Galois Field $GF(6)$
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The set called $GF(6)={0,1,2,3,4,5}$ also has the mathematical operations addition modulo $6$ and multiplication modulo $6$. We want to prove that this is not a field, since it does not match with the law of fields. In which one do we see the mistake?
field-theory
edited Oct 20 '18 at 11:42
José Carlos Santos
166k22132235
asked Oct 20 '18 at 9:57
M.PapapetrosM.Papapetros
246
$endgroup$
add a comment |
$begingroup$
The set called $GF(6)={0,1,2,3,4,5}$ also has the mathematical operations addition modulo $6$ and multiplication modulo $6$. We want to prove that this is not a field, since it does not match with the law of fields. In which one do we see the mistake?
field-theory
edited Oct 20 '18 at 11:42
José Carlos Santos
166k22132235
asked Oct 20 '18 at 9:57
M.PapapetrosM.Papapetros
246
$endgroup$
1
$begingroup$
Hint: $2cdot 3 = 6 = 0$.
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– lisyarus
Oct 20 '18 at 9:58
$begingroup$
why equal to 6?
$endgroup$
– M.Papapetros
Oct 20 '18 at 16:43
add a comment |
$begingroup$
The set called $GF(6)={0,1,2,3,4,5}$ also has the mathematical operations addition modulo $6$ and multiplication modulo $6$. We want to prove that this is not a field, since it does not match with the law of fields. In which one do we see the mistake?
field-theory
edited Oct 20 '18 at 11:42
José Carlos Santos
166k22132235
asked Oct 20 '18 at 9:57
M.PapapetrosM.Papapetros
246
$endgroup$
The set called $GF(6)={0,1,2,3,4,5}$ also has the mathematical operations addition modulo $6$ and multiplication modulo $6$. We want to prove that this is not a field, since it does not match with the law of fields. In which one do we see the mistake?
field-theory
field-theory
edited Oct 20 '18 at 11:42
José Carlos Santos
166k22132235
asked Oct 20 '18 at 9:57
M.PapapetrosM.Papapetros
246
edited Oct 20 '18 at 11:42
José Carlos Santos
166k22132235
asked Oct 20 '18 at 9:57
M.PapapetrosM.Papapetros
246
edited Oct 20 '18 at 11:42
José Carlos Santos
166k22132235
edited Oct 20 '18 at 11:42
José Carlos Santos
166k22132235
edited Oct 20 '18 at 11:42
José Carlos Santos
166k22132235
166k22132235
asked Oct 20 '18 at 9:57
M.PapapetrosM.Papapetros
246
asked Oct 20 '18 at 9:57
M.PapapetrosM.Papapetros
246
asked Oct 20 '18 at 9:57
M.PapapetrosM.Papapetros
246
246
1
$begingroup$
Hint: $2cdot 3 = 6 = 0$.
$endgroup$
– lisyarus
Oct 20 '18 at 9:58
$begingroup$
why equal to 6?
$endgroup$
– M.Papapetros
Oct 20 '18 at 16:43
add a comment |
1
$begingroup$
Hint: $2cdot 3 = 6 = 0$.
$endgroup$
– lisyarus
Oct 20 '18 at 9:58
$begingroup$
why equal to 6?
$endgroup$
– M.Papapetros
Oct 20 '18 at 16:43
1
1
$begingroup$
Hint: $2cdot 3 = 6 = 0$.
$endgroup$
– lisyarus
Oct 20 '18 at 9:58
$begingroup$
Hint: $2cdot 3 = 6 = 0$.
$endgroup$
– lisyarus
Oct 20 '18 at 9:58
$begingroup$
why equal to 6?
$endgroup$
– M.Papapetros
Oct 20 '18 at 16:43
$begingroup$
why equal to 6?
