Mixing
Elementary proof that all fields of four elements are isomorphic to each other
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A question in Rotman's Advanded Modern Algebra asks to prove the question in the title. I'm convinced of my proof, but a subquestion asked to prove that $1+1$ is zero, and for this I proceeded on a case by case basis; that is, assuming for the sake of contradiction that $1+1 ne 0$, given that we now know that the field $F = 0,1,1+1,a$, I proved that this structure cannot be a field.
However, I think this solution is quite ugly and I was wondering if is a nicer but still elementary solution (avoiding the fact that char($F$) = $2$).
PS I'm not sure if this is a suitable question. If it's not, feel free to remove it.
abstract-algebra proof-verification finite-fields
edited Sep 3 at 17:51
TonyK
38.7k348127
asked Sep 3 at 17:14
Jan De Meyer
586
add a comment |Â
up vote
10
down vote
favorite
A question in Rotman's Advanded Modern Algebra asks to prove the question in the title. I'm convinced of my proof, but a subquestion asked to prove that $1+1$ is zero, and for this I proceeded on a case by case basis; that is, assuming for the sake of contradiction that $1+1 ne 0$, given that we now know that the field $F = 0,1,1+1,a$, I proved that this structure cannot be a field.
However, I think this solution is quite ugly and I was wondering if is a nicer but still elementary solution (avoiding the fact that char($F$) = $2$).
PS I'm not sure if this is a suitable question. If it's not, feel free to remove it.
abstract-algebra proof-verification finite-fields
edited Sep 3 at 17:51
TonyK
38.7k348127
asked Sep 3 at 17:14
Jan De Meyer
586
Provided $1+1=0$, what are your arguments?
– Berci
Sep 3 at 17:40
add a comment |Â
up vote
10
down vote
favorite
up vote
10
down vote
favorite
A question in Rotman's Advanded Modern Algebra asks to prove the question in the title. I'm convinced of my proof, but a subquestion asked to prove that $1+1$ is zero, and for this I proceeded on a case by case basis; that is, assuming for the sake of contradiction that $1+1 ne 0$, given that we now know that the field $F = 0,1,1+1,a$, I proved that this structure cannot be a field.
However, I think this solution is quite ugly and I was wondering if is a nicer but still elementary solution (avoiding the fact that char($F$) = $2$).
PS I'm not sure if this is a suitable question. If it's not, feel free to remove it.
abstract-algebra proof-verification finite-fields
edited Sep 3 at 17:51
TonyK
38.7k348127
asked Sep 3 at 17:14
Jan De Meyer
586
A question in Rotman's Advanded Modern Algebra asks to prove the question in the title. I'm convinced of my proof, but a subquestion asked to prove that $1+1$ is zero, and for this I proceeded on a case by case basis; that is, assuming for the sake of contradiction that $1+1 ne 0$, given that we now know that the field $F = 0,1,1+1,a$, I proved that this structure cannot be a field.
However, I think this solution is quite ugly and I was wondering if is a nicer but still elementary solution (avoiding the fact that char($F$) = $2$).
PS I'm not sure if this is a suitable question. If it's not, feel free to remove it.
abstract-algebra proof-verification finite-fields
edited Sep 3 at 17:51
TonyK
38.7k348127
asked Sep 3 at 17:14
Jan De Meyer
586
edited Sep 3 at 17:51
TonyK
38.7k348127
edited Sep 3 at 17:51
TonyK
38.7k348127
edited Sep 3 at 17:51
TonyK
38.7k348127
38.7k348127
asked Sep 3 at 17:14
Jan De Meyer
586
asked Sep 3 at 17:14
Jan De Meyer
586
asked Sep 3 at 17:14
Jan De Meyer
586
586
Provided $1+1=0$, what are your arguments?
– Berci
Sep 3 at 17:40
add a comment |Â
Provided $1+1=0$, what are your arguments?
– Berci
Sep 3 at 17:40
Provided $1+1=0$, what are your arguments?
