Field with four elements
$begingroup$
If $F={0,1,a,b}$ is a field (where the four elements are distinct), then:
1.What is the characteristic of $F$?
2.Write $b$ in terms of the other elements.
3.What are the multiplication and addition tableau of these operations?
abstract-algebra
asked Feb 21 '16 at 11:12
CgomesCgomes
657411
$endgroup$
add a comment |
$begingroup$
If $F={0,1,a,b}$ is a field (where the four elements are distinct), then:
1.What is the characteristic of $F$?
2.Write $b$ in terms of the other elements.
3.What are the multiplication and addition tableau of these operations?
abstract-algebra
asked Feb 21 '16 at 11:12
CgomesCgomes
657411
$endgroup$
1
$begingroup$
can't be $4$. the characteristic of an finite integral domain is a prime.It is $2$
$endgroup$
– Upstart
Feb 21 '16 at 11:15
add a comment |
$begingroup$
If $F={0,1,a,b}$ is a field (where the four elements are distinct), then:
1.What is the characteristic of $F$?
2.Write $b$ in terms of the other elements.
3.What are the multiplication and addition tableau of these operations?
abstract-algebra
asked Feb 21 '16 at 11:12
CgomesCgomes
657411
$endgroup$
If $F={0,1,a,b}$ is a field (where the four elements are distinct), then:
1.What is the characteristic of $F$?
2.Write $b$ in terms of the other elements.
3.What are the multiplication and addition tableau of these operations?
abstract-algebra
abstract-algebra
asked Feb 21 '16 at 11:12
CgomesCgomes
657411
asked Feb 21 '16 at 11:12
CgomesCgomes
657411
edited Feb 21 '16 at 14:18
Cgomes
asked Feb 21 '16 at 11:12
CgomesCgomes
657411
asked Feb 21 '16 at 11:12
CgomesCgomes
657411
asked Feb 21 '16 at 11:12
CgomesCgomes
657411
657411
1
$begingroup$
can't be $4$. the characteristic of an finite integral domain is a prime.It is $2$
$endgroup$
– Upstart
Feb 21 '16 at 11:15
add a comment |
1
$begingroup$
can't be $4$. the characteristic of an finite integral domain is a prime.It is $2$
$endgroup$
– Upstart
Feb 21 '16 at 11:15
1
1
$begingroup$
can't be $4$. the characteristic of an finite integral domain is a prime.It is $2$
$endgroup$
– Upstart
Feb 21 '16 at 11:15
$begingroup$
can't be $4$. the characteristic of an finite integral domain is a prime.It is $2$
$endgroup$
– Upstart
Feb 21 '16 at 11:15
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
The addition: $;a+b=0implies a=-b=b;$ , since $;text{char}, F=2;$ . It also can't be $;a+b=a;,;;a+b=b;$ , else $;b=0;$ or $;a=0;$ . Thus it must be $$;a+b=1implies b=1-a=1+a;$$
Generalize the above and get the addition table.
As for multiplication $;abneq 0,a,b;$ , else $;a=0;$ or $;b=0;$ , or $;a,b=1;$ and none of this is true, thus it must be $;ab=1iff b=a^{-1};$ . Generalize now for the table.
answered Feb 21 '16 at 11:27
DonAntonioDonAntonio
180k1494233
$endgroup$
add a comment |
$begingroup$
A field has characteristic a prime number $p$ or $0$, (consider the homomorphism $mathbf Zrightarrow F, enspace nmapsto ncdot 1$.
Any finite field $mathbf F_{p^n}$ is an extension of degree $n$ of the prime field $mathbf F_p$ and it is a simple extension of $mathbf F_p$ generated by a root of any irreducible polynomial of degree $n$ in $mathbf F_p[x]$. In the present case the only irreducible polynomial of degree $2$ is $x^2+x+1$. Let $omega$ be one of its roots; the other root is its inverse, $omega+1$.
Here is its multiplication table:
$$begin{array}{c|cccc}
&0&1 &omega&omega+1\hline
0&0&0&0&0\hline
1&0&1 &omega&omega+1\hline
omega&0&omega&omega+1&1\hline
omega+1&0&omega+1&1&omega
end{array}$$
edited Jan 2 at 13:52
Dietrich Burde
81.6k648106
answered Feb 21 '16 at 11:35
BernardBernard
123k741117
$endgroup$
$begingroup$
The other answers give good ways to persuade yourself of the result, but this one is the right way to remember it once you know it.
