x^2+1$ $000=0, 100=1,010=lambda,001=lambda^2,101=lambda^3, 111=lambda^4,110=lambda^5,011=lambda^6$
The exampla say that $alpha$ and $lambda^3$ both have minimal polynomial $Pi_1$ and thus $alpha iff lambda^3$ form an isomorphism between the two version.
How can I see in a practical way this fact? Someone can explain me this concept?
abstract-algebra finite-fields cyclotomic-fields
asked Dec 19 '18 at 8:53
AlessarAlessar
313115
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marked as duplicate by Jyrki Lahtonen abstract-algebra
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Dec 19 '18 at 10:06
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.
1
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It may be easier to show that $alphamapsto 1/lambda$ gives an isomorphism. This follows from the fact that the two polynomials are reciprocals of each other: $$x^3Pi_1(frac1x)=Pi_2(x)$$ implying that $Pi_1(1/lambda)=0$. Gandalf61's solution is, of course, just fine.
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– Jyrki Lahtonen
Dec 19 '18 at 9:58
1
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Closing this as a duplicate because that other question studies exactly the same pair of extension fields. It may have been difficult for you to realize that this is likely the case. So, no harm intended. Just maintaining basic site hygiene.
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– Jyrki Lahtonen
Dec 19 '18 at 10:09
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This is a very interesting clarification! Thank you!
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– Alessar
Dec 19 '18 at 10:15
add a comment |
$begingroup$
This question already has an answer here:
Finding an isomorphism between these two finite fields
2 answers
I have $GF(2^3)$ generated by $Pi_1(alpha)=x^3+x+1$ and $GF(2^3)$ generated by $Pi_2(alpha)=x^3+x^2+1$.
$bullet$ $Pi_1(alpha)=x^3+x+1$ $000=0, 100=1,010=alpha,001=alpha^2,110=alpha^3, 011=alpha^4,111=alpha^5,101=alpha^6$
$bullet$ $Pi_2(alpha)=x^3+x^2+1$ $000=0, 100=1,010=lambda,001=lambda^2,101=lambda^3, 111=lambda^4,110=lambda^5,011=lambda^6$
The exampla say that $alpha$ and $lambda^3$ both have minimal polynomial $Pi_1$ and thus $alpha iff lambda^3$ form an isomorphism between the two version.
How can I see in a practical way this fact? Someone can explain me this concept?
abstract-algebra finite-fields cyclotomic-fields
asked Dec 19 '18 at 8:53
AlessarAlessar
313115
$endgroup$
This question already has an answer here:
Finding an isomorphism between these two finite fields
2 answers
I have $GF(2^3)$ generated by $Pi_1(alpha)=x^3+x+1$ and $GF(2^3)$ generated by $Pi_2(alpha)=x^3+x^2+1$.
$bullet$ $Pi_1(alpha)=x^3+x+1$ $000=0, 100=1,010=alpha,001=alpha^2,110=alpha^3, 011=alpha^4,111=alpha^5,101=alpha^6$
$bullet$ $Pi_2(alpha)=x^3+x^2+1$ $000=0, 100=1,010=lambda,001=lambda^2,101=lambda^3, 111=lambda^4,110=lambda^5,011=lambda^6$
The exampla say that $alpha$ and $lambda^3$ both have minimal polynomial $Pi_1$ and thus $alpha iff lambda^3$ form an isomorphism between the two version.
How can I see in a practical way this fact? Someone can explain me this concept?
This question already has an answer here:
Finding an isomorphism between these two finite fields
2 answers
abstract-algebra finite-fields cyclotomic-fields
abstract-algebra finite-fields cyclotomic-fields
asked Dec 19 '18 at 8:53
AlessarAlessar
313115
asked Dec 19 '18 at 8:53
AlessarAlessar
313115
asked Dec 19 '18 at 8:53
AlessarAlessar
313115
asked Dec 19 '18 at 8:53
AlessarAlessar
313115
asked Dec 19 '18 at 8:53
AlessarAlessar
313115
313115
marked as duplicate by Jyrki Lahtonen 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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Dec 19 '18 at 10:06
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 Jyrki Lahtonen abstract-algebra
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Dec 19 '18 at 10:06
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.
