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Galois Theory by Rotman, Exercise 60, a field of four elements by using Kroenecker's theorem and adjoining a root of $x^4-x$ to $Bbb Z_2$
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I've been working through Rotman's Galois Theory and am stumped by exercise 60:
- Use Kronecker's theorem to construct a field with four elements by adjoining a suitable root of $x^4 - x$ to $mathbbZ_2$.
I can split $x^4 - x$ in $mathbbC[x]$ (which is not a field but which is contained in the field $Frac(mathbbC[x])$).
The roots are $F = 0, 1, frac-1 - i sqrt3 2, frac-1 + i sqrt3 2 $.
So what does it mean to adjoin a suitable root to $mathbbZ_2$? I have been unable to construct a four element field using either complex root.
So I went and tried to use the proof of Thm 33 (Galois) which gives a construction. In this case, F is as above and has the required four elements with $p = 2$, $n = 2$, and $q = 4$, and indeed $F setminus 0$ behaves as desired for multiplication.
The problem is addition. How is addition defined? Normal addition is not closed, nor do I see how to define it to make it work. Further, given Kronecker's and Galois's theorems, I would assume that addition must be defined as it would be in the containing field.
abstract-algebra galois-theory finite-fields extension-field
edited Aug 5 at 23:35
Asaf Karagila♦
292k31403733
asked Aug 5 at 16:07
user230584
564
add a comment |Â
up vote
10
down vote
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I've been working through Rotman's Galois Theory and am stumped by exercise 60:
- Use Kronecker's theorem to construct a field with four elements by adjoining a suitable root of $x^4 - x$ to $mathbbZ_2$.
I can split $x^4 - x$ in $mathbbC[x]$ (which is not a field but which is contained in the field $Frac(mathbbC[x])$).
The roots are $F = 0, 1, frac-1 - i sqrt3 2, frac-1 + i sqrt3 2 $.
So what does it mean to adjoin a suitable root to $mathbbZ_2$? I have been unable to construct a four element field using either complex root.
So I went and tried to use the proof of Thm 33 (Galois) which gives a construction. In this case, F is as above and has the required four elements with $p = 2$, $n = 2$, and $q = 4$, and indeed $F setminus 0$ behaves as desired for multiplication.
The problem is addition. How is addition defined? Normal addition is not closed, nor do I see how to define it to make it work. Further, given Kronecker's and Galois's theorems, I would assume that addition must be defined as it would be in the containing field.
abstract-algebra galois-theory finite-fields extension-field
edited Aug 5 at 23:35
Asaf Karagila♦
292k31403733
asked Aug 5 at 16:07
user230584
564
What does Theorem 33 say?
– quid♦
Aug 5 at 16:18
1
$0$ and $1$ are already in $mathbbZ_2$. Therefore, you should adjoint a root of $x^2+x+1$. They wouldn't be written as $frac-1pm isqrt32$, since there is no $2$ in $mathbbZ_2$ and $3=i=1$. If you want you can call them $r_1=beginpmatrix0&1\1&1endpmatrix$ and $r_2=beginpmatrix1&1\1&0endpmatrix$. The resulting field is $mathbbZ_2[r_1]$ consists of the elements $a+br_1$ with $a,binmathbbZ_2$. Using that $r_1^2+r_1+1=0$ you can compute the multiplication table of the field.
– user580373
Aug 5 at 16:33
@spiralstotheleft: I'm not sure how you can say that $3=i=1$ and that there is no $2$ in $Bbb Z_2$ in the same sentence. Perhaps it'd be better to say that $2=0$ in $Bbb Z_2,$ at which point $3=i=1$ is completely immaterial.
– Cameron Buie
Aug 5 at 23:55
@CameronBuie Because it should be "there is no $1/2$ in $mathbbZ_2$", clearly.
– user580373
Aug 5 at 23:57
add a comment |Â
up vote
10
down vote
favorite
up vote
10
down vote
favorite
I've been working through Rotman's Galois Theory and am stumped by exercise 60:
- Use Kronecker's theorem to construct a field with four elements by adjoining a suitable root of $x^4 - x$ to $mathbbZ_2$.
I can split $x^4 - x$ in $mathbbC[x]$ (which is not a field but which is contained in the field $Frac(mathbbC[x])$).
The roots are $F = 0, 1, frac-1 - i sqrt3 2, frac-1 + i sqrt3 2 $.
So what does it mean to adjoin a suitable root to $mathbbZ_2$? I have been unable to construct a four element field using either complex root.
So I went and tried to use the proof of Thm 33 (Galois) which gives a construction. In this case, F is as above and has the required four elements with $p = 2$, $n = 2$, and $q = 4$, and indeed $F setminus 0$ behaves as desired for multiplication.