$endgroup$
– Colin McLarty
Feb 21 '16 at 17:49
add a comment |
$begingroup$
If the characteristic is not two, then WLOG you can say that $1+1=a$ and $1+1+1=b$. Then what is $a*a$? Can $F$ be a field in this case?
answered Feb 21 '16 at 11:27
Morgan RodgersMorgan Rodgers
9,85821440
$endgroup$
add a comment |
$begingroup$
For (1), suppose that the characteristic is $4$. Then, $1 + 1 neq 0$, but $$(1 + 1)(1 + 1) = 1 + 1 + 1 + 1 = 0,$$ so $1 + 1$ is a zero divisor, a contradiction, and hence the characteristic is $2$. One can use this sort of argument to show that the characteristic of any integral domain is either $0$ or prime.
Since the characteristic is $2$, for every $x in F$ we have $x + x = (1 + 1)x = 0$, so every element has order $leq 2$ under $+$, and this determines up to isomorphism the underlying group $(F, +)$, which in particular lets you solve (2) and the first part of (3). The second part of (3) follows quickly from this.
answered Feb 21 '16 at 11:27
TravisTravis
63.8k769151
$endgroup$
add a comment |
$begingroup$
Hint : What can u get of this $$F=frac {Z_2[x]}{<x^2+x+1>}$$
answered Feb 21 '16 at 11:20
UpstartUpstart
1,716617
$endgroup$
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The addition: $;a+b=0implies a=-b=b;$ , since $;text{char}, F=2;$ . It also can't be $;a+b=a;,;;a+b=b;$ , else $;b=0;$ or $;a=0;$ . Thus it must be $$;a+b=1implies b=1-a=1+a;$$
Generalize the above and get the addition table.
As for multiplication $;abneq 0,a,b;$ , else $;a=0;$ or $;b=0;$ , or $;a,b=1;$ and none of this is true, thus it must be $;ab=1iff b=a^{-1};$ . Generalize now for the table.
answered Feb 21 '16 at 11:27
DonAntonioDonAntonio
180k1494233
$endgroup$
add a comment |
$begingroup$
The addition: $;a+b=0implies a=-b=b;$ , since $;text{char}, F=2;$ . It also can't be $;a+b=a;,;;a+b=b;$ , else $;b=0;$ or $;a=0;$ . Thus it must be $$;a+b=1implies b=1-a=1+a;$$
Generalize the above and get the addition table.
As for multiplication $;abneq 0,a,b;$ , else $;a=0;$ or $;b=0;$ , or $;a,b=1;$ and none of this is true, thus it must be $;ab=1iff b=a^{-1};$ . Generalize now for the table.
answered Feb 21 '16 at 11:27
DonAntonioDonAntonio
180k1494233
$endgroup$
add a comment |
$begingroup$
The addition: $;a+b=0implies a=-b=b;$ , since $;text{char}, F=2;$ . It also can't be $;a+b=a;,;;a+b=b;$ , else $;b=0;$ or $;a=0;$ . Thus it must be $$;a+b=1implies b=1-a=1+a;$$
Generalize the above and get the addition table.
As for multiplication $;abneq 0,a,b;$ , else $;a=0;$ or $;b=0;$ , or $;a,b=1;$ and none of this is true, thus it must be $;ab=1iff b=a^{-1};$ . Generalize now for the table.
answered Feb 21 '16 at 11:27
DonAntonioDonAntonio
180k1494233
$endgroup$
The addition: $;a+b=0implies a=-b=b;$ , since $;text{char}, F=2;$ . It also can't be $;a+b=a;,;;a+b=b;$ , else $;b=0;$ or $;a=0;$ . Thus it must be $$;a+b=1implies b=1-a=1+a;$$
Generalize the above and get the addition table.
