Mixing
If $K= mathbb Z /2mathbb Z$ what is meant by $K^2$
$begingroup$
I realize that $K= mathbb Z /2mathbb Z$ is simply the set of equivalence classes. But I recently came across $K^{2}$ given $K= mathbb Z /2mathbb Z$
It is then stated that $K^{2}={{0,0},{1,0},{0,1},{1,1}}$
I have no idea what those vectors have to do with $mathbb Z /2mathbb Z$. Any ideas as to what I am missing here?
linear-algebra field-theory equivalence-relations
asked Jan 17 at 22:11
SABOYSABOY
656311
$endgroup$
add a comment |
$begingroup$
I realize that $K= mathbb Z /2mathbb Z$ is simply the set of equivalence classes. But I recently came across $K^{2}$ given $K= mathbb Z /2mathbb Z$
It is then stated that $K^{2}={{0,0},{1,0},{0,1},{1,1}}$
I have no idea what those vectors have to do with $mathbb Z /2mathbb Z$. Any ideas as to what I am missing here?
linear-algebra field-theory equivalence-relations
asked Jan 17 at 22:11
SABOYSABOY
656311
$endgroup$
1
$begingroup$
Would you understand it better if it was $Ktimes K$?
$endgroup$
– Arnaud D.
Jan 17 at 22:13
$begingroup$
Those are two-dimensional vectors whose entries are from (the field) $;Bbb Z/2Bbb Z;$ ...
$endgroup$
– DonAntonio
Jan 17 at 22:14
$begingroup$
$K^2$ is the ring product $Ktimes K$, which is the set of all pairs $(x,y)$ with $x,yin K$.
$endgroup$
– Dave
Jan 17 at 22:14
1
$begingroup$
If $K$ is a set (any set) then $K^2 = K times K = {(a,b)|a,bin K}=$ then set of all ordered pairs of elements of $K$. So $K = mathbb Z/2mathbb Z =$ set of equivalence classes modulo 2 = ${0, 1}$. So $K^2 = {(a,b)| a = 0, 1 b=0,1} = {(0,0),(0,1),(1,0),(1,1)}$.
$endgroup$
– fleablood
Jan 17 at 23:09
add a comment |
$begingroup$
I realize that $K= mathbb Z /2mathbb Z$ is simply the set of equivalence classes. But I recently came across $K^{2}$ given $K= mathbb Z /2mathbb Z$
It is then stated that $K^{2}={{0,0},{1,0},{0,1},{1,1}}$
I have no idea what those vectors have to do with $mathbb Z /2mathbb Z$. Any ideas as to what I am missing here?
linear-algebra field-theory equivalence-relations
asked Jan 17 at 22:11
SABOYSABOY
656311
$endgroup$
I realize that $K= mathbb Z /2mathbb Z$ is simply the set of equivalence classes. But I recently came across $K^{2}$ given $K= mathbb Z /2mathbb Z$
It is then stated that $K^{2}={{0,0},{1,0},{0,1},{1,1}}$
I have no idea what those vectors have to do with $mathbb Z /2mathbb Z$. Any ideas as to what I am missing here?
linear-algebra field-theory equivalence-relations
linear-algebra field-theory equivalence-relations
asked Jan 17 at 22:11
SABOYSABOY
656311
asked Jan 17 at 22:11
SABOYSABOY
656311
asked Jan 17 at 22:11
SABOYSABOY
656311
asked Jan 17 at 22:11
SABOYSABOY
656311
asked Jan 17 at 22:11
SABOYSABOY
656311
656311
1
$begingroup$
Would you understand it better if it was $Ktimes K$?
$endgroup$
– Arnaud D.
Jan 17 at 22:13
$begingroup$
Those are two-dimensional vectors whose entries are from (the field) $;Bbb Z/2Bbb Z;$ ...
$endgroup$
– DonAntonio
Jan 17 at 22:14
$begingroup$
$K^2$ is the ring product $Ktimes K$, which is the set of all pairs $(x,y)$ with $x,yin K$.