The problem is addition. How is addition defined? Normal addition is not closed, nor do I see how to define it to make it work. Further, given Kronecker's and Galois's theorems, I would assume that addition must be defined as it would be in the containing field.
abstract-algebra galois-theory finite-fields extension-field
edited Aug 5 at 23:35
Asaf Karagila♦
292k31403733
asked Aug 5 at 16:07
user230584
564
I've been working through Rotman's Galois Theory and am stumped by exercise 60:
- Use Kronecker's theorem to construct a field with four elements by adjoining a suitable root of $x^4 - x$ to $mathbbZ_2$.
I can split $x^4 - x$ in $mathbbC[x]$ (which is not a field but which is contained in the field $Frac(mathbbC[x])$).
The roots are $F = 0, 1, frac-1 - i sqrt3 2, frac-1 + i sqrt3 2 $.
So what does it mean to adjoin a suitable root to $mathbbZ_2$? I have been unable to construct a four element field using either complex root.
So I went and tried to use the proof of Thm 33 (Galois) which gives a construction. In this case, F is as above and has the required four elements with $p = 2$, $n = 2$, and $q = 4$, and indeed $F setminus 0$ behaves as desired for multiplication.
The problem is addition. How is addition defined? Normal addition is not closed, nor do I see how to define it to make it work. Further, given Kronecker's and Galois's theorems, I would assume that addition must be defined as it would be in the containing field.
abstract-algebra galois-theory finite-fields extension-field
edited Aug 5 at 23:35
Asaf Karagila♦
292k31403733
asked Aug 5 at 16:07
user230584
564
edited Aug 5 at 23:35
Asaf Karagila♦
292k31403733
edited Aug 5 at 23:35
Asaf Karagila♦
292k31403733
edited Aug 5 at 23:35
Asaf Karagila♦
292k31403733
292k31403733
asked Aug 5 at 16:07
user230584
564
asked Aug 5 at 16:07
user230584
564
asked Aug 5 at 16:07
user230584
564
564
What does Theorem 33 say?
– quid♦
Aug 5 at 16:18
1
$0$ and $1$ are already in $mathbbZ_2$. Therefore, you should adjoint a root of $x^2+x+1$. They wouldn't be written as $frac-1pm isqrt32$, since there is no $2$ in $mathbbZ_2$ and $3=i=1$. If you want you can call them $r_1=beginpmatrix0&1\1&1endpmatrix$ and $r_2=beginpmatrix1&1\1&0endpmatrix$. The resulting field is $mathbbZ_2[r_1]$ consists of the elements $a+br_1$ with $a,binmathbbZ_2$. Using that $r_1^2+r_1+1=0$ you can compute the multiplication table of the field.
– user580373
Aug 5 at 16:33
@spiralstotheleft: I'm not sure how you can say that $3=i=1$ and that there is no $2$ in $Bbb Z_2$ in the same sentence. Perhaps it'd be better to say that $2=0$ in $Bbb Z_2,$ at which point $3=i=1$ is completely immaterial.
– Cameron Buie
Aug 5 at 23:55
@CameronBuie Because it should be "there is no $1/2$ in $mathbbZ_2$", clearly.
– user580373
Aug 5 at 23:57
add a comment |Â
What does Theorem 33 say?
– quid♦
Aug 5 at 16:18
1
$0$ and $1$ are already in $mathbbZ_2$. Therefore, you should adjoint a root of $x^2+x+1$. They wouldn't be written as $frac-1pm isqrt32$, since there is no $2$ in $mathbbZ_2$ and $3=i=1$. If you want you can call them $r_1=beginpmatrix0&1\1&1endpmatrix$ and $r_2=beginpmatrix1&1\1&0endpmatrix$. The resulting field is $mathbbZ_2[r_1]$ consists of the elements $a+br_1$ with $a,binmathbbZ_2$. Using that $r_1^2+r_1+1=0$ you can compute the multiplication table of the field.
– user580373
Aug 5 at 16:33
@spiralstotheleft: I'm not sure how you can say that $3=i=1$ and that there is no $2$ in $Bbb Z_2$ in the same sentence. Perhaps it'd be better to say that $2=0$ in $Bbb Z_2,$ at which point $3=i=1$ is completely immaterial.
– Cameron Buie
Aug 5 at 23:55
@CameronBuie Because it should be "there is no $1/2$ in $mathbbZ_2$", clearly.
– user580373
Aug 5 at 23:57
What does Theorem 33 say?
– quid♦
Aug 5 at 16:18
What does Theorem 33 say?
– quid♦
Aug 5 at 16:18
1
1
$0$ and $1$ are already in $mathbbZ_2$. Therefore, you should adjoint a root of $x^2+x+1$. They wouldn't be written as $frac-1pm isqrt32$, since there is no $2$ in $mathbbZ_2$ and $3=i=1$. If you want you can call them $r_1=beginpmatrix0&1\1&1endpmatrix$ and $r_2=beginpmatrix1&1\1&0endpmatrix$. The resulting field is $mathbbZ_2[r_1]$ consists of the elements $a+br_1$ with $a,binmathbbZ_2$. Using that $r_1^2+r_1+1=0$ you can compute the multiplication table of the field.