As for multiplication $;abneq 0,a,b;$ , else $;a=0;$ or $;b=0;$ , or $;a,b=1;$ and none of this is true, thus it must be $;ab=1iff b=a^{-1};$ . Generalize now for the table.
answered Feb 21 '16 at 11:27
DonAntonioDonAntonio
180k1494233
edited Feb 21 '16 at 15:19
answered Feb 21 '16 at 11:27
DonAntonioDonAntonio
180k1494233
answered Feb 21 '16 at 11:27
DonAntonioDonAntonio
180k1494233
answered Feb 21 '16 at 11:27
DonAntonioDonAntonio
180k1494233
180k1494233
add a comment |
add a comment |
$begingroup$
A field has characteristic a prime number $p$ or $0$, (consider the homomorphism $mathbf Zrightarrow F, enspace nmapsto ncdot 1$.
Any finite field $mathbf F_{p^n}$ is an extension of degree $n$ of the prime field $mathbf F_p$ and it is a simple extension of $mathbf F_p$ generated by a root of any irreducible polynomial of degree $n$ in $mathbf F_p[x]$. In the present case the only irreducible polynomial of degree $2$ is $x^2+x+1$. Let $omega$ be one of its roots; the other root is its inverse, $omega+1$.
Here is its multiplication table:
$$begin{array}{c|cccc}
&0&1 &omega&omega+1\hline
0&0&0&0&0\hline
1&0&1 &omega&omega+1\hline
omega&0&omega&omega+1&1\hline
omega+1&0&omega+1&1&omega
end{array}$$
edited Jan 2 at 13:52
Dietrich Burde
81.6k648106
answered Feb 21 '16 at 11:35
BernardBernard
123k741117
$endgroup$
$begingroup$
The other answers give good ways to persuade yourself of the result, but this one is the right way to remember it once you know it.
$endgroup$
– Colin McLarty
Feb 21 '16 at 17:49
add a comment |
$begingroup$
A field has characteristic a prime number $p$ or $0$, (consider the homomorphism $mathbf Zrightarrow F, enspace nmapsto ncdot 1$.
Any finite field $mathbf F_{p^n}$ is an extension of degree $n$ of the prime field $mathbf F_p$ and it is a simple extension of $mathbf F_p$ generated by a root of any irreducible polynomial of degree $n$ in $mathbf F_p[x]$. In the present case the only irreducible polynomial of degree $2$ is $x^2+x+1$. Let $omega$ be one of its roots; the other root is its inverse, $omega+1$.
Here is its multiplication table:
$$begin{array}{c|cccc}
&0&1 &omega&omega+1\hline
0&0&0&0&0\hline
1&0&1 &omega&omega+1\hline
omega&0&omega&omega+1&1\hline
omega+1&0&omega+1&1&omega
end{array}$$
edited Jan 2 at 13:52
Dietrich Burde
81.6k648106
answered Feb 21 '16 at 11:35
BernardBernard
123k741117
$endgroup$
$begingroup$
The other answers give good ways to persuade yourself of the result, but this one is the right way to remember it once you know it.
$endgroup$
– Colin McLarty
Feb 21 '16 at 17:49
add a comment |
$begingroup$
A field has characteristic a prime number $p$ or $0$, (consider the homomorphism $mathbf Zrightarrow F, enspace nmapsto ncdot 1$.
Any finite field $mathbf F_{p^n}$ is an extension of degree $n$ of the prime field $mathbf F_p$ and it is a simple extension of $mathbf F_p$ generated by a root of any irreducible polynomial of degree $n$ in $mathbf F_p[x]$. In the present case the only irreducible polynomial of degree $2$ is $x^2+x+1$. Let $omega$ be one of its roots; the other root is its inverse, $omega+1$.
Here is its multiplication table:
$$begin{array}{c|cccc}
&0&1 &omega&omega+1\hline
0&0&0&0&0\hline
1&0&1 &omega&omega+1\hline
omega&0&omega&omega+1&1\hline
omega+1&0&omega+1&1&omega
end{array}$$
edited Jan 2 at 13:52
Dietrich Burde
81.6k648106
answered Feb 21 '16 at 11:35
BernardBernard
123k741117
$endgroup$
A field has characteristic a prime number $p$ or $0$, (consider the homomorphism $mathbf Zrightarrow F, enspace nmapsto ncdot 1$.