$endgroup$
– Dave
Jan 17 at 22:14
1
$begingroup$
If $K$ is a set (any set) then $K^2 = K times K = {(a,b)|a,bin K}=$ then set of all ordered pairs of elements of $K$. So $K = mathbb Z/2mathbb Z =$ set of equivalence classes modulo 2 = ${0, 1}$. So $K^2 = {(a,b)| a = 0, 1 b=0,1} = {(0,0),(0,1),(1,0),(1,1)}$.
$endgroup$
– fleablood
Jan 17 at 23:09
add a comment |
1
$begingroup$
Would you understand it better if it was $Ktimes K$?
$endgroup$
– Arnaud D.
Jan 17 at 22:13
$begingroup$
Those are two-dimensional vectors whose entries are from (the field) $;Bbb Z/2Bbb Z;$ ...
$endgroup$
– DonAntonio
Jan 17 at 22:14
$begingroup$
$K^2$ is the ring product $Ktimes K$, which is the set of all pairs $(x,y)$ with $x,yin K$.
$endgroup$
– Dave
Jan 17 at 22:14
1
$begingroup$
If $K$ is a set (any set) then $K^2 = K times K = {(a,b)|a,bin K}=$ then set of all ordered pairs of elements of $K$. So $K = mathbb Z/2mathbb Z =$ set of equivalence classes modulo 2 = ${0, 1}$. So $K^2 = {(a,b)| a = 0, 1 b=0,1} = {(0,0),(0,1),(1,0),(1,1)}$.
$endgroup$
– fleablood
Jan 17 at 23:09
1
1
$begingroup$
Would you understand it better if it was $Ktimes K$?
$endgroup$
– Arnaud D.
Jan 17 at 22:13
$begingroup$
Would you understand it better if it was $Ktimes K$?
$endgroup$
– Arnaud D.
Jan 17 at 22:13
$begingroup$
Those are two-dimensional vectors whose entries are from (the field) $;Bbb Z/2Bbb Z;$ ...
$endgroup$
– DonAntonio
Jan 17 at 22:14
$begingroup$
Those are two-dimensional vectors whose entries are from (the field) $;Bbb Z/2Bbb Z;$ ...
$endgroup$
– DonAntonio
Jan 17 at 22:14
$begingroup$
$K^2$ is the ring product $Ktimes K$, which is the set of all pairs $(x,y)$ with $x,yin K$.
$endgroup$
– Dave
Jan 17 at 22:14
$begingroup$
$K^2$ is the ring product $Ktimes K$, which is the set of all pairs $(x,y)$ with $x,yin K$.
$endgroup$
– Dave
Jan 17 at 22:14
1
1
$begingroup$
If $K$ is a set (any set) then $K^2 = K times K = {(a,b)|a,bin K}=$ then set of all ordered pairs of elements of $K$. So $K = mathbb Z/2mathbb Z =$ set of equivalence classes modulo 2 = ${0, 1}$. So $K^2 = {(a,b)| a = 0, 1 b=0,1} = {(0,0),(0,1),(1,0),(1,1)}$.
$endgroup$
– fleablood
Jan 17 at 23:09
$begingroup$
If $K$ is a set (any set) then $K^2 = K times K = {(a,b)|a,bin K}=$ then set of all ordered pairs of elements of $K$. So $K = mathbb Z/2mathbb Z =$ set of equivalence classes modulo 2 = ${0, 1}$. So $K^2 = {(a,b)| a = 0, 1 b=0,1} = {(0,0),(0,1),(1,0),(1,1)}$.