– Berci
Sep 3 at 17:40
Provided $1+1=0$, what are your arguments?
– Berci
Sep 3 at 17:40
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
8
down vote
Since there is an odd number of non-zero elements in the field, at least one of them must be its own additive inverse (because those who aren't come in pairs). So let's call that element $a$.
Since $ane 0$, it has a multiplicative inverse $a^-1$. But then we have
$$1 + 1 = a a^-1 + a a^-1 = (a+a) a^-1 = 0 a^-1 = 0$$
Note that this in turn implies that all elements of the field are their own additive inverse, since
$$x+x = 1x + 1x = (1+1)x = 0x = 0$$
Or in short, any finite field with an even number of elements must be of characteristic $2$.
edited Sep 4 at 3:01
Joonas Ilmavirta
20.3k84181
answered Sep 3 at 19:55
celtschk
28.6k75497
add a comment |Â
up vote
6
down vote
The underlying Abelian group $(F, +)$ has 4 elements, so the additive order of $1$ divides 4. If it were 4, the element $1+1$ would not be invertible.
answered Sep 3 at 17:38
Berci
57.1k23570
I thought $1+1=0$ is not invertible.
– Chris Custer
Sep 3 at 19:34
Yes of course. But besides $0$, in a field, there is no noninvertible element.
– Berci
Sep 3 at 19:42
Of course. You are arguing by contradiction. My mistake. An inverse for $1+1$ would require a new element...
– Chris Custer
Sep 3 at 20:01
6
We could adjust this argument like this: $(1+1)^2=(1+1)(1+1) = 1+1+1+1 = 0$ where the last equality must hold since order of the underlying additive group is $4$.
– Ennar
Sep 3 at 20:12
add a comment |Â
up vote
3
down vote
Another path is to consider the multiplicative group $F^times$. Since it has order $3$, it has to be cyclic. Thus, the multiplication table is fixed. This will also fix the addition table.
answered Sep 3 at 18:13
lhf
156k9161372
add a comment |Â
up vote
3
down vote
We need $0$ and $1$. Call the other two elements $a$ and $b$. Then $ab=1, a^2=b$ and $b^2=a$ (otherwise $a^2=1 implies a=b$ etc...) So the multiplication table is fixed.
But so is the addition table: $1+a=b$, necessarily. (Also, $b+1=a$.) For if $1+a=a,0$ or $1$ we get a contradiction. For instance, $1+a=0implies a=-1implies a=1$, since $1+1not =1,a$ or $b$. Again, we would have a contradiction: say $1+1=a$, then $1+b=b,0$ or $1$, in which case $1=0$ or $b=a$ or $0$.
So our hand winds up being forced.
answered Sep 3 at 19:32
Chris Custer
6,3162622
I'm not following $a^2 = b^2 = 1 Rightarrow a = b = 1$. (After showing $ab=1$, I do see $a^2 = 1 Rightarrow a = b$, which is enough of a contradiction.) The second paragraph gets rather confusing with nested suppositions and contradictions.
– aschepler
Sep 4 at 0:38
Well we do have $ab=1$, because the other choices are not feasible. I regretted how complicated it is; but the OP wanted an elementary proof, and it's the most elementary way I could see to approach it...
– Chris Custer
Sep 4 at 0:50
@aschepler I adjusted it. Thanks.
– Chris Custer
Sep 4 at 1:07
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
Since there is an odd number of non-zero elements in the field, at least one of them must be its own additive inverse (because those who aren't come in pairs). So let's call that element $a$.
Since $ane 0$, it has a multiplicative inverse $a^-1$. But then we have
$$1 + 1 = a a^-1 + a a^-1 = (a+a) a^-1 = 0 a^-1 = 0$$
Note that this in turn implies that all elements of the field are their own additive inverse, since
$$x+x = 1x + 1x = (1+1)x = 0x = 0$$
Or in short, any finite field with an even number of elements must be of characteristic $2$.
edited Sep 4 at 3:01
Joonas Ilmavirta
20.3k84181
answered Sep 3 at 19:55
celtschk
28.6k75497
add a comment |Â
up vote
8
down vote
Since there is an odd number of non-zero elements in the field, at least one of them must be its own additive inverse (because those who aren't come in pairs). So let's call that element $a$.