Any finite field $mathbf F_{p^n}$ is an extension of degree $n$ of the prime field $mathbf F_p$ and it is a simple extension of $mathbf F_p$ generated by a root of any irreducible polynomial of degree $n$ in $mathbf F_p[x]$. In the present case the only irreducible polynomial of degree $2$ is $x^2+x+1$. Let $omega$ be one of its roots; the other root is its inverse, $omega+1$.
Here is its multiplication table:
$$begin{array}{c|cccc}
&0&1 &omega&omega+1\hline
0&0&0&0&0\hline
1&0&1 &omega&omega+1\hline
omega&0&omega&omega+1&1\hline
omega+1&0&omega+1&1&omega
end{array}$$
edited Jan 2 at 13:52
Dietrich Burde
81.6k648106
answered Feb 21 '16 at 11:35
BernardBernard
123k741117
edited Jan 2 at 13:52
Dietrich Burde
81.6k648106
edited Jan 2 at 13:52
Dietrich Burde
81.6k648106
edited Jan 2 at 13:52
Dietrich Burde
81.6k648106
81.6k648106
answered Feb 21 '16 at 11:35
BernardBernard
123k741117
answered Feb 21 '16 at 11:35
BernardBernard
123k741117
answered Feb 21 '16 at 11:35
BernardBernard
123k741117
123k741117
$begingroup$
The other answers give good ways to persuade yourself of the result, but this one is the right way to remember it once you know it.
$endgroup$
– Colin McLarty
Feb 21 '16 at 17:49
add a comment |
$begingroup$
The other answers give good ways to persuade yourself of the result, but this one is the right way to remember it once you know it.
$endgroup$
– Colin McLarty
Feb 21 '16 at 17:49
$begingroup$
The other answers give good ways to persuade yourself of the result, but this one is the right way to remember it once you know it.
$endgroup$
– Colin McLarty
Feb 21 '16 at 17:49
$begingroup$
The other answers give good ways to persuade yourself of the result, but this one is the right way to remember it once you know it.
$endgroup$
– Colin McLarty
Feb 21 '16 at 17:49
add a comment |
$begingroup$
If the characteristic is not two, then WLOG you can say that $1+1=a$ and $1+1+1=b$. Then what is $a*a$? Can $F$ be a field in this case?
answered Feb 21 '16 at 11:27
Morgan RodgersMorgan Rodgers
9,85821440
$endgroup$
add a comment |
$begingroup$
If the characteristic is not two, then WLOG you can say that $1+1=a$ and $1+1+1=b$. Then what is $a*a$? Can $F$ be a field in this case?
answered Feb 21 '16 at 11:27
Morgan RodgersMorgan Rodgers
9,85821440
$endgroup$
add a comment |
$begingroup$
If the characteristic is not two, then WLOG you can say that $1+1=a$ and $1+1+1=b$. Then what is $a*a$? Can $F$ be a field in this case?
answered Feb 21 '16 at 11:27
Morgan RodgersMorgan Rodgers
9,85821440
$endgroup$
If the characteristic is not two, then WLOG you can say that $1+1=a$ and $1+1+1=b$. Then what is $a*a$? Can $F$ be a field in this case?
answered Feb 21 '16 at 11:27
Morgan RodgersMorgan Rodgers
9,85821440
answered Feb 21 '16 at 11:27
Morgan RodgersMorgan Rodgers
9,85821440
answered Feb 21 '16 at 11:27
Morgan RodgersMorgan Rodgers
9,85821440
answered Feb 21 '16 at 11:27
Morgan RodgersMorgan Rodgers
9,85821440
9,85821440
add a comment |
add a comment |
$begingroup$
For (1), suppose that the characteristic is $4$. Then, $1 + 1 neq 0$, but $$(1 + 1)(1 + 1) = 1 + 1 + 1 + 1 = 0,$$ so $1 + 1$ is a zero divisor, a contradiction, and hence the characteristic is $2$. One can use this sort of argument to show that the characteristic of any integral domain is either $0$ or prime.