$endgroup$
– fleablood
Jan 17 at 23:09
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Yes, $K = mathbb{Z}/2mathbb{Z}$ contains two equivalences classes:
$$ mathbb{Z}/2mathbb{Z} = {[0],[1]}$$
where $0$ and $1$ are used to act as the representative of those equivlances classes. As you know, $[0]$ is the set of all even integers, and $[1]$ is the set of all odd integers. So the Cartesian product $(mathbb{Z}/2mathbb{Z})times(mathbb{Z}/2mathbb{Z})$ is the set
$${([0],[0]),([0],[1]),([1],[0]),([1],[1])}$$
This is pretty cluttered. Since there is a canonical isomorphism between $mathbb{Z}/2mathbb{Z}$ and the subgroup ${0,1}$ of $mathbb{Z}$, there is then a canonical isomorphism between the groups $(mathbb{Z}/2mathbb{Z})times(mathbb{Z}/2mathbb{Z})$ and ${0,1}times{0,1}subseteq mathbb{Z}timesmathbb{Z}$. Since $mathbb{Z}/2mathbb{Z}$ is usually identified with ${0,1}$, the group $K^2 = K times K =(mathbb{Z}/2mathbb{Z})times(mathbb{Z}/2mathbb{Z})$ is usually just identified with ${0,1}times{0,1}$:
$${(0,0),(0,1),(1,0),(1,1)}$$
Note that you used sets rather than 2-tuples in your expression, which isn't correct, since, for example, ${1,1} = {1}$ by Extensionality.
answered Jan 18 at 16:58
MetricMetric
1,23649
$endgroup$
add a comment |
$begingroup$
If you follow the definition, $K= mathbb Z /2mathbb Z$ consists of two classes: even and odd integers. It is natural to map the class of even integers into $0$ and the class of odd integers into $1$. This bijection in turn becomes a ring isomorphism between $mathbb Z /2mathbb Z$ and ${0,1}$. Strictly speaking, these sets are isomorphic, but more loosely they are often identified. $K^2=K times K$ is the Cartesian product of $K$ with itself: $K^{2}={(0,0),(1,0),(0,1),(1,1)}$
answered Jan 17 at 23:05
David TarandekDavid Tarandek
234
$endgroup$
add a comment |
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2 Answers
2
active
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes, $K = mathbb{Z}/2mathbb{Z}$ contains two equivalences classes:
$$ mathbb{Z}/2mathbb{Z} = {[0],[1]}$$
where $0$ and $1$ are used to act as the representative of those equivlances classes. As you know, $[0]$ is the set of all even integers, and $[1]$ is the set of all odd integers. So the Cartesian product $(mathbb{Z}/2mathbb{Z})times(mathbb{Z}/2mathbb{Z})$ is the set
$${([0],[0]),([0],[1]),([1],[0]),([1],[1])}$$
This is pretty cluttered. Since there is a canonical isomorphism between $mathbb{Z}/2mathbb{Z}$ and the subgroup ${0,1}$ of $mathbb{Z}$, there is then a canonical isomorphism between the groups $(mathbb{Z}/2mathbb{Z})times(mathbb{Z}/2mathbb{Z})$ and ${0,1}times{0,1}subseteq mathbb{Z}timesmathbb{Z}$. Since $mathbb{Z}/2mathbb{Z}$ is usually identified with ${0,1}$, the group $K^2 = K times K =(mathbb{Z}/2mathbb{Z})times(mathbb{Z}/2mathbb{Z})$ is usually just identified with ${0,1}times{0,1}$:
$${(0,0),(0,1),(1,0),(1,1)}$$
Note that you used sets rather than 2-tuples in your expression, which isn't correct, since, for example, ${1,1} = {1}$ by Extensionality.
