Is $Kcong (mathbb{Z}/3mathbb{Z})times(mathbb{Z}/3mathbb{Z})$? [closed]
Is the following statement true/false?
Let $(mathbb{F},+,cdot)$ be the finite field with $9$ elements. Let $K=(mathbb{F},+)$ denote the underlying additive group; then
$$Kcong (mathbb{Z}/3mathbb{Z})times(mathbb{Z}/3mathbb{Z})$$
My attempt: I thinks this statement is false because $gcd(3,3)= 3 neq 1$
Is my logic correct?
Any hints/solution will be appreciated
abstract-algebra
edited Dec 3 '18 at 8:38
egreg
178k1484201
asked Dec 3 '18 at 8:31
jasmine
1,535416
closed as off-topic by user21820, Jyrki Lahtonen, Saad, Brahadeesh, ancientmathematician Dec 8 '18 at 15:19
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Jyrki Lahtonen, Saad, Brahadeesh, ancientmathematician
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
Is the following statement true/false?
Let $(mathbb{F},+,cdot)$ be the finite field with $9$ elements. Let $K=(mathbb{F},+)$ denote the underlying additive group; then
$$Kcong (mathbb{Z}/3mathbb{Z})times(mathbb{Z}/3mathbb{Z})$$
My attempt: I thinks this statement is false because $gcd(3,3)= 3 neq 1$
Is my logic correct?
Any hints/solution will be appreciated
abstract-algebra
edited Dec 3 '18 at 8:38
egreg
178k1484201
asked Dec 3 '18 at 8:31
jasmine
1,535416
closed as off-topic by user21820, Jyrki Lahtonen, Saad, Brahadeesh, ancientmathematician Dec 8 '18 at 15:19
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Jyrki Lahtonen, Saad, Brahadeesh, ancientmathematician
If this question can be reworded to fit the rules in the help center, please edit the question.
1
Why would $gcd(3, 3)neq 1$ prove the result?
– Arthur
Dec 3 '18 at 8:39
add a comment |
Is the following statement true/false?
Let $(mathbb{F},+,cdot)$ be the finite field with $9$ elements. Let $K=(mathbb{F},+)$ denote the underlying additive group; then
$$Kcong (mathbb{Z}/3mathbb{Z})times(mathbb{Z}/3mathbb{Z})$$
My attempt: I thinks this statement is false because $gcd(3,3)= 3 neq 1$
Is my logic correct?
Any hints/solution will be appreciated
abstract-algebra
edited Dec 3 '18 at 8:38
egreg
178k1484201
asked Dec 3 '18 at 8:31
jasmine
1,535416
Is the following statement true/false?
Let $(mathbb{F},+,cdot)$ be the finite field with $9$ elements. Let $K=(mathbb{F},+)$ denote the underlying additive group; then
$$Kcong (mathbb{Z}/3mathbb{Z})times(mathbb{Z}/3mathbb{Z})$$
My attempt: I thinks this statement is false because $gcd(3,3)= 3 neq 1$
Is my logic correct?
Any hints/solution will be appreciated
abstract-algebra
abstract-algebra
edited Dec 3 '18 at 8:38
egreg
178k1484201
asked Dec 3 '18 at 8:31
jasmine
1,535416
edited Dec 3 '18 at 8:38
egreg
178k1484201
asked Dec 3 '18 at 8:31
jasmine
1,535416
edited Dec 3 '18 at 8:38
egreg
178k1484201
edited Dec 3 '18 at 8:38
egreg
178k1484201
edited Dec 3 '18 at 8:38
egreg
178k1484201
178k1484201
asked Dec 3 '18 at 8:31
jasmine
1,535416
asked Dec 3 '18 at 8:31
jasmine
1,535416
asked Dec 3 '18 at 8:31
jasmine
1,535416
1,535416
closed as off-topic by user21820, Jyrki Lahtonen, Saad, Brahadeesh, ancientmathematician Dec 8 '18 at 15:19
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Jyrki Lahtonen, Saad, Brahadeesh, ancientmathematician
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by user21820, Jyrki Lahtonen, Saad, Brahadeesh, ancientmathematician Dec 8 '18 at 15:19
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Jyrki Lahtonen, Saad, Brahadeesh, ancientmathematician
If this question can be reworded to fit the rules in the help center, please edit the question.
1
Why would $gcd(3, 3)neq 1$ prove the result?
– Arthur
Dec 3 '18 at 8:39
add a comment |
1
Why would $gcd(3, 3)neq 1$ prove the result?
– Arthur
Dec 3 '18 at 8:39
1
1
Why would $gcd(3, 3)neq 1$ prove the result?
– Arthur
Dec 3 '18 at 8:39
Why would $gcd(3, 3)neq 1$ prove the result?
– Arthur
Dec 3 '18 at 8:39
add a comment |
2 Answers
2
active
oldest
votes
The statement is true; $mathbb{F}$ is a vector space over the three-element field and has dimension $2$. The same holds for $(mathbb{Z}/3mathbb{Z})times(mathbb{Z}/3mathbb{Z})$.