Since the characteristic is $2$, for every $x in F$ we have $x + x = (1 + 1)x = 0$, so every element has order $leq 2$ under $+$, and this determines up to isomorphism the underlying group $(F, +)$, which in particular lets you solve (2) and the first part of (3). The second part of (3) follows quickly from this.
answered Feb 21 '16 at 11:27
TravisTravis
63.8k769151
$endgroup$
add a comment |
$begingroup$
For (1), suppose that the characteristic is $4$. Then, $1 + 1 neq 0$, but $$(1 + 1)(1 + 1) = 1 + 1 + 1 + 1 = 0,$$ so $1 + 1$ is a zero divisor, a contradiction, and hence the characteristic is $2$. One can use this sort of argument to show that the characteristic of any integral domain is either $0$ or prime.
Since the characteristic is $2$, for every $x in F$ we have $x + x = (1 + 1)x = 0$, so every element has order $leq 2$ under $+$, and this determines up to isomorphism the underlying group $(F, +)$, which in particular lets you solve (2) and the first part of (3). The second part of (3) follows quickly from this.
answered Feb 21 '16 at 11:27
TravisTravis
63.8k769151
$endgroup$
add a comment |
$begingroup$
For (1), suppose that the characteristic is $4$. Then, $1 + 1 neq 0$, but $$(1 + 1)(1 + 1) = 1 + 1 + 1 + 1 = 0,$$ so $1 + 1$ is a zero divisor, a contradiction, and hence the characteristic is $2$. One can use this sort of argument to show that the characteristic of any integral domain is either $0$ or prime.
Since the characteristic is $2$, for every $x in F$ we have $x + x = (1 + 1)x = 0$, so every element has order $leq 2$ under $+$, and this determines up to isomorphism the underlying group $(F, +)$, which in particular lets you solve (2) and the first part of (3). The second part of (3) follows quickly from this.
answered Feb 21 '16 at 11:27
TravisTravis
63.8k769151
$endgroup$
For (1), suppose that the characteristic is $4$. Then, $1 + 1 neq 0$, but $$(1 + 1)(1 + 1) = 1 + 1 + 1 + 1 = 0,$$ so $1 + 1$ is a zero divisor, a contradiction, and hence the characteristic is $2$. One can use this sort of argument to show that the characteristic of any integral domain is either $0$ or prime.
Since the characteristic is $2$, for every $x in F$ we have $x + x = (1 + 1)x = 0$, so every element has order $leq 2$ under $+$, and this determines up to isomorphism the underlying group $(F, +)$, which in particular lets you solve (2) and the first part of (3). The second part of (3) follows quickly from this.
answered Feb 21 '16 at 11:27
TravisTravis
63.8k769151
edited Feb 21 '16 at 11:36
answered Feb 21 '16 at 11:27
TravisTravis
63.8k769151
answered Feb 21 '16 at 11:27
TravisTravis
63.8k769151
answered Feb 21 '16 at 11:27
TravisTravis
63.8k769151
63.8k769151
add a comment |
add a comment |
$begingroup$
Hint : What can u get of this $$F=frac {Z_2[x]}{<x^2+x+1>}$$
answered Feb 21 '16 at 11:20
UpstartUpstart
1,716617
$endgroup$
add a comment |
$begingroup$
Hint : What can u get of this $$F=frac {Z_2[x]}{<x^2+x+1>}$$
answered Feb 21 '16 at 11:20
UpstartUpstart
1,716617
$endgroup$
add a comment |
$begingroup$
Hint : What can u get of this $$F=frac {Z_2[x]}{<x^2+x+1>}$$
answered Feb 21 '16 at 11:20
UpstartUpstart
1,716617
$endgroup$
Hint : What can u get of this $$F=frac {Z_2[x]}{<x^2+x+1>}$$
answered Feb 21 '16 at 11:20
UpstartUpstart
1,716617
answered Feb 21 '16 at 11:20
UpstartUpstart
1,716617
answered Feb 21 '16 at 11:20
UpstartUpstart
1,716617
answered Feb 21 '16 at 11:20
UpstartUpstart
1,716617
1,716617
add a comment |
add a comment |
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can't be $4$. the characteristic of an finite integral domain is a prime.It is $2$
$endgroup$
– Upstart
Feb 21 '16 at 11:15