Are the rings $mathbb{R}^2$ and $mathbb{R}^3$ isomorphic?
up vote
4
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Are the rings $mathbb{R}^2$ and $mathbb{R}^3$ isomorphic, where $mathbb{R}^2=mathbb{R}+mathbb{R}$ and is the set of all pairs $(a,b)$ with $a,b in mathbb{R}$, and $mathbb{R}^3=mathbb{R}+mathbb{R}+mathbb{R}$ is the set of all triples $(a,b,c)$ with $a,b,c in mathbb{R}$, using component wise addition and multiplication?
abstract-algebra ring-isomorphism
asked Dec 3 at 4:54
JM23
212
|
show 3 more comments
up vote
4
down vote
favorite
Are the rings $mathbb{R}^2$ and $mathbb{R}^3$ isomorphic, where $mathbb{R}^2=mathbb{R}+mathbb{R}$ and is the set of all pairs $(a,b)$ with $a,b in mathbb{R}$, and $mathbb{R}^3=mathbb{R}+mathbb{R}+mathbb{R}$ is the set of all triples $(a,b,c)$ with $a,b,c in mathbb{R}$, using component wise addition and multiplication?
abstract-algebra ring-isomorphism
asked Dec 3 at 4:54
JM23
212
What are your thoughts on this matter?
– MJD
Dec 3 at 4:59
2
What happens when $R$ is finite?
– D_S
Dec 3 at 5:04
What is the definition of isomorphic for a pair of rings? How might the extra degree of freedom of $R^3$ contradict that definition?
– AlexanderJ93
Dec 3 at 5:13
@D_S If its finite than R^3>R^2. I do not see how this helps however, since both rings are infinite. Can I still prove some how that R^2<R^3?
– JM23
Dec 3 at 5:42
1
When you say "I thought they were isomorphic because each ring is an infinite ring", are you saying that all infinite rings are isomorphic?
– bof
Dec 3 at 5:56
|
show 3 more comments
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Are the rings $mathbb{R}^2$ and $mathbb{R}^3$ isomorphic, where $mathbb{R}^2=mathbb{R}+mathbb{R}$ and is the set of all pairs $(a,b)$ with $a,b in mathbb{R}$, and $mathbb{R}^3=mathbb{R}+mathbb{R}+mathbb{R}$ is the set of all triples $(a,b,c)$ with $a,b,c in mathbb{R}$, using component wise addition and multiplication?
abstract-algebra ring-isomorphism
asked Dec 3 at 4:54
JM23
212
Are the rings $mathbb{R}^2$ and $mathbb{R}^3$ isomorphic, where $mathbb{R}^2=mathbb{R}+mathbb{R}$ and is the set of all pairs $(a,b)$ with $a,b in mathbb{R}$, and $mathbb{R}^3=mathbb{R}+mathbb{R}+mathbb{R}$ is the set of all triples $(a,b,c)$ with $a,b,c in mathbb{R}$, using component wise addition and multiplication?
abstract-algebra ring-isomorphism
abstract-algebra ring-isomorphism
asked Dec 3 at 4:54
JM23
212
asked Dec 3 at 4:54
JM23
212
edited Dec 3 at 5:46
asked Dec 3 at 4:54
JM23
212
asked Dec 3 at 4:54
JM23
212
asked Dec 3 at 4:54
JM23
212
212
What are your thoughts on this matter?
– MJD
Dec 3 at 4:59
2
What happens when $R$ is finite?
– D_S
Dec 3 at 5:04
What is the definition of isomorphic for a pair of rings? How might the extra degree of freedom of $R^3$ contradict that definition?
– AlexanderJ93
Dec 3 at 5:13
@D_S If its finite than R^3>R^2. I do not see how this helps however, since both rings are infinite. Can I still prove some how that R^2<R^3?
– JM23
Dec 3 at 5:42
1
When you say "I thought they were isomorphic because each ring is an infinite ring", are you saying that all infinite rings are isomorphic?
