Mixing
Existence of homomorphisms between finite fields
Clash Royale CLAN TAG#URR8PPP
up vote
6
down vote
favorite
Let $F$ and $E$ be the fields of order $8$ and $32$ respectively. Construct a ring homomorphism $Fto E$ or prove that one cannot exist.
Any element $x$ of $F$ satisfies $x^8=x$ and any nonzero element $xin F$ satisfies $x^7=1$. Let $f:Fto E$ be a homomorphism. Then $1=f(1)=f(x^7)=f(x)^7$ for all nonzero $xin F$. Also $0=f(0)=f(2)=f(1+1)=f(1)+f(1)=0$ (here $2=1+1in F$). Now on the one hand, $f(2^7)=f(2)^7=1$. On the other hand, $f(2^7)$ is $f$ applied to $2+dots+2$ ($2^6$ terms), so $f(2^7)=f(2cdot 2^6)=f(2)+dots+f(2) text ($2^6$ terms) =0+dots+0=0$. This is a contradiction.
Is this reasoning correct? Are there other ways to solve this? In general, for which finite fields there exists a homomorphism between them?
As I pointed out in the comments, the above argument is incorrect. How do I solve the problem then?
abstract-algebra ring-theory field-theory finite-fields ring-homomorphism
asked Jul 16 at 22:48
user437309
556212
add a comment |Â
up vote
6
down vote
favorite
Let $F$ and $E$ be the fields of order $8$ and $32$ respectively. Construct a ring homomorphism $Fto E$ or prove that one cannot exist.
Any element $x$ of $F$ satisfies $x^8=x$ and any nonzero element $xin F$ satisfies $x^7=1$. Let $f:Fto E$ be a homomorphism. Then $1=f(1)=f(x^7)=f(x)^7$ for all nonzero $xin F$. Also $0=f(0)=f(2)=f(1+1)=f(1)+f(1)=0$ (here $2=1+1in F$). Now on the one hand, $f(2^7)=f(2)^7=1$. On the other hand, $f(2^7)$ is $f$ applied to $2+dots+2$ ($2^6$ terms), so $f(2^7)=f(2cdot 2^6)=f(2)+dots+f(2) text ($2^6$ terms) =0+dots+0=0$. This is a contradiction.
Is this reasoning correct? Are there other ways to solve this? In general, for which finite fields there exists a homomorphism between them?
As I pointed out in the comments, the above argument is incorrect. How do I solve the problem then?
abstract-algebra ring-theory field-theory finite-fields ring-homomorphism
asked Jul 16 at 22:48
user437309
556212
5
As always, right after posting the question, I found a flaw in my argument: I use that $2^7=1$, but $x^7=1$ is valid only for $xne 0$ and $2=0$.
– user437309
Jul 16 at 22:51
1
Good call on that flaw (and the exact same flaw in my answer, effectively)!
– Cameron Buie
Jul 17 at 0:07
Does the definition of "ring homomorphism" you work with include that $f(1) =1$? All answers assume this and it is a natural thing to do. However, without that assumption, the zero map $f(x)=0$ for all $x$ is the only possibility.
– Torsten Schoeneberg
Jul 17 at 17:00
@TorstenSchoeneberg Yes it does.
– user437309
Jul 17 at 17:26
add a comment |Â
up vote
6
down vote
favorite
up vote
6
down vote
favorite
Let $F$ and $E$ be the fields of order $8$ and $32$ respectively. Construct a ring homomorphism $Fto E$ or prove that one cannot exist.
Any element $x$ of $F$ satisfies $x^8=x$ and any nonzero element $xin F$ satisfies $x^7=1$. Let $f:Fto E$ be a homomorphism. Then $1=f(1)=f(x^7)=f(x)^7$ for all nonzero $xin F$. Also $0=f(0)=f(2)=f(1+1)=f(1)+f(1)=0$ (here $2=1+1in F$). Now on the one hand, $f(2^7)=f(2)^7=1$. On the other hand, $f(2^7)$ is $f$ applied to $2+dots+2$ ($2^6$ terms), so $f(2^7)=f(2cdot 2^6)=f(2)+dots+f(2) text ($2^6$ terms) =0+dots+0=0$. This is a contradiction.
Is this reasoning correct? Are there other ways to solve this? In general, for which finite fields there exists a homomorphism between them?
As I pointed out in the comments, the above argument is incorrect. How do I solve the problem then?
abstract-algebra ring-theory field-theory finite-fields ring-homomorphism
asked Jul 16 at 22:48
user437309
556212
Let $F$ and $E$ be the fields of order $8$ and $32$ respectively. Construct a ring homomorphism $Fto E$ or prove that one cannot exist.