– user580373
Aug 5 at 16:33
$0$ and $1$ are already in $mathbbZ_2$. Therefore, you should adjoint a root of $x^2+x+1$. They wouldn't be written as $frac-1pm isqrt32$, since there is no $2$ in $mathbbZ_2$ and $3=i=1$. If you want you can call them $r_1=beginpmatrix0&1\1&1endpmatrix$ and $r_2=beginpmatrix1&1\1&0endpmatrix$. The resulting field is $mathbbZ_2[r_1]$ consists of the elements $a+br_1$ with $a,binmathbbZ_2$. Using that $r_1^2+r_1+1=0$ you can compute the multiplication table of the field.
– user580373
Aug 5 at 16:33
@spiralstotheleft: I'm not sure how you can say that $3=i=1$ and that there is no $2$ in $Bbb Z_2$ in the same sentence. Perhaps it'd be better to say that $2=0$ in $Bbb Z_2,$ at which point $3=i=1$ is completely immaterial.
– Cameron Buie
Aug 5 at 23:55
@spiralstotheleft: I'm not sure how you can say that $3=i=1$ and that there is no $2$ in $Bbb Z_2$ in the same sentence. Perhaps it'd be better to say that $2=0$ in $Bbb Z_2,$ at which point $3=i=1$ is completely immaterial.
– Cameron Buie
Aug 5 at 23:55
@CameronBuie Because it should be "there is no $1/2$ in $mathbbZ_2$", clearly.
– user580373
Aug 5 at 23:57
@CameronBuie Because it should be "there is no $1/2$ in $mathbbZ_2$", clearly.
– user580373
Aug 5 at 23:57
add a comment |Â
6 Answers
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up vote
7
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accepted
In the exercise, you're asked to work on the field $mathbb Z_2$ which is of characteristic $2$. So you can't split $p(x) = x^4-x$ on $mathbb C$.
Now you can write on $mathbb Z_2$:
$$p(x) = x(x^3-1)=x(x+1)(x^2+x+1)$$
If $zeta$ is a root of $q(x) = x^2+x+1$ (which is irreducible on $mathbb Z_2$) in a splitting field, the other one is $1+zeta$ as $(1+zeta)^2+(1+zeta) + 1=1+zeta^2+ 1 +zeta +1 = 1+1=0$.
Regarding your problem which is addition
So the four elements of a splitting field of $p$ over $mathbb Z_2$ are $z_1=0, z_2=1, z_3=zeta, z_4=1+ zeta$. And regarding addition, you have
$$beginmatrix
+ & 0 & 1 & zeta & 1+zeta\
0 & 0 & 1 & zeta & 1+zeta\
1 & 1 & 0 & 1+zeta & zeta\
zeta & zeta & 1+zeta & 0 & 1\
1+zeta & 1+zeta & zeta & 1 & 0\
endmatrix$$
answered Aug 5 at 16:37
mathcounterexamples.net
24.6k21653
I think this is an excellent answer. It gives an explicit field, with explicit elements in it. This should be done in every single course teaching splitting fields
– dEmigOd
Aug 5 at 20:23
add a comment |Â
up vote
7
down vote
You are not required to adjoin a complex root to $mathbbZ_2$. You can't do that even if you try because $mathbbC$ and $mathbbZ_2$ have different characteristic.
Instead, the hint is to use Kronecker's theorem, so let's do that.
First, what are the obvious roots of $f(x) = x^4 - x$ in $mathbbZ_2$? Clearly, $bar0$ and $bar1$ are both roots of $f$. So, we can factor out $x$ and $x - 1$ to get $f(x) = x(x-1)(x^2+x+1)$. Note that $x^2 + x + 1$ is irreducible over $mathbbZ_2$. By Kronecker's theorem, there exists an extension field of $mathbbZ_2$ that contains a root of $x^2+x+1$, which is also a root of $f(x)$. The degree of the extension is equal to the degree of the polynomial $x^2 + x + 1$, since it is irreducible over $mathbbZ_2$. A degree two extension of $mathbbZ_2$ has to contain $4$ elements (do you see why?) and so we are done.
answered Aug 5 at 16:21
Brahadeesh
3,74931449
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Hint:
Over any commutative ring, you have the factorisation
$$x^4-x=x(x^3-1)=x(x-1)(x^2+x+1).$$
Now over the field with two elements $mathbf F_2$, it is easy to check
$x^2+x+1$ is irreducible, so the quotient $;mathbf F_2[x]/(x^2+x+1)$ is a field. Denoting $xi=xbmod x^2+x+1$, you have by construction
$xi^2+xi+1=0$, hence $;xi^4-xi=0$.