– bof
Dec 3 at 5:56
|
show 3 more comments
What are your thoughts on this matter?
– MJD
Dec 3 at 4:59
2
What happens when $R$ is finite?
– D_S
Dec 3 at 5:04
What is the definition of isomorphic for a pair of rings? How might the extra degree of freedom of $R^3$ contradict that definition?
– AlexanderJ93
Dec 3 at 5:13
@D_S If its finite than R^3>R^2. I do not see how this helps however, since both rings are infinite. Can I still prove some how that R^2<R^3?
– JM23
Dec 3 at 5:42
1
When you say "I thought they were isomorphic because each ring is an infinite ring", are you saying that all infinite rings are isomorphic?
– bof
Dec 3 at 5:56
What are your thoughts on this matter?
– MJD
Dec 3 at 4:59
What are your thoughts on this matter?
– MJD
Dec 3 at 4:59
2
2
What happens when $R$ is finite?
– D_S
Dec 3 at 5:04
What happens when $R$ is finite?
– D_S
Dec 3 at 5:04
What is the definition of isomorphic for a pair of rings? How might the extra degree of freedom of $R^3$ contradict that definition?
– AlexanderJ93
Dec 3 at 5:13
What is the definition of isomorphic for a pair of rings? How might the extra degree of freedom of $R^3$ contradict that definition?
– AlexanderJ93
Dec 3 at 5:13
@D_S If its finite than R^3>R^2. I do not see how this helps however, since both rings are infinite. Can I still prove some how that R^2<R^3?
– JM23
Dec 3 at 5:42
@D_S If its finite than R^3>R^2. I do not see how this helps however, since both rings are infinite. Can I still prove some how that R^2<R^3?
– JM23
Dec 3 at 5:42
1
1
When you say "I thought they were isomorphic because each ring is an infinite ring", are you saying that all infinite rings are isomorphic?
– bof
Dec 3 at 5:56
When you say "I thought they were isomorphic because each ring is an infinite ring", are you saying that all infinite rings are isomorphic?
– bof
Dec 3 at 5:56
|
show 3 more comments
2 Answers
2
active
oldest
votes
up vote
4
down vote
Consider idempotents in each ring, that is solutions of $e^2=e$. In your
first ring you are solving $(a^2,b^2)=(a,b)$, and in the second, $(a^2,b^2,c^2)=(a,b,c)$. How may solutions have you in each case?
answered Dec 3 at 7:01
Lord Shark the Unknown
98.7k958131
1
For $mathbb{R}^2$ would you have $infty^2$ solutions and for $mathbb{R}$ you would have $infty^3$ solutions, and since $infty^2<infty^3$ then $mathbb{R}^2$ and $mathbb{R}^3$ are not isomorphic?
– JM23
Dec 3 at 7:59
@JM23: think again, por favor: if $(a^2, b^2) = (a, b)$, then $a^2 = a$, $b^2 = b$. Since $a, b in Bbb R$, the solutions are $a = 0, 1 = b$. There are thus $4$ idempotents in $Bbb R^2$ und so weiter . . .
– Robert Lewis
Dec 3 at 8:20
add a comment |
up vote
1
down vote
HINT. If they are going to be isomorphic, it should work for any ring. Think of the special case where the ring is actually a field, i.e. $R=mathbb{k}$. Can you think of some theorems from basic Linear Algebra to say whether or not the rings $R^2=mathbb{k}^2$ and $R^3=mathbb{k}^3$ are isomorphic? Maybe focus on the only difference between them...
answered Dec 3 at 5:02
mathematics2x2life
8,03221738
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
Consider idempotents in each ring, that is solutions of $e^2=e$. In your
first ring you are solving $(a^2,b^2)=(a,b)$, and in the second, $(a^2,b^2,c^2)=(a,b,c)$. How may solutions have you in each case?
answered Dec 3 at 7:01
Lord Shark the Unknown
98.7k958131
1
For $mathbb{R}^2$ would you have $infty^2$ solutions and for $mathbb{R}$ you would have $infty^3$ solutions, and since $infty^2<infty^3$ then $mathbb{R}^2$ and $mathbb{R}^3$ are not isomorphic?