answered Jan 18 at 16:58
MetricMetric
1,23649
$endgroup$
add a comment |
$begingroup$
Yes, $K = mathbb{Z}/2mathbb{Z}$ contains two equivalences classes:
$$ mathbb{Z}/2mathbb{Z} = {[0],[1]}$$
where $0$ and $1$ are used to act as the representative of those equivlances classes. As you know, $[0]$ is the set of all even integers, and $[1]$ is the set of all odd integers. So the Cartesian product $(mathbb{Z}/2mathbb{Z})times(mathbb{Z}/2mathbb{Z})$ is the set
$${([0],[0]),([0],[1]),([1],[0]),([1],[1])}$$
This is pretty cluttered. Since there is a canonical isomorphism between $mathbb{Z}/2mathbb{Z}$ and the subgroup ${0,1}$ of $mathbb{Z}$, there is then a canonical isomorphism between the groups $(mathbb{Z}/2mathbb{Z})times(mathbb{Z}/2mathbb{Z})$ and ${0,1}times{0,1}subseteq mathbb{Z}timesmathbb{Z}$. Since $mathbb{Z}/2mathbb{Z}$ is usually identified with ${0,1}$, the group $K^2 = K times K =(mathbb{Z}/2mathbb{Z})times(mathbb{Z}/2mathbb{Z})$ is usually just identified with ${0,1}times{0,1}$:
$${(0,0),(0,1),(1,0),(1,1)}$$
Note that you used sets rather than 2-tuples in your expression, which isn't correct, since, for example, ${1,1} = {1}$ by Extensionality.
answered Jan 18 at 16:58
MetricMetric
1,23649
$endgroup$
add a comment |
$begingroup$
Yes, $K = mathbb{Z}/2mathbb{Z}$ contains two equivalences classes:
$$ mathbb{Z}/2mathbb{Z} = {[0],[1]}$$
where $0$ and $1$ are used to act as the representative of those equivlances classes. As you know, $[0]$ is the set of all even integers, and $[1]$ is the set of all odd integers. So the Cartesian product $(mathbb{Z}/2mathbb{Z})times(mathbb{Z}/2mathbb{Z})$ is the set
$${([0],[0]),([0],[1]),([1],[0]),([1],[1])}$$
This is pretty cluttered. Since there is a canonical isomorphism between $mathbb{Z}/2mathbb{Z}$ and the subgroup ${0,1}$ of $mathbb{Z}$, there is then a canonical isomorphism between the groups $(mathbb{Z}/2mathbb{Z})times(mathbb{Z}/2mathbb{Z})$ and ${0,1}times{0,1}subseteq mathbb{Z}timesmathbb{Z}$. Since $mathbb{Z}/2mathbb{Z}$ is usually identified with ${0,1}$, the group $K^2 = K times K =(mathbb{Z}/2mathbb{Z})times(mathbb{Z}/2mathbb{Z})$ is usually just identified with ${0,1}times{0,1}$:
$${(0,0),(0,1),(1,0),(1,1)}$$
Note that you used sets rather than 2-tuples in your expression, which isn't correct, since, for example, ${1,1} = {1}$ by Extensionality.
answered Jan 18 at 16:58
MetricMetric
1,23649
$endgroup$
Yes, $K = mathbb{Z}/2mathbb{Z}$ contains two equivalences classes:
$$ mathbb{Z}/2mathbb{Z} = {[0],[1]}$$
where $0$ and $1$ are used to act as the representative of those equivlances classes. As you know, $[0]$ is the set of all even integers, and $[1]$ is the set of all odd integers. So the Cartesian product $(mathbb{Z}/2mathbb{Z})times(mathbb{Z}/2mathbb{Z})$ is the set
$${([0],[0]),([0],[1]),([1],[0]),([1],[1])}$$
This is pretty cluttered. Since there is a canonical isomorphism between $mathbb{Z}/2mathbb{Z}$ and the subgroup ${0,1}$ of $mathbb{Z}$, there is then a canonical isomorphism between the groups $(mathbb{Z}/2mathbb{Z})times(mathbb{Z}/2mathbb{Z})$ and ${0,1}times{0,1}subseteq mathbb{Z}timesmathbb{Z}$. Since $mathbb{Z}/2mathbb{Z}$ is usually identified with ${0,1}$, the group $K^2 = K times K =(mathbb{Z}/2mathbb{Z})times(mathbb{Z}/2mathbb{Z})$ is usually just identified with ${0,1}times{0,1}$:
$${(0,0),(0,1),(1,0),(1,1)}$$
Note that you used sets rather than 2-tuples in your expression, which isn't correct, since, for example, ${1,1} = {1}$ by Extensionality.