Any element $x$ of $F$ satisfies $x^8=x$ and any nonzero element $xin F$ satisfies $x^7=1$. Let $f:Fto E$ be a homomorphism. Then $1=f(1)=f(x^7)=f(x)^7$ for all nonzero $xin F$. Also $0=f(0)=f(2)=f(1+1)=f(1)+f(1)=0$ (here $2=1+1in F$). Now on the one hand, $f(2^7)=f(2)^7=1$. On the other hand, $f(2^7)$ is $f$ applied to $2+dots+2$ ($2^6$ terms), so $f(2^7)=f(2cdot 2^6)=f(2)+dots+f(2) text ($2^6$ terms) =0+dots+0=0$. This is a contradiction.
Is this reasoning correct? Are there other ways to solve this? In general, for which finite fields there exists a homomorphism between them?
As I pointed out in the comments, the above argument is incorrect. How do I solve the problem then?
abstract-algebra ring-theory field-theory finite-fields ring-homomorphism
asked Jul 16 at 22:48
user437309
556212
edited Jul 16 at 23:14
asked Jul 16 at 22:48
user437309
556212
asked Jul 16 at 22:48
user437309
556212
asked Jul 16 at 22:48
user437309
556212
556212
5
As always, right after posting the question, I found a flaw in my argument: I use that $2^7=1$, but $x^7=1$ is valid only for $xne 0$ and $2=0$.
– user437309
Jul 16 at 22:51
1
Good call on that flaw (and the exact same flaw in my answer, effectively)!
– Cameron Buie
Jul 17 at 0:07
Does the definition of "ring homomorphism" you work with include that $f(1) =1$? All answers assume this and it is a natural thing to do. However, without that assumption, the zero map $f(x)=0$ for all $x$ is the only possibility.
– Torsten Schoeneberg
Jul 17 at 17:00
@TorstenSchoeneberg Yes it does.
– user437309
Jul 17 at 17:26
add a comment |Â
5
As always, right after posting the question, I found a flaw in my argument: I use that $2^7=1$, but $x^7=1$ is valid only for $xne 0$ and $2=0$.
– user437309
Jul 16 at 22:51
1
Good call on that flaw (and the exact same flaw in my answer, effectively)!
– Cameron Buie
Jul 17 at 0:07
Does the definition of "ring homomorphism" you work with include that $f(1) =1$? All answers assume this and it is a natural thing to do. However, without that assumption, the zero map $f(x)=0$ for all $x$ is the only possibility.
– Torsten Schoeneberg
Jul 17 at 17:00
@TorstenSchoeneberg Yes it does.
– user437309
Jul 17 at 17:26
5
5
As always, right after posting the question, I found a flaw in my argument: I use that $2^7=1$, but $x^7=1$ is valid only for $xne 0$ and $2=0$.
– user437309
Jul 16 at 22:51
As always, right after posting the question, I found a flaw in my argument: I use that $2^7=1$, but $x^7=1$ is valid only for $xne 0$ and $2=0$.
– user437309
Jul 16 at 22:51
1
1
Good call on that flaw (and the exact same flaw in my answer, effectively)!
– Cameron Buie
Jul 17 at 0:07
Good call on that flaw (and the exact same flaw in my answer, effectively)!
– Cameron Buie
Jul 17 at 0:07
Does the definition of "ring homomorphism" you work with include that $f(1) =1$? All answers assume this and it is a natural thing to do. However, without that assumption, the zero map $f(x)=0$ for all $x$ is the only possibility.
– Torsten Schoeneberg
Jul 17 at 17:00
Does the definition of "ring homomorphism" you work with include that $f(1) =1$? All answers assume this and it is a natural thing to do. However, without that assumption, the zero map $f(x)=0$ for all $x$ is the only possibility.
– Torsten Schoeneberg
Jul 17 at 17:00
@TorstenSchoeneberg Yes it does.
– user437309
Jul 17 at 17:26
@TorstenSchoeneberg Yes it does.