Since $ane 0$, it has a multiplicative inverse $a^-1$. But then we have
$$1 + 1 = a a^-1 + a a^-1 = (a+a) a^-1 = 0 a^-1 = 0$$
Note that this in turn implies that all elements of the field are their own additive inverse, since
$$x+x = 1x + 1x = (1+1)x = 0x = 0$$
Or in short, any finite field with an even number of elements must be of characteristic $2$.
edited Sep 4 at 3:01
Joonas Ilmavirta
20.3k84181
answered Sep 3 at 19:55
celtschk
28.6k75497
add a comment |Â
up vote
8
down vote
up vote
8
down vote
Since there is an odd number of non-zero elements in the field, at least one of them must be its own additive inverse (because those who aren't come in pairs). So let's call that element $a$.
Since $ane 0$, it has a multiplicative inverse $a^-1$. But then we have
$$1 + 1 = a a^-1 + a a^-1 = (a+a) a^-1 = 0 a^-1 = 0$$
Note that this in turn implies that all elements of the field are their own additive inverse, since
$$x+x = 1x + 1x = (1+1)x = 0x = 0$$
Or in short, any finite field with an even number of elements must be of characteristic $2$.
edited Sep 4 at 3:01
Joonas Ilmavirta
20.3k84181
answered Sep 3 at 19:55
celtschk
28.6k75497
Since there is an odd number of non-zero elements in the field, at least one of them must be its own additive inverse (because those who aren't come in pairs). So let's call that element $a$.
Since $ane 0$, it has a multiplicative inverse $a^-1$. But then we have
$$1 + 1 = a a^-1 + a a^-1 = (a+a) a^-1 = 0 a^-1 = 0$$
Note that this in turn implies that all elements of the field are their own additive inverse, since
$$x+x = 1x + 1x = (1+1)x = 0x = 0$$
Or in short, any finite field with an even number of elements must be of characteristic $2$.
edited Sep 4 at 3:01
Joonas Ilmavirta
20.3k84181
answered Sep 3 at 19:55
celtschk
28.6k75497
edited Sep 4 at 3:01
Joonas Ilmavirta
20.3k84181
edited Sep 4 at 3:01
Joonas Ilmavirta
20.3k84181
edited Sep 4 at 3:01
Joonas Ilmavirta
20.3k84181
20.3k84181
answered Sep 3 at 19:55
celtschk
28.6k75497
answered Sep 3 at 19:55
celtschk
28.6k75497
answered Sep 3 at 19:55
celtschk
28.6k75497
28.6k75497
add a comment |Â
add a comment |Â
up vote
6
down vote
The underlying Abelian group $(F, +)$ has 4 elements, so the additive order of $1$ divides 4. If it were 4, the element $1+1$ would not be invertible.
answered Sep 3 at 17:38
Berci
57.1k23570
I thought $1+1=0$ is not invertible.
– Chris Custer
Sep 3 at 19:34
Yes of course. But besides $0$, in a field, there is no noninvertible element.
– Berci
Sep 3 at 19:42
Of course. You are arguing by contradiction. My mistake. An inverse for $1+1$ would require a new element...
– Chris Custer
Sep 3 at 20:01
6
We could adjust this argument like this: $(1+1)^2=(1+1)(1+1) = 1+1+1+1 = 0$ where the last equality must hold since order of the underlying additive group is $4$.
– Ennar
Sep 3 at 20:12
add a comment |Â
up vote
6
down vote
The underlying Abelian group $(F, +)$ has 4 elements, so the additive order of $1$ divides 4. If it were 4, the element $1+1$ would not be invertible.
answered Sep 3 at 17:38
Berci
57.1k23570
I thought $1+1=0$ is not invertible.