This field is a $mathbf F_2$-vector space with dimension $2$, so it has $2^2=4$ elements.
answered Aug 5 at 16:30
Bernard
110k635103
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I don't understand why you are talking about the field $mathbb C$. It has nothing to do with this exercise.
Anyway, $x^4-x=x(x-1)(x^2+x+1)$ and $x^2+x+1$ has no roots in $mathbbZ_2$. So, consider the field $mathbbZ_2[x]/langle x^2+x+1rangle$, which has exactly $4$ elements.
edited Aug 5 at 16:23
egreg
164k1180187
answered Aug 5 at 16:22
José Carlos Santos
115k1698177
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The field is $mathbb Z_2(xi)$, where $xi^2+xi+1=0$. It is a $mathbb Z_2$ -vector space of dimension $2$, with basis $1,xi$. Thus has $4$ elements.
It can also be written as $fracmathbb Z_2[x](x^2+x+1)$.
answered Aug 5 at 17:16
Chris Custer
5,5382622
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The kicker, here, is that $0,1$ are already roots of $f,$ since $x$ and $x-1$ are factors of $f(x).$ In particular, $f(x)=x(x-1)(x^2+x+1),$ but neither $0$ nor $1$ are roots of $x^2+x+1,$ so we need to adjoin elements that are roots to this.
To accomplish this, we define $$Bbb Z_2[alpha]:=Bbb Z_2[t]/(t^2+t+1),$$ where $(t^2+t+1)$ indicates the ideal of $Bbb Z_2[t]$ generated by $t^2+t+1.$ Letting $alpha=overline t,$ we find that $alpha$ and $alpha+1$ are elements of $Bbb Z_2[alpha]$ that are roots of $x^2+x+1.$ At that point, we simply need to verify that $Bbb Z_2[alpha]$ is a field of four elements, which is straightforward.
answered Aug 5 at 16:26
Cameron Buie
83.5k771153
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
In the exercise, you're asked to work on the field $mathbb Z_2$ which is of characteristic $2$. So you can't split $p(x) = x^4-x$ on $mathbb C$.
Now you can write on $mathbb Z_2$:
$$p(x) = x(x^3-1)=x(x+1)(x^2+x+1)$$
If $zeta$ is a root of $q(x) = x^2+x+1$ (which is irreducible on $mathbb Z_2$) in a splitting field, the other one is $1+zeta$ as $(1+zeta)^2+(1+zeta) + 1=1+zeta^2+ 1 +zeta +1 = 1+1=0$.
Regarding your problem which is addition
So the four elements of a splitting field of $p$ over $mathbb Z_2$ are $z_1=0, z_2=1, z_3=zeta, z_4=1+ zeta$. And regarding addition, you have
$$beginmatrix
+ & 0 & 1 & zeta & 1+zeta\
0 & 0 & 1 & zeta & 1+zeta\
1 & 1 & 0 & 1+zeta & zeta\
zeta & zeta & 1+zeta & 0 & 1\
1+zeta & 1+zeta & zeta & 1 & 0\
endmatrix$$
answered Aug 5 at 16:37
mathcounterexamples.net
24.6k21653
I think this is an excellent answer. It gives an explicit field, with explicit elements in it. This should be done in every single course teaching splitting fields
– dEmigOd
Aug 5 at 20:23
add a comment |Â
up vote
7
down vote
accepted
In the exercise, you're asked to work on the field $mathbb Z_2$ which is of characteristic $2$. So you can't split $p(x) = x^4-x$ on $mathbb C$.
Now you can write on $mathbb Z_2$:
$$p(x) = x(x^3-1)=x(x+1)(x^2+x+1)$$
If $zeta$ is a root of $q(x) = x^2+x+1$ (which is irreducible on $mathbb Z_2$) in a splitting field, the other one is $1+zeta$ as $(1+zeta)^2+(1+zeta) + 1=1+zeta^2+ 1 +zeta +1 = 1+1=0$.
Regarding your problem which is addition
So the four elements of a splitting field of $p$ over $mathbb Z_2$ are $z_1=0, z_2=1, z_3=zeta, z_4=1+ zeta$. And regarding addition, you have
$$beginmatrix
+ & 0 & 1 & zeta & 1+zeta\
0 & 0 & 1 & zeta & 1+zeta\
1 & 1 & 0 & 1+zeta & zeta\
zeta & zeta & 1+zeta & 0 & 1\
1+zeta & 1+zeta & zeta & 1 & 0\
endmatrix$$
answered Aug 5 at 16:37
mathcounterexamples.net
24.6k21653
I think this is an excellent answer. It gives an explicit field, with explicit elements in it. This should be done in every single course teaching splitting fields
– dEmigOd
Aug 5 at 20:23
add a comment |Â
up vote
7
down vote
accepted
up vote
7
down vote
accepted
In the exercise, you're asked to work on the field $mathbb Z_2$ which is of characteristic $2$. So you can't split $p(x) = x^4-x$ on $mathbb C$.