– user437309
Jul 17 at 17:26
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
2
down vote
Here is a direct proof similar to your attempt. Let $x notin0,1$ be an element of $Bbb F_8$ and let $f: Bbb F_8 rightarrow Bbb F_32$ be a (unital) ring homomorphism. If $f(x) neq 0$, we have $f(x)^31 = 1$; on the other hand, $f(x)^7 = f(x^7) =f(1)=1$. But since $gcd(7,31)=1$, this means $f(x)=1$. Consequently $f(x-1)=f(x)-f(1)=0$, but since we chose $x-1 neq 0$, this means that $f$ is the zero map (why?), contradiction. (If $f(x) =0$, it follows immediately that $f$ is the zero map.)
answered Jul 17 at 2:42
Torsten Schoeneberg
2,6021732
add a comment |Â
up vote
2
down vote
A ring homomorphism $Fto E$ induces a group homomorphism $F^times to E^times$. These groups have orders $7$ and $31$ and so this homomorphism must be the trivial one. On the other hand, a ring homomorphism from a field is injective.
answered Jul 17 at 11:56
lhf
156k9160367
add a comment |Â
up vote
1
down vote
Since such a homomorphism $phi$ would be injective (the kernel is an ideal, and we are assuming $phi$ nontrivial), it is in fact an embedding, and $phi (textGF(2^3))cong textGF(2^3)$ would be a subfield of $textGF(2^5)$; but since $3notmid5$, this is impossible .
answered Jul 17 at 1:46
Chris Custer
5,4582622
add a comment |Â
up vote
1
down vote
Let $F$ be the field with only two elements. If $F_8$ and $F_32$ are fields of order $8$ and $32$ respectively, then they are both realizable as algebraic extensions of $F.$ Any homomorphism $phi : F_8 longrightarrow F_32$ must be injective, as all fields are simple. That is, if we assume that $F_8 subset F_32,$ by way of identifying $F_8$ with $textImg(phi)$ then this is to say that $F_32$ must be realizable as an algebraic extension of $F_8.$ So we get the following inclusions $Fsubset F_8 subset F_32.$ Since these are all finite extensions, and since every finite extension of a finite field is Galois, then the Galois correspondence gives us corresponding subgroups of orders $1,$ $3$ and $5$ respectively. Since any group of order $5$ has no subgroup of order $3$, then the existence of an embedding $F_8subset F_32$ must be impossible.
Conclude that no such homomorphism exists.
edited Jul 17 at 5:23
Chris Custer
5,4582622
answered Jul 17 at 0:17
Chickenmancer
3,017621
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Here is a direct proof similar to your attempt. Let $x notin0,1$ be an element of $Bbb F_8$ and let $f: Bbb F_8 rightarrow Bbb F_32$ be a (unital) ring homomorphism. If $f(x) neq 0$, we have $f(x)^31 = 1$; on the other hand, $f(x)^7 = f(x^7) =f(1)=1$. But since $gcd(7,31)=1$, this means $f(x)=1$. Consequently $f(x-1)=f(x)-f(1)=0$, but since we chose $x-1 neq 0$, this means that $f$ is the zero map (why?), contradiction. (If $f(x) =0$, it follows immediately that $f$ is the zero map.)
answered Jul 17 at 2:42
Torsten Schoeneberg
2,6021732
add a comment |Â
up vote
2
down vote
Here is a direct proof similar to your attempt. Let $x notin0,1$ be an element of $Bbb F_8$ and let $f: Bbb F_8 rightarrow Bbb F_32$ be a (unital) ring homomorphism. If $f(x) neq 0$, we have $f(x)^31 = 1$; on the other hand, $f(x)^7 = f(x^7) =f(1)=1$. But since $gcd(7,31)=1$, this means $f(x)=1$. Consequently $f(x-1)=f(x)-f(1)=0$, but since we chose $x-1 neq 0$, this means that $f$ is the zero map (why?), contradiction. (If $f(x) =0$, it follows immediately that $f$ is the zero map.)
answered Jul 17 at 2:42
Torsten Schoeneberg
2,6021732
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Here is a direct proof similar to your attempt. Let $x notin0,1$ be an element of $Bbb F_8$ and let $f: Bbb F_8 rightarrow Bbb F_32$ be a (unital) ring homomorphism. If $f(x) neq 0$, we have $f(x)^31 = 1$; on the other hand, $f(x)^7 = f(x^7) =f(1)=1$. But since $gcd(7,31)=1$, this means $f(x)=1$. Consequently $f(x-1)=f(x)-f(1)=0$, but since we chose $x-1 neq 0$, this means that $f$ is the zero map (why?), contradiction. (If $f(x) =0$, it follows immediately that $f$ is the zero map.)
answered Jul 17 at 2:42
Torsten Schoeneberg
2,6021732
Here is a direct proof similar to your attempt. Let $x notin0,1$ be an element of $Bbb F_8$ and let $f: Bbb F_8 rightarrow Bbb F_32$ be a (unital) ring homomorphism. If $f(x) neq 0$, we have $f(x)^31 = 1$; on the other hand, $f(x)^7 = f(x^7) =f(1)=1$. But since $gcd(7,31)=1$, this means $f(x)=1$. Consequently $f(x-1)=f(x)-f(1)=0$, but since we chose $x-1 neq 0$, this means that $f$ is the zero map (why?), contradiction. (If $f(x) =0$, it follows immediately that $f$ is the zero map.)