1
$begingroup$
It may be easier to show that $alphamapsto 1/lambda$ gives an isomorphism. This follows from the fact that the two polynomials are reciprocals of each other: $$x^3Pi_1(frac1x)=Pi_2(x)$$ implying that $Pi_1(1/lambda)=0$. Gandalf61's solution is, of course, just fine.
$endgroup$
– Jyrki Lahtonen
Dec 19 '18 at 9:58
1
$begingroup$
Closing this as a duplicate because that other question studies exactly the same pair of extension fields. It may have been difficult for you to realize that this is likely the case. So, no harm intended. Just maintaining basic site hygiene.
$endgroup$
– Jyrki Lahtonen
Dec 19 '18 at 10:09
$begingroup$
This is a very interesting clarification! Thank you!
$endgroup$
– Alessar
Dec 19 '18 at 10:15
add a comment |
1
$begingroup$
It may be easier to show that $alphamapsto 1/lambda$ gives an isomorphism. This follows from the fact that the two polynomials are reciprocals of each other: $$x^3Pi_1(frac1x)=Pi_2(x)$$ implying that $Pi_1(1/lambda)=0$. Gandalf61's solution is, of course, just fine.
$endgroup$
– Jyrki Lahtonen
Dec 19 '18 at 9:58
1
$begingroup$
Closing this as a duplicate because that other question studies exactly the same pair of extension fields. It may have been difficult for you to realize that this is likely the case. So, no harm intended. Just maintaining basic site hygiene.
$endgroup$
– Jyrki Lahtonen
Dec 19 '18 at 10:09
$begingroup$
This is a very interesting clarification! Thank you!
$endgroup$
– Alessar
Dec 19 '18 at 10:15
1
1
$begingroup$
It may be easier to show that $alphamapsto 1/lambda$ gives an isomorphism. This follows from the fact that the two polynomials are reciprocals of each other: $$x^3Pi_1(frac1x)=Pi_2(x)$$ implying that $Pi_1(1/lambda)=0$. Gandalf61's solution is, of course, just fine.
$endgroup$
– Jyrki Lahtonen
Dec 19 '18 at 9:58
$begingroup$
It may be easier to show that $alphamapsto 1/lambda$ gives an isomorphism. This follows from the fact that the two polynomials are reciprocals of each other: $$x^3Pi_1(frac1x)=Pi_2(x)$$ implying that $Pi_1(1/lambda)=0$. Gandalf61's solution is, of course, just fine.
$endgroup$
– Jyrki Lahtonen
Dec 19 '18 at 9:58
1
1
$begingroup$
Closing this as a duplicate because that other question studies exactly the same pair of extension fields. It may have been difficult for you to realize that this is likely the case. So, no harm intended. Just maintaining basic site hygiene.
$endgroup$
– Jyrki Lahtonen
Dec 19 '18 at 10:09
$begingroup$
Closing this as a duplicate because that other question studies exactly the same pair of extension fields. It may have been difficult for you to realize that this is likely the case. So, no harm intended. Just maintaining basic site hygiene.
$endgroup$
– Jyrki Lahtonen
Dec 19 '18 at 10:09
$begingroup$
This is a very interesting clarification! Thank you!
$endgroup$
– Alessar
Dec 19 '18 at 10:15
$begingroup$
This is a very interesting clarification! Thank you!