Now you can write on $mathbb Z_2$:
$$p(x) = x(x^3-1)=x(x+1)(x^2+x+1)$$
If $zeta$ is a root of $q(x) = x^2+x+1$ (which is irreducible on $mathbb Z_2$) in a splitting field, the other one is $1+zeta$ as $(1+zeta)^2+(1+zeta) + 1=1+zeta^2+ 1 +zeta +1 = 1+1=0$.
Regarding your problem which is addition
So the four elements of a splitting field of $p$ over $mathbb Z_2$ are $z_1=0, z_2=1, z_3=zeta, z_4=1+ zeta$. And regarding addition, you have
$$beginmatrix
+ & 0 & 1 & zeta & 1+zeta\
0 & 0 & 1 & zeta & 1+zeta\
1 & 1 & 0 & 1+zeta & zeta\
zeta & zeta & 1+zeta & 0 & 1\
1+zeta & 1+zeta & zeta & 1 & 0\
endmatrix$$
answered Aug 5 at 16:37
mathcounterexamples.net
24.6k21653
In the exercise, you're asked to work on the field $mathbb Z_2$ which is of characteristic $2$. So you can't split $p(x) = x^4-x$ on $mathbb C$.
Now you can write on $mathbb Z_2$:
$$p(x) = x(x^3-1)=x(x+1)(x^2+x+1)$$
If $zeta$ is a root of $q(x) = x^2+x+1$ (which is irreducible on $mathbb Z_2$) in a splitting field, the other one is $1+zeta$ as $(1+zeta)^2+(1+zeta) + 1=1+zeta^2+ 1 +zeta +1 = 1+1=0$.
Regarding your problem which is addition
So the four elements of a splitting field of $p$ over $mathbb Z_2$ are $z_1=0, z_2=1, z_3=zeta, z_4=1+ zeta$. And regarding addition, you have
$$beginmatrix
+ & 0 & 1 & zeta & 1+zeta\
0 & 0 & 1 & zeta & 1+zeta\
1 & 1 & 0 & 1+zeta & zeta\
zeta & zeta & 1+zeta & 0 & 1\
1+zeta & 1+zeta & zeta & 1 & 0\
endmatrix$$
answered Aug 5 at 16:37
mathcounterexamples.net
24.6k21653
answered Aug 5 at 16:37
mathcounterexamples.net
24.6k21653
answered Aug 5 at 16:37
mathcounterexamples.net
24.6k21653
answered Aug 5 at 16:37
mathcounterexamples.net
24.6k21653
24.6k21653
I think this is an excellent answer. It gives an explicit field, with explicit elements in it. This should be done in every single course teaching splitting fields
– dEmigOd
Aug 5 at 20:23
add a comment |Â
I think this is an excellent answer. It gives an explicit field, with explicit elements in it. This should be done in every single course teaching splitting fields
– dEmigOd
Aug 5 at 20:23
I think this is an excellent answer. It gives an explicit field, with explicit elements in it. This should be done in every single course teaching splitting fields
– dEmigOd
Aug 5 at 20:23
I think this is an excellent answer. It gives an explicit field, with explicit elements in it. This should be done in every single course teaching splitting fields
– dEmigOd
Aug 5 at 20:23
add a comment |Â
up vote
7
down vote
You are not required to adjoin a complex root to $mathbbZ_2$. You can't do that even if you try because $mathbbC$ and $mathbbZ_2$ have different characteristic.
Instead, the hint is to use Kronecker's theorem, so let's do that.
First, what are the obvious roots of $f(x) = x^4 - x$ in $mathbbZ_2$? Clearly, $bar0$ and $bar1$ are both roots of $f$. So, we can factor out $x$ and $x - 1$ to get $f(x) = x(x-1)(x^2+x+1)$. Note that $x^2 + x + 1$ is irreducible over $mathbbZ_2$. By Kronecker's theorem, there exists an extension field of $mathbbZ_2$ that contains a root of $x^2+x+1$, which is also a root of $f(x)$. The degree of the extension is equal to the degree of the polynomial $x^2 + x + 1$, since it is irreducible over $mathbbZ_2$. A degree two extension of $mathbbZ_2$ has to contain $4$ elements (do you see why?) and so we are done.
answered Aug 5 at 16:21
Brahadeesh
3,74931449
add a comment |Â
up vote
7
down vote
You are not required to adjoin a complex root to $mathbbZ_2$. You can't do that even if you try because $mathbbC$ and $mathbbZ_2$ have different characteristic.
Instead, the hint is to use Kronecker's theorem, so let's do that.