answered Jul 17 at 2:42
Torsten Schoeneberg
2,6021732
answered Jul 17 at 2:42
Torsten Schoeneberg
2,6021732
answered Jul 17 at 2:42
Torsten Schoeneberg
2,6021732
answered Jul 17 at 2:42
Torsten Schoeneberg
2,6021732
2,6021732
add a comment |Â
add a comment |Â
up vote
2
down vote
A ring homomorphism $Fto E$ induces a group homomorphism $F^times to E^times$. These groups have orders $7$ and $31$ and so this homomorphism must be the trivial one. On the other hand, a ring homomorphism from a field is injective.
answered Jul 17 at 11:56
lhf
156k9160367
add a comment |Â
up vote
2
down vote
A ring homomorphism $Fto E$ induces a group homomorphism $F^times to E^times$. These groups have orders $7$ and $31$ and so this homomorphism must be the trivial one. On the other hand, a ring homomorphism from a field is injective.
answered Jul 17 at 11:56
lhf
156k9160367
add a comment |Â
up vote
2
down vote
up vote
2
down vote
A ring homomorphism $Fto E$ induces a group homomorphism $F^times to E^times$. These groups have orders $7$ and $31$ and so this homomorphism must be the trivial one. On the other hand, a ring homomorphism from a field is injective.
answered Jul 17 at 11:56
lhf
156k9160367
A ring homomorphism $Fto E$ induces a group homomorphism $F^times to E^times$. These groups have orders $7$ and $31$ and so this homomorphism must be the trivial one. On the other hand, a ring homomorphism from a field is injective.
answered Jul 17 at 11:56
lhf
156k9160367
answered Jul 17 at 11:56
lhf
156k9160367
answered Jul 17 at 11:56
lhf
156k9160367
answered Jul 17 at 11:56
lhf
156k9160367
156k9160367
add a comment |Â
add a comment |Â
up vote
1
down vote
Since such a homomorphism $phi$ would be injective (the kernel is an ideal, and we are assuming $phi$ nontrivial), it is in fact an embedding, and $phi (textGF(2^3))cong textGF(2^3)$ would be a subfield of $textGF(2^5)$; but since $3notmid5$, this is impossible .
answered Jul 17 at 1:46
Chris Custer
5,4582622
add a comment |Â
up vote
1
down vote
Since such a homomorphism $phi$ would be injective (the kernel is an ideal, and we are assuming $phi$ nontrivial), it is in fact an embedding, and $phi (textGF(2^3))cong textGF(2^3)$ would be a subfield of $textGF(2^5)$; but since $3notmid5$, this is impossible .
answered Jul 17 at 1:46
Chris Custer
5,4582622
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Since such a homomorphism $phi$ would be injective (the kernel is an ideal, and we are assuming $phi$ nontrivial), it is in fact an embedding, and $phi (textGF(2^3))cong textGF(2^3)$ would be a subfield of $textGF(2^5)$; but since $3notmid5$, this is impossible .
answered Jul 17 at 1:46
Chris Custer
5,4582622
Since such a homomorphism $phi$ would be injective (the kernel is an ideal, and we are assuming $phi$ nontrivial), it is in fact an embedding, and $phi (textGF(2^3))cong textGF(2^3)$ would be a subfield of $textGF(2^5)$; but since $3notmid5$, this is impossible .
answered Jul 17 at 1:46
Chris Custer
5,4582622
edited Jul 17 at 2:11
answered Jul 17 at 1:46
Chris Custer
5,4582622
answered Jul 17 at 1:46
Chris Custer
5,4582622
answered Jul 17 at 1:46
Chris Custer
5,4582622
5,4582622
add a comment |Â
add a comment |Â
up vote
1
down vote
Let $F$ be the field with only two elements. If $F_8$ and $F_32$ are fields of order $8$ and $32$ respectively, then they are both realizable as algebraic extensions of $F.$ Any homomorphism $phi : F_8 longrightarrow F_32$ must be injective, as all fields are simple. That is, if we assume that $F_8 subset F_32,$ by way of identifying $F_8$ with $textImg(phi)$ then this is to say that $F_32$ must be realizable as an algebraic extension of $F_8.$ So we get the following inclusions $Fsubset F_8 subset F_32.$ Since these are all finite extensions, and since every finite extension of a finite field is Galois, then the Galois correspondence gives us corresponding subgroups of orders $1,$ $3$ and $5$ respectively. Since any group of order $5$ has no subgroup of order $3$, then the existence of an embedding $F_8subset F_32$ must be impossible.