– Chris Custer
Sep 3 at 19:34
Yes of course. But besides $0$, in a field, there is no noninvertible element.
– Berci
Sep 3 at 19:42
Of course. You are arguing by contradiction. My mistake. An inverse for $1+1$ would require a new element...
– Chris Custer
Sep 3 at 20:01
6
We could adjust this argument like this: $(1+1)^2=(1+1)(1+1) = 1+1+1+1 = 0$ where the last equality must hold since order of the underlying additive group is $4$.
– Ennar
Sep 3 at 20:12
add a comment |Â
up vote
6
down vote
up vote
6
down vote
The underlying Abelian group $(F, +)$ has 4 elements, so the additive order of $1$ divides 4. If it were 4, the element $1+1$ would not be invertible.
answered Sep 3 at 17:38
Berci
57.1k23570
The underlying Abelian group $(F, +)$ has 4 elements, so the additive order of $1$ divides 4. If it were 4, the element $1+1$ would not be invertible.
answered Sep 3 at 17:38
Berci
57.1k23570
answered Sep 3 at 17:38
Berci
57.1k23570
answered Sep 3 at 17:38
Berci
57.1k23570
answered Sep 3 at 17:38
Berci
57.1k23570
57.1k23570
I thought $1+1=0$ is not invertible.
– Chris Custer
Sep 3 at 19:34
Yes of course. But besides $0$, in a field, there is no noninvertible element.
– Berci
Sep 3 at 19:42
Of course. You are arguing by contradiction. My mistake. An inverse for $1+1$ would require a new element...
– Chris Custer
Sep 3 at 20:01
6
We could adjust this argument like this: $(1+1)^2=(1+1)(1+1) = 1+1+1+1 = 0$ where the last equality must hold since order of the underlying additive group is $4$.
– Ennar
Sep 3 at 20:12
add a comment |Â
I thought $1+1=0$ is not invertible.
– Chris Custer
Sep 3 at 19:34
Yes of course. But besides $0$, in a field, there is no noninvertible element.
– Berci
Sep 3 at 19:42
Of course. You are arguing by contradiction. My mistake. An inverse for $1+1$ would require a new element...
– Chris Custer
Sep 3 at 20:01
6
We could adjust this argument like this: $(1+1)^2=(1+1)(1+1) = 1+1+1+1 = 0$ where the last equality must hold since order of the underlying additive group is $4$.
– Ennar
Sep 3 at 20:12
I thought $1+1=0$ is not invertible.
– Chris Custer
Sep 3 at 19:34
I thought $1+1=0$ is not invertible.
– Chris Custer
Sep 3 at 19:34
Yes of course. But besides $0$, in a field, there is no noninvertible element.
– Berci
Sep 3 at 19:42
Yes of course. But besides $0$, in a field, there is no noninvertible element.
– Berci
Sep 3 at 19:42
Of course. You are arguing by contradiction. My mistake. An inverse for $1+1$ would require a new element...
– Chris Custer
Sep 3 at 20:01
Of course. You are arguing by contradiction. My mistake. An inverse for $1+1$ would require a new element...
– Chris Custer
Sep 3 at 20:01
6
6
We could adjust this argument like this: $(1+1)^2=(1+1)(1+1) = 1+1+1+1 = 0$ where the last equality must hold since order of the underlying additive group is $4$.
– Ennar
Sep 3 at 20:12
We could adjust this argument like this: $(1+1)^2=(1+1)(1+1) = 1+1+1+1 = 0$ where the last equality must hold since order of the underlying additive group is $4$.