First, what are the obvious roots of $f(x) = x^4 - x$ in $mathbbZ_2$? Clearly, $bar0$ and $bar1$ are both roots of $f$. So, we can factor out $x$ and $x - 1$ to get $f(x) = x(x-1)(x^2+x+1)$. Note that $x^2 + x + 1$ is irreducible over $mathbbZ_2$. By Kronecker's theorem, there exists an extension field of $mathbbZ_2$ that contains a root of $x^2+x+1$, which is also a root of $f(x)$. The degree of the extension is equal to the degree of the polynomial $x^2 + x + 1$, since it is irreducible over $mathbbZ_2$. A degree two extension of $mathbbZ_2$ has to contain $4$ elements (do you see why?) and so we are done.
answered Aug 5 at 16:21
Brahadeesh
3,74931449
add a comment |Â
up vote
7
down vote
up vote
7
down vote
You are not required to adjoin a complex root to $mathbbZ_2$. You can't do that even if you try because $mathbbC$ and $mathbbZ_2$ have different characteristic.
Instead, the hint is to use Kronecker's theorem, so let's do that.
First, what are the obvious roots of $f(x) = x^4 - x$ in $mathbbZ_2$? Clearly, $bar0$ and $bar1$ are both roots of $f$. So, we can factor out $x$ and $x - 1$ to get $f(x) = x(x-1)(x^2+x+1)$. Note that $x^2 + x + 1$ is irreducible over $mathbbZ_2$. By Kronecker's theorem, there exists an extension field of $mathbbZ_2$ that contains a root of $x^2+x+1$, which is also a root of $f(x)$. The degree of the extension is equal to the degree of the polynomial $x^2 + x + 1$, since it is irreducible over $mathbbZ_2$. A degree two extension of $mathbbZ_2$ has to contain $4$ elements (do you see why?) and so we are done.
answered Aug 5 at 16:21
Brahadeesh
3,74931449
You are not required to adjoin a complex root to $mathbbZ_2$. You can't do that even if you try because $mathbbC$ and $mathbbZ_2$ have different characteristic.
Instead, the hint is to use Kronecker's theorem, so let's do that.
First, what are the obvious roots of $f(x) = x^4 - x$ in $mathbbZ_2$? Clearly, $bar0$ and $bar1$ are both roots of $f$. So, we can factor out $x$ and $x - 1$ to get $f(x) = x(x-1)(x^2+x+1)$. Note that $x^2 + x + 1$ is irreducible over $mathbbZ_2$. By Kronecker's theorem, there exists an extension field of $mathbbZ_2$ that contains a root of $x^2+x+1$, which is also a root of $f(x)$. The degree of the extension is equal to the degree of the polynomial $x^2 + x + 1$, since it is irreducible over $mathbbZ_2$. A degree two extension of $mathbbZ_2$ has to contain $4$ elements (do you see why?) and so we are done.
answered Aug 5 at 16:21
Brahadeesh
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answered Aug 5 at 16:21
Brahadeesh
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answered Aug 5 at 16:21
Brahadeesh
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answered Aug 5 at 16:21
Brahadeesh
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Hint:
Over any commutative ring, you have the factorisation
$$x^4-x=x(x^3-1)=x(x-1)(x^2+x+1).$$
Now over the field with two elements $mathbf F_2$, it is easy to check
$x^2+x+1$ is irreducible, so the quotient $;mathbf F_2[x]/(x^2+x+1)$ is a field. Denoting $xi=xbmod x^2+x+1$, you have by construction
$xi^2+xi+1=0$, hence $;xi^4-xi=0$.
This field is a $mathbf F_2$-vector space with dimension $2$, so it has $2^2=4$ elements.
answered Aug 5 at 16:30
Bernard
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Hint:
Over any commutative ring, you have the factorisation
$$x^4-x=x(x^3-1)=x(x-1)(x^2+x+1).$$
Now over the field with two elements $mathbf F_2$, it is easy to check
$x^2+x+1$ is irreducible, so the quotient $;mathbf F_2[x]/(x^2+x+1)$ is a field. Denoting $xi=xbmod x^2+x+1$, you have by construction
$xi^2+xi+1=0$, hence $;xi^4-xi=0$.
This field is a $mathbf F_2$-vector space with dimension $2$, so it has $2^2=4$ elements.
answered Aug 5 at 16:30
Bernard
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up vote
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Hint:
Over any commutative ring, you have the factorisation
$$x^4-x=x(x^3-1)=x(x-1)(x^2+x+1).$$
Now over the field with two elements $mathbf F_2$, it is easy to check
$x^2+x+1$ is irreducible, so the quotient $;mathbf F_2[x]/(x^2+x+1)$ is a field. Denoting $xi=xbmod x^2+x+1$, you have by construction
$xi^2+xi+1=0$, hence $;xi^4-xi=0$.
