Prove or disprove this table is a field
$begingroup$
Any help will be appreciated on how to approach this and get started.
abstract-algebra discrete-mathematics
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add a comment |
$begingroup$
Any help will be appreciated on how to approach this and get started.
abstract-algebra discrete-mathematics
$endgroup$
4
$begingroup$
Hint: removal of $0$ should yield a group under multiplication.
$endgroup$
– Randall
2 hours ago
add a comment |
$begingroup$
Any help will be appreciated on how to approach this and get started.
abstract-algebra discrete-mathematics
$endgroup$
Any help will be appreciated on how to approach this and get started.
abstract-algebra discrete-mathematics
abstract-algebra discrete-mathematics
asked 2 hours ago
Wade KemmsiesWade Kemmsies
263
263
4
$begingroup$
Hint: removal of $0$ should yield a group under multiplication.
$endgroup$
– Randall
2 hours ago
add a comment |
4
$begingroup$
Hint: removal of $0$ should yield a group under multiplication.
$endgroup$
– Randall
2 hours ago
4
4
$begingroup$
Hint: removal of $0$ should yield a group under multiplication.
$endgroup$
– Randall
2 hours ago
$begingroup$
Hint: removal of $0$ should yield a group under multiplication.
$endgroup$
– Randall
2 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
A field has no non-zero divisors of zero.
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add a comment |
$begingroup$
Note from the multiplication table, the one for the $cdot$ operation, that
$2 cdot 2 = 0; tag 1$
thus $2$ is nilpotent; but a field $F$ can have no non-zero nilpotents (note $2 ne 0$ since $3 + 2 = 1 ne 3$).
If $F$ is a field and $a in F$ were nilpotent, that is, if
$a^k = 0, ; 2 le k in Bbb N, tag 2$
then
$a = a^{1 - k} a^k = a^{1 - k} = 0; tag 3$
which shows a field has no non-zero nilpotent elements.
$endgroup$
1
$begingroup$
Did you mean $a = a^{1-k} a^k = a^{1-k} times 0 =0$ ?
$endgroup$
– J. W. Tanner
16 mins ago
$begingroup$
@J.W.Tanner: yuppers, sure did!
$endgroup$
– Robert Lewis
2 mins ago
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A field has no non-zero divisors of zero.
$endgroup$
add a comment |
$begingroup$
A field has no non-zero divisors of zero.
$endgroup$
add a comment |
$begingroup$
A field has no non-zero divisors of zero.
$endgroup$
A field has no non-zero divisors of zero.
answered 2 hours ago
J. W. TannerJ. W. Tanner
937
937
add a comment |
add a comment |
$begingroup$
Note from the multiplication table, the one for the $cdot$ operation, that
$2 cdot 2 = 0; tag 1$
thus $2$ is nilpotent; but a field $F$ can have no non-zero nilpotents (note $2 ne 0$ since $3 + 2 = 1 ne 3$).
If $F$ is a field and $a in F$ were nilpotent, that is, if
$a^k = 0, ; 2 le k in Bbb N, tag 2$
then
$a = a^{1 - k} a^k = a^{1 - k} = 0; tag 3$
which shows a field has no non-zero nilpotent elements.
$endgroup$
1
$begingroup$
Did you mean $a = a^{1-k} a^k = a^{1-k} times 0 =0$ ?
$endgroup$
– J. W. Tanner
16 mins ago
$begingroup$
@J.W.Tanner: yuppers, sure did!
$endgroup$
– Robert Lewis
2 mins ago
add a comment |
$begingroup$
Note from the multiplication table, the one for the $cdot$ operation, that
$2 cdot 2 = 0; tag 1$
thus $2$ is nilpotent; but a field $F$ can have no non-zero nilpotents (note $2 ne 0$ since $3 + 2 = 1 ne 3$).
If $F$ is a field and $a in F$ were nilpotent, that is, if
$a^k = 0, ; 2 le k in Bbb N, tag 2$
then
$a = a^{1 - k} a^k = a^{1 - k} = 0; tag 3$
which shows a field has no non-zero nilpotent elements.
$endgroup$
1
$begingroup$
Did you mean $a = a^{1-k} a^k = a^{1-k} times 0 =0$ ?
$endgroup$
– J. W. Tanner
16 mins ago
$begingroup$
@J.W.Tanner: yuppers, sure did!
$endgroup$
– Robert Lewis
2 mins ago
add a comment |
$begingroup$
Note from the multiplication table, the one for the $cdot$ operation, that
$2 cdot 2 = 0; tag 1$
thus $2$ is nilpotent; but a field $F$ can have no non-zero nilpotents (note $2 ne 0$ since $3 + 2 = 1 ne 3$).
If $F$ is a field and $a in F$ were nilpotent, that is, if
$a^k = 0, ; 2 le k in Bbb N, tag 2$
then
$a = a^{1 - k} a^k = a^{1 - k} = 0; tag 3$
which shows a field has no non-zero nilpotent elements.
$endgroup$
Note from the multiplication table, the one for the $cdot$ operation, that
$2 cdot 2 = 0; tag 1$
thus $2$ is nilpotent; but a field $F$ can have no non-zero nilpotents (note $2 ne 0$ since $3 + 2 = 1 ne 3$).
If $F$ is a field and $a in F$ were nilpotent, that is, if
$a^k = 0, ; 2 le k in Bbb N, tag 2$
then
$a = a^{1 - k} a^k = a^{1 - k} = 0; tag 3$
which shows a field has no non-zero nilpotent elements.
edited 2 hours ago
answered 2 hours ago
Robert LewisRobert Lewis
44.5k22964
44.5k22964
1
$begingroup$
Did you mean $a = a^{1-k} a^k = a^{1-k} times 0 =0$ ?
$endgroup$
– J. W. Tanner
16 mins ago
$begingroup$
@J.W.Tanner: yuppers, sure did!
$endgroup$
– Robert Lewis
2 mins ago
add a comment |
1
$begingroup$
Did you mean $a = a^{1-k} a^k = a^{1-k} times 0 =0$ ?
$endgroup$
– J. W. Tanner
16 mins ago
$begingroup$
@J.W.Tanner: yuppers, sure did!
$endgroup$
– Robert Lewis
2 mins ago
1
1
$begingroup$
Did you mean $a = a^{1-k} a^k = a^{1-k} times 0 =0$ ?
$endgroup$
– J. W. Tanner
16 mins ago
$begingroup$
Did you mean $a = a^{1-k} a^k = a^{1-k} times 0 =0$ ?
$endgroup$
– J. W. Tanner
16 mins ago
$begingroup$
@J.W.Tanner: yuppers, sure did!
$endgroup$
– Robert Lewis
2 mins ago
$begingroup$
@J.W.Tanner: yuppers, sure did!
$endgroup$
– Robert Lewis
2 mins ago
add a comment |
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4
$begingroup$
Hint: removal of $0$ should yield a group under multiplication.
$endgroup$
– Randall
2 hours ago