– Ennar
Sep 3 at 20:12
add a comment |Â
up vote
3
down vote
Another path is to consider the multiplicative group $F^times$. Since it has order $3$, it has to be cyclic. Thus, the multiplication table is fixed. This will also fix the addition table.
answered Sep 3 at 18:13
lhf
156k9161372
add a comment |Â
up vote
3
down vote
Another path is to consider the multiplicative group $F^times$. Since it has order $3$, it has to be cyclic. Thus, the multiplication table is fixed. This will also fix the addition table.
answered Sep 3 at 18:13
lhf
156k9161372
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Another path is to consider the multiplicative group $F^times$. Since it has order $3$, it has to be cyclic. Thus, the multiplication table is fixed. This will also fix the addition table.
answered Sep 3 at 18:13
lhf
156k9161372
Another path is to consider the multiplicative group $F^times$. Since it has order $3$, it has to be cyclic. Thus, the multiplication table is fixed. This will also fix the addition table.
answered Sep 3 at 18:13
lhf
156k9161372
answered Sep 3 at 18:13
lhf
156k9161372
answered Sep 3 at 18:13
lhf
156k9161372
answered Sep 3 at 18:13
lhf
156k9161372
156k9161372
add a comment |Â
add a comment |Â
up vote
3
down vote
We need $0$ and $1$. Call the other two elements $a$ and $b$. Then $ab=1, a^2=b$ and $b^2=a$ (otherwise $a^2=1 implies a=b$ etc...) So the multiplication table is fixed.
But so is the addition table: $1+a=b$, necessarily. (Also, $b+1=a$.) For if $1+a=a,0$ or $1$ we get a contradiction. For instance, $1+a=0implies a=-1implies a=1$, since $1+1not =1,a$ or $b$. Again, we would have a contradiction: say $1+1=a$, then $1+b=b,0$ or $1$, in which case $1=0$ or $b=a$ or $0$.
So our hand winds up being forced.
answered Sep 3 at 19:32
Chris Custer
6,3162622
I'm not following $a^2 = b^2 = 1 Rightarrow a = b = 1$. (After showing $ab=1$, I do see $a^2 = 1 Rightarrow a = b$, which is enough of a contradiction.) The second paragraph gets rather confusing with nested suppositions and contradictions.
– aschepler
Sep 4 at 0:38
Well we do have $ab=1$, because the other choices are not feasible. I regretted how complicated it is; but the OP wanted an elementary proof, and it's the most elementary way I could see to approach it...
– Chris Custer
Sep 4 at 0:50
@aschepler I adjusted it. Thanks.
– Chris Custer
Sep 4 at 1:07
add a comment |Â
up vote
3
down vote
We need $0$ and $1$. Call the other two elements $a$ and $b$. Then $ab=1, a^2=b$ and $b^2=a$ (otherwise $a^2=1 implies a=b$ etc...) So the multiplication table is fixed.
But so is the addition table: $1+a=b$, necessarily. (Also, $b+1=a$.) For if $1+a=a,0$ or $1$ we get a contradiction. For instance, $1+a=0implies a=-1implies a=1$, since $1+1not =1,a$ or $b$. Again, we would have a contradiction: say $1+1=a$, then $1+b=b,0$ or $1$, in which case $1=0$ or $b=a$ or $0$.
So our hand winds up being forced.
answered Sep 3 at 19:32
Chris Custer
6,3162622
I'm not following $a^2 = b^2 = 1 Rightarrow a = b = 1$. (After showing $ab=1$, I do see $a^2 = 1 Rightarrow a = b$, which is enough of a contradiction.) The second paragraph gets rather confusing with nested suppositions and contradictions.
– aschepler
Sep 4 at 0:38
Well we do have $ab=1$, because the other choices are not feasible. I regretted how complicated it is; but the OP wanted an elementary proof, and it's the most elementary way I could see to approach it...
– Chris Custer
Sep 4 at 0:50
@aschepler I adjusted it. Thanks.
– Chris Custer
Sep 4 at 1:07
add a comment |Â
up vote
3
down vote
up vote
3
down vote
We need $0$ and $1$. Call the other two elements $a$ and $b$. Then $ab=1, a^2=b$ and $b^2=a$ (otherwise $a^2=1 implies a=b$ etc...) So the multiplication table is fixed.