$endgroup$
– Alessar
Dec 19 '18 at 10:15
add a comment |
1 Answer
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$begingroup$
$Pi_1(lambda^3) = (lambda^3)^3 + lambda^3 + 1 = lambda^9 + lambda^3 + 1$
But you know that $lambda^7=1$. So $lambda^9=lambda^2$ and
$Pi_1(lambda^3) = lambda^3 + lambda^2 + 1 = Pi_2(lambda) = 0$
The complete isomorphism is as follows:
$0 leftrightarrow 0 \ 1 leftrightarrow 1 \ alpha leftrightarrow lambda^3 \ alpha^2 leftrightarrow lambda^6 \ alpha^3 leftrightarrow lambda^2 \ alpha^4 leftrightarrow lambda^5 \ alpha^5 leftrightarrow lambda \ alpha^6 leftrightarrow lambda^4$
answered Dec 19 '18 at 9:21
gandalf61gandalf61
8,826725
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$Pi_1(lambda^3) = (lambda^3)^3 + lambda^3 + 1 = lambda^9 + lambda^3 + 1$
But you know that $lambda^7=1$. So $lambda^9=lambda^2$ and
$Pi_1(lambda^3) = lambda^3 + lambda^2 + 1 = Pi_2(lambda) = 0$
The complete isomorphism is as follows:
$0 leftrightarrow 0 \ 1 leftrightarrow 1 \ alpha leftrightarrow lambda^3 \ alpha^2 leftrightarrow lambda^6 \ alpha^3 leftrightarrow lambda^2 \ alpha^4 leftrightarrow lambda^5 \ alpha^5 leftrightarrow lambda \ alpha^6 leftrightarrow lambda^4$
answered Dec 19 '18 at 9:21
gandalf61gandalf61
8,826725
$endgroup$
add a comment |
$begingroup$
$Pi_1(lambda^3) = (lambda^3)^3 + lambda^3 + 1 = lambda^9 + lambda^3 + 1$
But you know that $lambda^7=1$. So $lambda^9=lambda^2$ and
$Pi_1(lambda^3) = lambda^3 + lambda^2 + 1 = Pi_2(lambda) = 0$
The complete isomorphism is as follows:
$0 leftrightarrow 0 \ 1 leftrightarrow 1 \ alpha leftrightarrow lambda^3 \ alpha^2 leftrightarrow lambda^6 \ alpha^3 leftrightarrow lambda^2 \ alpha^4 leftrightarrow lambda^5 \ alpha^5 leftrightarrow lambda \ alpha^6 leftrightarrow lambda^4$
answered Dec 19 '18 at 9:21
gandalf61gandalf61
8,826725
$endgroup$
add a comment |
$begingroup$
$Pi_1(lambda^3) = (lambda^3)^3 + lambda^3 + 1 = lambda^9 + lambda^3 + 1$
But you know that $lambda^7=1$. So $lambda^9=lambda^2$ and
$Pi_1(lambda^3) = lambda^3 + lambda^2 + 1 = Pi_2(lambda) = 0$
The complete isomorphism is as follows:
$0 leftrightarrow 0 \ 1 leftrightarrow 1 \ alpha leftrightarrow lambda^3 \ alpha^2 leftrightarrow lambda^6 \ alpha^3 leftrightarrow lambda^2 \ alpha^4 leftrightarrow lambda^5 \ alpha^5 leftrightarrow lambda \ alpha^6 leftrightarrow lambda^4$
answered Dec 19 '18 at 9:21
gandalf61gandalf61
8,826725
$endgroup$
$Pi_1(lambda^3) = (lambda^3)^3 + lambda^3 + 1 = lambda^9 + lambda^3 + 1$
But you know that $lambda^7=1$. So $lambda^9=lambda^2$ and
$Pi_1(lambda^3) = lambda^3 + lambda^2 + 1 = Pi_2(lambda) = 0$
The complete isomorphism is as follows:
$0 leftrightarrow 0 \ 1 leftrightarrow 1 \ alpha leftrightarrow lambda^3 \ alpha^2 leftrightarrow lambda^6 \ alpha^3 leftrightarrow lambda^2 \ alpha^4 leftrightarrow lambda^5 \ alpha^5 leftrightarrow lambda \ alpha^6 leftrightarrow lambda^4$
answered Dec 19 '18 at 9:21
gandalf61gandalf61
8,826725
answered Dec 19 '18 at 9:21
gandalf61gandalf61
8,826725
answered Dec 19 '18 at 9:21
gandalf61gandalf61
8,826725
answered Dec 19 '18 at 9:21
gandalf61gandalf61
8,826725
8,826725
add a comment |
add a comment |
1
$begingroup$
It may be easier to show that $alphamapsto 1/lambda$ gives an isomorphism. This follows from the fact that the two polynomials are reciprocals of each other: $$x^3Pi_1(frac1x)=Pi_2(x)$$ implying that $Pi_1(1/lambda)=0$. Gandalf61's solution is, of course, just fine.
$endgroup$
– Jyrki Lahtonen
Dec 19 '18 at 9:58
1
$begingroup$
Closing this as a duplicate because that other question studies exactly the same pair of extension fields. It may have been difficult for you to realize that this is likely the case. So, no harm intended. Just maintaining basic site hygiene.
$endgroup$
– Jyrki Lahtonen
Dec 19 '18 at 10:09
$begingroup$
This is a very interesting clarification! Thank you!
$endgroup$
– Alessar
Dec 19 '18 at 10:15