This field is a $mathbf F_2$-vector space with dimension $2$, so it has $2^2=4$ elements.
answered Aug 5 at 16:30
Bernard
110k635103
Hint:
Over any commutative ring, you have the factorisation
$$x^4-x=x(x^3-1)=x(x-1)(x^2+x+1).$$
Now over the field with two elements $mathbf F_2$, it is easy to check
$x^2+x+1$ is irreducible, so the quotient $;mathbf F_2[x]/(x^2+x+1)$ is a field. Denoting $xi=xbmod x^2+x+1$, you have by construction
$xi^2+xi+1=0$, hence $;xi^4-xi=0$.
This field is a $mathbf F_2$-vector space with dimension $2$, so it has $2^2=4$ elements.
answered Aug 5 at 16:30
Bernard
110k635103
edited Aug 5 at 19:18
answered Aug 5 at 16:30
Bernard
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answered Aug 5 at 16:30
Bernard
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answered Aug 5 at 16:30
Bernard
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I don't understand why you are talking about the field $mathbb C$. It has nothing to do with this exercise.
Anyway, $x^4-x=x(x-1)(x^2+x+1)$ and $x^2+x+1$ has no roots in $mathbbZ_2$. So, consider the field $mathbbZ_2[x]/langle x^2+x+1rangle$, which has exactly $4$ elements.
edited Aug 5 at 16:23
egreg
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answered Aug 5 at 16:22
José Carlos Santos
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I don't understand why you are talking about the field $mathbb C$. It has nothing to do with this exercise.
Anyway, $x^4-x=x(x-1)(x^2+x+1)$ and $x^2+x+1$ has no roots in $mathbbZ_2$. So, consider the field $mathbbZ_2[x]/langle x^2+x+1rangle$, which has exactly $4$ elements.
edited Aug 5 at 16:23
egreg
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answered Aug 5 at 16:22
José Carlos Santos
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I don't understand why you are talking about the field $mathbb C$. It has nothing to do with this exercise.
Anyway, $x^4-x=x(x-1)(x^2+x+1)$ and $x^2+x+1$ has no roots in $mathbbZ_2$. So, consider the field $mathbbZ_2[x]/langle x^2+x+1rangle$, which has exactly $4$ elements.
edited Aug 5 at 16:23
egreg
164k1180187
answered Aug 5 at 16:22
José Carlos Santos
115k1698177
I don't understand why you are talking about the field $mathbb C$. It has nothing to do with this exercise.
Anyway, $x^4-x=x(x-1)(x^2+x+1)$ and $x^2+x+1$ has no roots in $mathbbZ_2$. So, consider the field $mathbbZ_2[x]/langle x^2+x+1rangle$, which has exactly $4$ elements.
edited Aug 5 at 16:23
egreg
164k1180187
answered Aug 5 at 16:22
José Carlos Santos
115k1698177
edited Aug 5 at 16:23
egreg
164k1180187
edited Aug 5 at 16:23
egreg
164k1180187
edited Aug 5 at 16:23
egreg
164k1180187
164k1180187
answered Aug 5 at 16:22
José Carlos Santos
115k1698177
answered Aug 5 at 16:22
José Carlos Santos
115k1698177
answered Aug 5 at 16:22
José Carlos Santos
115k1698177
115k1698177
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The field is $mathbb Z_2(xi)$, where $xi^2+xi+1=0$. It is a $mathbb Z_2$ -vector space of dimension $2$, with basis $1,xi$. Thus has $4$ elements.
It can also be written as $fracmathbb Z_2[x](x^2+x+1)$.
answered Aug 5 at 17:16
Chris Custer
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The field is $mathbb Z_2(xi)$, where $xi^2+xi+1=0$. It is a $mathbb Z_2$ -vector space of dimension $2$, with basis $1,xi$. Thus has $4$ elements.
It can also be written as $fracmathbb Z_2[x](x^2+x+1)$.
answered Aug 5 at 17:16
Chris Custer
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up vote
2
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up vote
2
down vote
The field is $mathbb Z_2(xi)$, where $xi^2+xi+1=0$. It is a $mathbb Z_2$ -vector space of dimension $2$, with basis $1,xi$. Thus has $4$ elements.
It can also be written as $fracmathbb Z_2[x](x^2+x+1)$.
answered Aug 5 at 17:16
Chris Custer
5,5382622
The field is $mathbb Z_2(xi)$, where $xi^2+xi+1=0$. It is a $mathbb Z_2$ -vector space of dimension $2$, with basis $1,xi$. Thus has $4$ elements.
It can also be written as $fracmathbb Z_2[x](x^2+x+1)$.
answered Aug 5 at 17:16
Chris Custer
5,5382622
answered Aug 5 at 17:16
Chris Custer
5,5382622
answered Aug 5 at 17:16
Chris Custer
5,5382622
answered Aug 5 at 17:16
Chris Custer
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5,5382622
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1
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The kicker, here, is that $0,1$ are already roots of $f,$ since $x$ and $x-1$ are factors of $f(x).$ In particular, $f(x)=x(x-1)(x^2+x+1),$ but neither $0$ nor $1$ are roots of $x^2+x+1,$ so we need to adjoin elements that are roots to this.