But so is the addition table: $1+a=b$, necessarily. (Also, $b+1=a$.) For if $1+a=a,0$ or $1$ we get a contradiction. For instance, $1+a=0implies a=-1implies a=1$, since $1+1not =1,a$ or $b$. Again, we would have a contradiction: say $1+1=a$, then $1+b=b,0$ or $1$, in which case $1=0$ or $b=a$ or $0$.
So our hand winds up being forced.
answered Sep 3 at 19:32
Chris Custer
6,3162622
We need $0$ and $1$. Call the other two elements $a$ and $b$. Then $ab=1, a^2=b$ and $b^2=a$ (otherwise $a^2=1 implies a=b$ etc...) So the multiplication table is fixed.
But so is the addition table: $1+a=b$, necessarily. (Also, $b+1=a$.) For if $1+a=a,0$ or $1$ we get a contradiction. For instance, $1+a=0implies a=-1implies a=1$, since $1+1not =1,a$ or $b$. Again, we would have a contradiction: say $1+1=a$, then $1+b=b,0$ or $1$, in which case $1=0$ or $b=a$ or $0$.
So our hand winds up being forced.
answered Sep 3 at 19:32
Chris Custer
6,3162622
edited Sep 4 at 7:47
answered Sep 3 at 19:32
Chris Custer
6,3162622
answered Sep 3 at 19:32
Chris Custer
6,3162622
answered Sep 3 at 19:32
Chris Custer
6,3162622
6,3162622
I'm not following $a^2 = b^2 = 1 Rightarrow a = b = 1$. (After showing $ab=1$, I do see $a^2 = 1 Rightarrow a = b$, which is enough of a contradiction.) The second paragraph gets rather confusing with nested suppositions and contradictions.
– aschepler
Sep 4 at 0:38
Well we do have $ab=1$, because the other choices are not feasible. I regretted how complicated it is; but the OP wanted an elementary proof, and it's the most elementary way I could see to approach it...
– Chris Custer
Sep 4 at 0:50
@aschepler I adjusted it. Thanks.
– Chris Custer
Sep 4 at 1:07
add a comment |Â
I'm not following $a^2 = b^2 = 1 Rightarrow a = b = 1$. (After showing $ab=1$, I do see $a^2 = 1 Rightarrow a = b$, which is enough of a contradiction.) The second paragraph gets rather confusing with nested suppositions and contradictions.
– aschepler
Sep 4 at 0:38
Well we do have $ab=1$, because the other choices are not feasible. I regretted how complicated it is; but the OP wanted an elementary proof, and it's the most elementary way I could see to approach it...
– Chris Custer
Sep 4 at 0:50
@aschepler I adjusted it. Thanks.
– Chris Custer
Sep 4 at 1:07
I'm not following $a^2 = b^2 = 1 Rightarrow a = b = 1$. (After showing $ab=1$, I do see $a^2 = 1 Rightarrow a = b$, which is enough of a contradiction.) The second paragraph gets rather confusing with nested suppositions and contradictions.
– aschepler
Sep 4 at 0:38
I'm not following $a^2 = b^2 = 1 Rightarrow a = b = 1$. (After showing $ab=1$, I do see $a^2 = 1 Rightarrow a = b$, which is enough of a contradiction.) The second paragraph gets rather confusing with nested suppositions and contradictions.
– aschepler
Sep 4 at 0:38
Well we do have $ab=1$, because the other choices are not feasible. I regretted how complicated it is; but the OP wanted an elementary proof, and it's the most elementary way I could see to approach it...
– Chris Custer
Sep 4 at 0:50
Well we do have $ab=1$, because the other choices are not feasible. I regretted how complicated it is; but the OP wanted an elementary proof, and it's the most elementary way I could see to approach it...
– Chris Custer
Sep 4 at 0:50
@aschepler I adjusted it. Thanks.
– Chris Custer
Sep 4 at 1:07
@aschepler I adjusted it. Thanks.
– Chris Custer
Sep 4 at 1:07
add a comment |Â
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Provided $1+1=0$, what are your arguments?
– Berci
Sep 3 at 17:40