$endgroup$
– M.Papapetros
Oct 20 '18 at 16:43
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: In this one:$$(forall xin Ksetminus{0})(exists yin K):xy=yx=1.$$
answered Oct 20 '18 at 10:01
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Oct 21 '18 at 5:50
add a comment |
$begingroup$
For $GF$ to be a field it has to have for every element (except for $0$) an multiplicative inverse. Hence only $1$ and $5$ has one and the others don’t, it’s not an field. For every finite field such as it has an number of elements which is not prime $p$ many or $p^n$ for $n$ is natural, it can’t be an field.
answered Jan 24 at 22:48
Vala. D.Vala. D.
1
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add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Hint: In this one:$$(forall xin Ksetminus{0})(exists yin K):xy=yx=1.$$
answered Oct 20 '18 at 10:01
José Carlos SantosJosé Carlos Santos
166k22132235
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Comments are not for extended discussion; this conversation has been moved to chat.
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– Aloizio Macedo♦
Oct 21 '18 at 5:50
add a comment |
$begingroup$
Hint: In this one:$$(forall xin Ksetminus{0})(exists yin K):xy=yx=1.$$
answered Oct 20 '18 at 10:01
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Oct 21 '18 at 5:50
add a comment |
$begingroup$
Hint: In this one:$$(forall xin Ksetminus{0})(exists yin K):xy=yx=1.$$
answered Oct 20 '18 at 10:01
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
Hint: In this one:$$(forall xin Ksetminus{0})(exists yin K):xy=yx=1.$$
answered Oct 20 '18 at 10:01
José Carlos SantosJosé Carlos Santos
166k22132235
answered Oct 20 '18 at 10:01
José Carlos SantosJosé Carlos Santos
166k22132235
answered Oct 20 '18 at 10:01
José Carlos SantosJosé Carlos Santos
166k22132235
answered Oct 20 '18 at 10:01
José Carlos SantosJosé Carlos Santos
166k22132235
166k22132235
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Oct 21 '18 at 5:50
add a comment |
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Oct 21 '18 at 5:50
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Oct 21 '18 at 5:50
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Oct 21 '18 at 5:50
add a comment |
$begingroup$
For $GF$ to be a field it has to have for every element (except for $0$) an multiplicative inverse. Hence only $1$ and $5$ has one and the others don’t, it’s not an field. For every finite field such as it has an number of elements which is not prime $p$ many or $p^n$ for $n$ is natural, it can’t be an field.
answered Jan 24 at 22:48
Vala. D.Vala. D.
1
$endgroup$
add a comment |
$begingroup$
For $GF$ to be a field it has to have for every element (except for $0$) an multiplicative inverse. Hence only $1$ and $5$ has one and the others don’t, it’s not an field. For every finite field such as it has an number of elements which is not prime $p$ many or $p^n$ for $n$ is natural, it can’t be an field.
answered Jan 24 at 22:48
Vala. D.Vala. D.
1
$endgroup$
add a comment |
$begingroup$
For $GF$ to be a field it has to have for every element (except for $0$) an multiplicative inverse. Hence only $1$ and $5$ has one and the others don’t, it’s not an field. For every finite field such as it has an number of elements which is not prime $p$ many or $p^n$ for $n$ is natural, it can’t be an field.
answered Jan 24 at 22:48
Vala. D.Vala. D.
1
$endgroup$
For $GF$ to be a field it has to have for every element (except for $0$) an multiplicative inverse. Hence only $1$ and $5$ has one and the others don’t, it’s not an field. For every finite field such as it has an number of elements which is not prime $p$ many or $p^n$ for $n$ is natural, it can’t be an field.
answered Jan 24 at 22:48
Vala. D.Vala. D.
1
answered Jan 24 at 22:48
Vala. D.Vala. D.
1
answered Jan 24 at 22:48
Vala. D.Vala. D.
1
answered Jan 24 at 22:48
Vala. D.Vala. D.
1
1
add a comment |
add a comment |
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1
$begingroup$
Hint: $2cdot 3 = 6 = 0$.
$endgroup$
– lisyarus
Oct 20 '18 at 9:58
$begingroup$
why equal to 6?
$endgroup$
– M.Papapetros
Oct 20 '18 at 16:43