To accomplish this, we define $$Bbb Z_2[alpha]:=Bbb Z_2[t]/(t^2+t+1),$$ where $(t^2+t+1)$ indicates the ideal of $Bbb Z_2[t]$ generated by $t^2+t+1.$ Letting $alpha=overline t,$ we find that $alpha$ and $alpha+1$ are elements of $Bbb Z_2[alpha]$ that are roots of $x^2+x+1.$ At that point, we simply need to verify that $Bbb Z_2[alpha]$ is a field of four elements, which is straightforward.
answered Aug 5 at 16:26
Cameron Buie
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up vote
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The kicker, here, is that $0,1$ are already roots of $f,$ since $x$ and $x-1$ are factors of $f(x).$ In particular, $f(x)=x(x-1)(x^2+x+1),$ but neither $0$ nor $1$ are roots of $x^2+x+1,$ so we need to adjoin elements that are roots to this.
To accomplish this, we define $$Bbb Z_2[alpha]:=Bbb Z_2[t]/(t^2+t+1),$$ where $(t^2+t+1)$ indicates the ideal of $Bbb Z_2[t]$ generated by $t^2+t+1.$ Letting $alpha=overline t,$ we find that $alpha$ and $alpha+1$ are elements of $Bbb Z_2[alpha]$ that are roots of $x^2+x+1.$ At that point, we simply need to verify that $Bbb Z_2[alpha]$ is a field of four elements, which is straightforward.
answered Aug 5 at 16:26
Cameron Buie
83.5k771153
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The kicker, here, is that $0,1$ are already roots of $f,$ since $x$ and $x-1$ are factors of $f(x).$ In particular, $f(x)=x(x-1)(x^2+x+1),$ but neither $0$ nor $1$ are roots of $x^2+x+1,$ so we need to adjoin elements that are roots to this.
To accomplish this, we define $$Bbb Z_2[alpha]:=Bbb Z_2[t]/(t^2+t+1),$$ where $(t^2+t+1)$ indicates the ideal of $Bbb Z_2[t]$ generated by $t^2+t+1.$ Letting $alpha=overline t,$ we find that $alpha$ and $alpha+1$ are elements of $Bbb Z_2[alpha]$ that are roots of $x^2+x+1.$ At that point, we simply need to verify that $Bbb Z_2[alpha]$ is a field of four elements, which is straightforward.
answered Aug 5 at 16:26
Cameron Buie
83.5k771153
The kicker, here, is that $0,1$ are already roots of $f,$ since $x$ and $x-1$ are factors of $f(x).$ In particular, $f(x)=x(x-1)(x^2+x+1),$ but neither $0$ nor $1$ are roots of $x^2+x+1,$ so we need to adjoin elements that are roots to this.
To accomplish this, we define $$Bbb Z_2[alpha]:=Bbb Z_2[t]/(t^2+t+1),$$ where $(t^2+t+1)$ indicates the ideal of $Bbb Z_2[t]$ generated by $t^2+t+1.$ Letting $alpha=overline t,$ we find that $alpha$ and $alpha+1$ are elements of $Bbb Z_2[alpha]$ that are roots of $x^2+x+1.$ At that point, we simply need to verify that $Bbb Z_2[alpha]$ is a field of four elements, which is straightforward.
answered Aug 5 at 16:26
Cameron Buie
83.5k771153
answered Aug 5 at 16:26
Cameron Buie
83.5k771153
answered Aug 5 at 16:26
Cameron Buie
83.5k771153
answered Aug 5 at 16:26
Cameron Buie
83.5k771153
83.5k771153
add a comment |Â
add a comment |Â
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What does Theorem 33 say?
– quid♦
Aug 5 at 16:18
1
$0$ and $1$ are already in $mathbbZ_2$. Therefore, you should adjoint a root of $x^2+x+1$. They wouldn't be written as $frac-1pm isqrt32$, since there is no $2$ in $mathbbZ_2$ and $3=i=1$. If you want you can call them $r_1=beginpmatrix0&1\1&1endpmatrix$ and $r_2=beginpmatrix1&1\1&0endpmatrix$. The resulting field is $mathbbZ_2[r_1]$ consists of the elements $a+br_1$ with $a,binmathbbZ_2$. Using that $r_1^2+r_1+1=0$ you can compute the multiplication table of the field.
– user580373
Aug 5 at 16:33
@spiralstotheleft: I'm not sure how you can say that $3=i=1$ and that there is no $2$ in $Bbb Z_2$ in the same sentence. Perhaps it'd be better to say that $2=0$ in $Bbb Z_2,$ at which point $3=i=1$ is completely immaterial.
– Cameron Buie
Aug 5 at 23:55
@CameronBuie Because it should be "there is no $1/2$ in $mathbbZ_2$", clearly.
– user580373
Aug 5 at 23:57