– JM23
Dec 3 at 7:59
@JM23: think again, por favor: if $(a^2, b^2) = (a, b)$, then $a^2 = a$, $b^2 = b$. Since $a, b in Bbb R$, the solutions are $a = 0, 1 = b$. There are thus $4$ idempotents in $Bbb R^2$ und so weiter . . .
– Robert Lewis
Dec 3 at 8:20
add a comment |
up vote
4
down vote
Consider idempotents in each ring, that is solutions of $e^2=e$. In your
first ring you are solving $(a^2,b^2)=(a,b)$, and in the second, $(a^2,b^2,c^2)=(a,b,c)$. How may solutions have you in each case?
answered Dec 3 at 7:01
Lord Shark the Unknown
98.7k958131
1
For $mathbb{R}^2$ would you have $infty^2$ solutions and for $mathbb{R}$ you would have $infty^3$ solutions, and since $infty^2<infty^3$ then $mathbb{R}^2$ and $mathbb{R}^3$ are not isomorphic?
– JM23
Dec 3 at 7:59
@JM23: think again, por favor: if $(a^2, b^2) = (a, b)$, then $a^2 = a$, $b^2 = b$. Since $a, b in Bbb R$, the solutions are $a = 0, 1 = b$. There are thus $4$ idempotents in $Bbb R^2$ und so weiter . . .
– Robert Lewis
Dec 3 at 8:20
add a comment |
up vote
4
down vote
up vote
4
down vote
Consider idempotents in each ring, that is solutions of $e^2=e$. In your
first ring you are solving $(a^2,b^2)=(a,b)$, and in the second, $(a^2,b^2,c^2)=(a,b,c)$. How may solutions have you in each case?
answered Dec 3 at 7:01
Lord Shark the Unknown
98.7k958131
Consider idempotents in each ring, that is solutions of $e^2=e$. In your
first ring you are solving $(a^2,b^2)=(a,b)$, and in the second, $(a^2,b^2,c^2)=(a,b,c)$. How may solutions have you in each case?
answered Dec 3 at 7:01
Lord Shark the Unknown
98.7k958131
answered Dec 3 at 7:01
Lord Shark the Unknown
98.7k958131
answered Dec 3 at 7:01
Lord Shark the Unknown
98.7k958131
answered Dec 3 at 7:01
Lord Shark the Unknown
98.7k958131
98.7k958131
1
For $mathbb{R}^2$ would you have $infty^2$ solutions and for $mathbb{R}$ you would have $infty^3$ solutions, and since $infty^2<infty^3$ then $mathbb{R}^2$ and $mathbb{R}^3$ are not isomorphic?
– JM23
Dec 3 at 7:59
@JM23: think again, por favor: if $(a^2, b^2) = (a, b)$, then $a^2 = a$, $b^2 = b$. Since $a, b in Bbb R$, the solutions are $a = 0, 1 = b$. There are thus $4$ idempotents in $Bbb R^2$ und so weiter . . .
– Robert Lewis
Dec 3 at 8:20
add a comment |
1
For $mathbb{R}^2$ would you have $infty^2$ solutions and for $mathbb{R}$ you would have $infty^3$ solutions, and since $infty^2<infty^3$ then $mathbb{R}^2$ and $mathbb{R}^3$ are not isomorphic?
– JM23
Dec 3 at 7:59
@JM23: think again, por favor: if $(a^2, b^2) = (a, b)$, then $a^2 = a$, $b^2 = b$. Since $a, b in Bbb R$, the solutions are $a = 0, 1 = b$. There are thus $4$ idempotents in $Bbb R^2$ und so weiter . . .