It is false that $mathbb{F}cong(mathbb{Z}/3mathbb{Z})times(mathbb{Z}/3mathbb{Z})$ as rings.
answered Dec 3 '18 at 8:40
egreg
178k1484201
@@@egreq that mean $mathbb{Z} /3mathbb{Z} timesmathbb{Z} /3mathbb{Z}= mathbb{Z_9} $ or $mathbb{F_9} $ ? which one u r saying
– jasmine
Dec 3 '18 at 8:42
@jasmine It is not true that $mathbb{Z}/3mathbb{Z}timesmathbb{Z}/3mathbb{Z}simeqmathbb{Z}_9$.
– freakish
Dec 3 '18 at 8:47
@freakish that mean $mathbb{F_9}$ is True ??
– jasmine
Dec 3 '18 at 8:48
1
@jasmine The statement in the question (isomorphism as abelian group) is true. The statement that $(mathbb{Z}/3mathbb{Z})times(mathbb{Z}/3mathbb{Z})congmathbb{Z}/9mathbb{Z}$ is false.
– egreg
Dec 3 '18 at 8:53
@egreg thanks u .
– jasmine
Dec 3 '18 at 8:56
add a comment |
It's an abelian group with nine elements, where each element has order $3$. There is only one such group (which I suppose is what you actually need to prove).
answered Dec 3 '18 at 8:40
Arthur
110k7105186
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
The statement is true; $mathbb{F}$ is a vector space over the three-element field and has dimension $2$. The same holds for $(mathbb{Z}/3mathbb{Z})times(mathbb{Z}/3mathbb{Z})$.
It is false that $mathbb{F}cong(mathbb{Z}/3mathbb{Z})times(mathbb{Z}/3mathbb{Z})$ as rings.
answered Dec 3 '18 at 8:40
egreg
178k1484201
@@@egreq that mean $mathbb{Z} /3mathbb{Z} timesmathbb{Z} /3mathbb{Z}= mathbb{Z_9} $ or $mathbb{F_9} $ ? which one u r saying
– jasmine
Dec 3 '18 at 8:42
@jasmine It is not true that $mathbb{Z}/3mathbb{Z}timesmathbb{Z}/3mathbb{Z}simeqmathbb{Z}_9$.
– freakish
Dec 3 '18 at 8:47
@freakish that mean $mathbb{F_9}$ is True ??
– jasmine
Dec 3 '18 at 8:48
1
@jasmine The statement in the question (isomorphism as abelian group) is true. The statement that $(mathbb{Z}/3mathbb{Z})times(mathbb{Z}/3mathbb{Z})congmathbb{Z}/9mathbb{Z}$ is false.
– egreg
Dec 3 '18 at 8:53
@egreg thanks u .
– jasmine
Dec 3 '18 at 8:56
add a comment |
The statement is true; $mathbb{F}$ is a vector space over the three-element field and has dimension $2$. The same holds for $(mathbb{Z}/3mathbb{Z})times(mathbb{Z}/3mathbb{Z})$.
It is false that $mathbb{F}cong(mathbb{Z}/3mathbb{Z})times(mathbb{Z}/3mathbb{Z})$ as rings.
answered Dec 3 '18 at 8:40
egreg
178k1484201
@@@egreq that mean $mathbb{Z} /3mathbb{Z} timesmathbb{Z} /3mathbb{Z}= mathbb{Z_9} $ or $mathbb{F_9} $ ? which one u r saying
– jasmine
Dec 3 '18 at 8:42
@jasmine It is not true that $mathbb{Z}/3mathbb{Z}timesmathbb{Z}/3mathbb{Z}simeqmathbb{Z}_9$.
– freakish
Dec 3 '18 at 8:47
@freakish that mean $mathbb{F_9}$ is True ??
– jasmine
Dec 3 '18 at 8:48
1
@jasmine The statement in the question (isomorphism as abelian group) is true. The statement that $(mathbb{Z}/3mathbb{Z})times(mathbb{Z}/3mathbb{Z})congmathbb{Z}/9mathbb{Z}$ is false.
– egreg
Dec 3 '18 at 8:53
@egreg thanks u .
– jasmine
Dec 3 '18 at 8:56
add a comment |
The statement is true; $mathbb{F}$ is a vector space over the three-element field and has dimension $2$. The same holds for $(mathbb{Z}/3mathbb{Z})times(mathbb{Z}/3mathbb{Z})$.
It is false that $mathbb{F}cong(mathbb{Z}/3mathbb{Z})times(mathbb{Z}/3mathbb{Z})$ as rings.
answered Dec 3 '18 at 8:40
egreg
178k1484201
The statement is true; $mathbb{F}$ is a vector space over the three-element field and has dimension $2$. The same holds for $(mathbb{Z}/3mathbb{Z})times(mathbb{Z}/3mathbb{Z})$.