answered Jan 18 at 16:58
MetricMetric
1,23649
answered Jan 18 at 16:58
MetricMetric
1,23649
answered Jan 18 at 16:58
MetricMetric
1,23649
answered Jan 18 at 16:58
MetricMetric
1,23649
1,23649
add a comment |
add a comment |
$begingroup$
If you follow the definition, $K= mathbb Z /2mathbb Z$ consists of two classes: even and odd integers. It is natural to map the class of even integers into $0$ and the class of odd integers into $1$. This bijection in turn becomes a ring isomorphism between $mathbb Z /2mathbb Z$ and ${0,1}$. Strictly speaking, these sets are isomorphic, but more loosely they are often identified. $K^2=K times K$ is the Cartesian product of $K$ with itself: $K^{2}={(0,0),(1,0),(0,1),(1,1)}$
answered Jan 17 at 23:05
David TarandekDavid Tarandek
234
$endgroup$
add a comment |
$begingroup$
If you follow the definition, $K= mathbb Z /2mathbb Z$ consists of two classes: even and odd integers. It is natural to map the class of even integers into $0$ and the class of odd integers into $1$. This bijection in turn becomes a ring isomorphism between $mathbb Z /2mathbb Z$ and ${0,1}$. Strictly speaking, these sets are isomorphic, but more loosely they are often identified. $K^2=K times K$ is the Cartesian product of $K$ with itself: $K^{2}={(0,0),(1,0),(0,1),(1,1)}$
answered Jan 17 at 23:05
David TarandekDavid Tarandek
234
$endgroup$
add a comment |
$begingroup$
If you follow the definition, $K= mathbb Z /2mathbb Z$ consists of two classes: even and odd integers. It is natural to map the class of even integers into $0$ and the class of odd integers into $1$. This bijection in turn becomes a ring isomorphism between $mathbb Z /2mathbb Z$ and ${0,1}$. Strictly speaking, these sets are isomorphic, but more loosely they are often identified. $K^2=K times K$ is the Cartesian product of $K$ with itself: $K^{2}={(0,0),(1,0),(0,1),(1,1)}$
answered Jan 17 at 23:05
David TarandekDavid Tarandek
234
$endgroup$
If you follow the definition, $K= mathbb Z /2mathbb Z$ consists of two classes: even and odd integers. It is natural to map the class of even integers into $0$ and the class of odd integers into $1$. This bijection in turn becomes a ring isomorphism between $mathbb Z /2mathbb Z$ and ${0,1}$. Strictly speaking, these sets are isomorphic, but more loosely they are often identified. $K^2=K times K$ is the Cartesian product of $K$ with itself: $K^{2}={(0,0),(1,0),(0,1),(1,1)}$
answered Jan 17 at 23:05
David TarandekDavid Tarandek
234
answered Jan 17 at 23:05
David TarandekDavid Tarandek
234
answered Jan 17 at 23:05
David TarandekDavid Tarandek
234
answered Jan 17 at 23:05
David TarandekDavid Tarandek
234
234
add a comment |
add a comment |
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1
$begingroup$
Would you understand it better if it was $Ktimes K$?
$endgroup$
– Arnaud D.
Jan 17 at 22:13
$begingroup$
Those are two-dimensional vectors whose entries are from (the field) $;Bbb Z/2Bbb Z;$ ...
$endgroup$
– DonAntonio
Jan 17 at 22:14
$begingroup$
$K^2$ is the ring product $Ktimes K$, which is the set of all pairs $(x,y)$ with $x,yin K$.
$endgroup$
– Dave
Jan 17 at 22:14
1
$begingroup$
If $K$ is a set (any set) then $K^2 = K times K = {(a,b)|a,bin K}=$ then set of all ordered pairs of elements of $K$. So $K = mathbb Z/2mathbb Z =$ set of equivalence classes modulo 2 = ${0, 1}$. So $K^2 = {(a,b)| a = 0, 1 b=0,1} = {(0,0),(0,1),(1,0),(1,1)}$.
$endgroup$
– fleablood
Jan 17 at 23:09