Conclude that no such homomorphism exists.
edited Jul 17 at 5:23
Chris Custer
5,4582622
answered Jul 17 at 0:17
Chickenmancer
3,017621
add a comment |Â
up vote
1
down vote
Let $F$ be the field with only two elements. If $F_8$ and $F_32$ are fields of order $8$ and $32$ respectively, then they are both realizable as algebraic extensions of $F.$ Any homomorphism $phi : F_8 longrightarrow F_32$ must be injective, as all fields are simple. That is, if we assume that $F_8 subset F_32,$ by way of identifying $F_8$ with $textImg(phi)$ then this is to say that $F_32$ must be realizable as an algebraic extension of $F_8.$ So we get the following inclusions $Fsubset F_8 subset F_32.$ Since these are all finite extensions, and since every finite extension of a finite field is Galois, then the Galois correspondence gives us corresponding subgroups of orders $1,$ $3$ and $5$ respectively. Since any group of order $5$ has no subgroup of order $3$, then the existence of an embedding $F_8subset F_32$ must be impossible.
Conclude that no such homomorphism exists.
edited Jul 17 at 5:23
Chris Custer
5,4582622
answered Jul 17 at 0:17
Chickenmancer
3,017621
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Let $F$ be the field with only two elements. If $F_8$ and $F_32$ are fields of order $8$ and $32$ respectively, then they are both realizable as algebraic extensions of $F.$ Any homomorphism $phi : F_8 longrightarrow F_32$ must be injective, as all fields are simple. That is, if we assume that $F_8 subset F_32,$ by way of identifying $F_8$ with $textImg(phi)$ then this is to say that $F_32$ must be realizable as an algebraic extension of $F_8.$ So we get the following inclusions $Fsubset F_8 subset F_32.$ Since these are all finite extensions, and since every finite extension of a finite field is Galois, then the Galois correspondence gives us corresponding subgroups of orders $1,$ $3$ and $5$ respectively. Since any group of order $5$ has no subgroup of order $3$, then the existence of an embedding $F_8subset F_32$ must be impossible.
Conclude that no such homomorphism exists.
edited Jul 17 at 5:23
Chris Custer
5,4582622
answered Jul 17 at 0:17
Chickenmancer
3,017621
Let $F$ be the field with only two elements. If $F_8$ and $F_32$ are fields of order $8$ and $32$ respectively, then they are both realizable as algebraic extensions of $F.$ Any homomorphism $phi : F_8 longrightarrow F_32$ must be injective, as all fields are simple. That is, if we assume that $F_8 subset F_32,$ by way of identifying $F_8$ with $textImg(phi)$ then this is to say that $F_32$ must be realizable as an algebraic extension of $F_8.$ So we get the following inclusions $Fsubset F_8 subset F_32.$ Since these are all finite extensions, and since every finite extension of a finite field is Galois, then the Galois correspondence gives us corresponding subgroups of orders $1,$ $3$ and $5$ respectively. Since any group of order $5$ has no subgroup of order $3$, then the existence of an embedding $F_8subset F_32$ must be impossible.
Conclude that no such homomorphism exists.
edited Jul 17 at 5:23
Chris Custer
5,4582622
answered Jul 17 at 0:17
Chickenmancer
3,017621
edited Jul 17 at 5:23
Chris Custer
5,4582622
edited Jul 17 at 5:23
Chris Custer
5,4582622
edited Jul 17 at 5:23
Chris Custer
5,4582622
5,4582622
answered Jul 17 at 0:17
Chickenmancer
3,017621
answered Jul 17 at 0:17
Chickenmancer
3,017621
answered Jul 17 at 0:17
Chickenmancer
3,017621
3,017621
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2853934%2fexistence-of-homomorphisms-between-finite-fields%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
5
As always, right after posting the question, I found a flaw in my argument: I use that $2^7=1$, but $x^7=1$ is valid only for $xne 0$ and $2=0$.
– user437309
Jul 16 at 22:51
1
Good call on that flaw (and the exact same flaw in my answer, effectively)!
– Cameron Buie
Jul 17 at 0:07
Does the definition of "ring homomorphism" you work with include that $f(1) =1$? All answers assume this and it is a natural thing to do. However, without that assumption, the zero map $f(x)=0$ for all $x$ is the only possibility.
– Torsten Schoeneberg
Jul 17 at 17:00
@TorstenSchoeneberg Yes it does.
– user437309
Jul 17 at 17:26