It is false that $mathbb{F}cong(mathbb{Z}/3mathbb{Z})times(mathbb{Z}/3mathbb{Z})$ as rings.
answered Dec 3 '18 at 8:40
egreg
178k1484201
answered Dec 3 '18 at 8:40
egreg
178k1484201
answered Dec 3 '18 at 8:40
egreg
178k1484201
answered Dec 3 '18 at 8:40
egreg
178k1484201
178k1484201
@@@egreq that mean $mathbb{Z} /3mathbb{Z} timesmathbb{Z} /3mathbb{Z}= mathbb{Z_9} $ or $mathbb{F_9} $ ? which one u r saying
– jasmine
Dec 3 '18 at 8:42
@jasmine It is not true that $mathbb{Z}/3mathbb{Z}timesmathbb{Z}/3mathbb{Z}simeqmathbb{Z}_9$.
– freakish
Dec 3 '18 at 8:47
@freakish that mean $mathbb{F_9}$ is True ??
– jasmine
Dec 3 '18 at 8:48
1
@jasmine The statement in the question (isomorphism as abelian group) is true. The statement that $(mathbb{Z}/3mathbb{Z})times(mathbb{Z}/3mathbb{Z})congmathbb{Z}/9mathbb{Z}$ is false.
– egreg
Dec 3 '18 at 8:53
@egreg thanks u .
– jasmine
Dec 3 '18 at 8:56
add a comment |
@@@egreq that mean $mathbb{Z} /3mathbb{Z} timesmathbb{Z} /3mathbb{Z}= mathbb{Z_9} $ or $mathbb{F_9} $ ? which one u r saying
– jasmine
Dec 3 '18 at 8:42
@jasmine It is not true that $mathbb{Z}/3mathbb{Z}timesmathbb{Z}/3mathbb{Z}simeqmathbb{Z}_9$.
– freakish
Dec 3 '18 at 8:47
@freakish that mean $mathbb{F_9}$ is True ??
– jasmine
Dec 3 '18 at 8:48
1
@jasmine The statement in the question (isomorphism as abelian group) is true. The statement that $(mathbb{Z}/3mathbb{Z})times(mathbb{Z}/3mathbb{Z})congmathbb{Z}/9mathbb{Z}$ is false.
– egreg
Dec 3 '18 at 8:53
@egreg thanks u .
– jasmine
Dec 3 '18 at 8:56
@@@egreq that mean $mathbb{Z} /3mathbb{Z} timesmathbb{Z} /3mathbb{Z}= mathbb{Z_9} $ or $mathbb{F_9} $ ? which one u r saying
– jasmine
Dec 3 '18 at 8:42
@@@egreq that mean $mathbb{Z} /3mathbb{Z} timesmathbb{Z} /3mathbb{Z}= mathbb{Z_9} $ or $mathbb{F_9} $ ? which one u r saying
– jasmine
Dec 3 '18 at 8:42
@jasmine It is not true that $mathbb{Z}/3mathbb{Z}timesmathbb{Z}/3mathbb{Z}simeqmathbb{Z}_9$.
– freakish
Dec 3 '18 at 8:47
@jasmine It is not true that $mathbb{Z}/3mathbb{Z}timesmathbb{Z}/3mathbb{Z}simeqmathbb{Z}_9$.
– freakish
Dec 3 '18 at 8:47
@freakish that mean $mathbb{F_9}$ is True ??
– jasmine
Dec 3 '18 at 8:48
@freakish that mean $mathbb{F_9}$ is True ??
– jasmine
Dec 3 '18 at 8:48
1
1
@jasmine The statement in the question (isomorphism as abelian group) is true. The statement that $(mathbb{Z}/3mathbb{Z})times(mathbb{Z}/3mathbb{Z})congmathbb{Z}/9mathbb{Z}$ is false.
– egreg
Dec 3 '18 at 8:53
@jasmine The statement in the question (isomorphism as abelian group) is true. The statement that $(mathbb{Z}/3mathbb{Z})times(mathbb{Z}/3mathbb{Z})congmathbb{Z}/9mathbb{Z}$ is false.
– egreg
Dec 3 '18 at 8:53
@egreg thanks u .
– jasmine
Dec 3 '18 at 8:56
@egreg thanks u .
– jasmine
Dec 3 '18 at 8:56
add a comment |
It's an abelian group with nine elements, where each element has order $3$. There is only one such group (which I suppose is what you actually need to prove).
answered Dec 3 '18 at 8:40
Arthur
110k7105186
add a comment |
It's an abelian group with nine elements, where each element has order $3$. There is only one such group (which I suppose is what you actually need to prove).
answered Dec 3 '18 at 8:40
Arthur
110k7105186
add a comment |
It's an abelian group with nine elements, where each element has order $3$. There is only one such group (which I suppose is what you actually need to prove).
answered Dec 3 '18 at 8:40
Arthur
110k7105186
It's an abelian group with nine elements, where each element has order $3$. There is only one such group (which I suppose is what you actually need to prove).
answered Dec 3 '18 at 8:40
Arthur
110k7105186
answered Dec 3 '18 at 8:40
Arthur
110k7105186
answered Dec 3 '18 at 8:40
Arthur
110k7105186
answered Dec 3 '18 at 8:40
Arthur
110k7105186
110k7105186
add a comment |
add a comment |
1
Why would $gcd(3, 3)neq 1$ prove the result?
– Arthur
Dec 3 '18 at 8:39