– Robert Lewis
Dec 3 at 8:20
1
1
For $mathbb{R}^2$ would you have $infty^2$ solutions and for $mathbb{R}$ you would have $infty^3$ solutions, and since $infty^2<infty^3$ then $mathbb{R}^2$ and $mathbb{R}^3$ are not isomorphic?
– JM23
Dec 3 at 7:59
For $mathbb{R}^2$ would you have $infty^2$ solutions and for $mathbb{R}$ you would have $infty^3$ solutions, and since $infty^2<infty^3$ then $mathbb{R}^2$ and $mathbb{R}^3$ are not isomorphic?
– JM23
Dec 3 at 7:59
@JM23: think again, por favor: if $(a^2, b^2) = (a, b)$, then $a^2 = a$, $b^2 = b$. Since $a, b in Bbb R$, the solutions are $a = 0, 1 = b$. There are thus $4$ idempotents in $Bbb R^2$ und so weiter . . .
– Robert Lewis
Dec 3 at 8:20
@JM23: think again, por favor: if $(a^2, b^2) = (a, b)$, then $a^2 = a$, $b^2 = b$. Since $a, b in Bbb R$, the solutions are $a = 0, 1 = b$. There are thus $4$ idempotents in $Bbb R^2$ und so weiter . . .
– Robert Lewis
Dec 3 at 8:20
add a comment |
up vote
1
down vote
HINT. If they are going to be isomorphic, it should work for any ring. Think of the special case where the ring is actually a field, i.e. $R=mathbb{k}$. Can you think of some theorems from basic Linear Algebra to say whether or not the rings $R^2=mathbb{k}^2$ and $R^3=mathbb{k}^3$ are isomorphic? Maybe focus on the only difference between them...
answered Dec 3 at 5:02
mathematics2x2life
8,03221738
add a comment |
up vote
1
down vote
HINT. If they are going to be isomorphic, it should work for any ring. Think of the special case where the ring is actually a field, i.e. $R=mathbb{k}$. Can you think of some theorems from basic Linear Algebra to say whether or not the rings $R^2=mathbb{k}^2$ and $R^3=mathbb{k}^3$ are isomorphic? Maybe focus on the only difference between them...
answered Dec 3 at 5:02
mathematics2x2life
8,03221738
add a comment |
up vote
1
down vote
up vote
1
down vote
HINT. If they are going to be isomorphic, it should work for any ring. Think of the special case where the ring is actually a field, i.e. $R=mathbb{k}$. Can you think of some theorems from basic Linear Algebra to say whether or not the rings $R^2=mathbb{k}^2$ and $R^3=mathbb{k}^3$ are isomorphic? Maybe focus on the only difference between them...
answered Dec 3 at 5:02
mathematics2x2life
8,03221738
HINT. If they are going to be isomorphic, it should work for any ring. Think of the special case where the ring is actually a field, i.e. $R=mathbb{k}$. Can you think of some theorems from basic Linear Algebra to say whether or not the rings $R^2=mathbb{k}^2$ and $R^3=mathbb{k}^3$ are isomorphic? Maybe focus on the only difference between them...
answered Dec 3 at 5:02
mathematics2x2life
8,03221738
answered Dec 3 at 5:02
mathematics2x2life
8,03221738
answered Dec 3 at 5:02
mathematics2x2life
8,03221738
answered Dec 3 at 5:02
mathematics2x2life
8,03221738
8,03221738
add a comment |
add a comment |
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What are your thoughts on this matter?
– MJD
Dec 3 at 4:59
2
What happens when $R$ is finite?
– D_S
Dec 3 at 5:04
What is the definition of isomorphic for a pair of rings? How might the extra degree of freedom of $R^3$ contradict that definition?
– AlexanderJ93
Dec 3 at 5:13
@D_S If its finite than R^3>R^2. I do not see how this helps however, since both rings are infinite. Can I still prove some how that R^2<R^3?
– JM23
Dec 3 at 5:42
1
When you say "I thought they were isomorphic because each ring is an infinite ring", are you saying that all infinite rings are isomorphic?
– bof
Dec 3 at 5:56