Mixing
Ring with additive group isomorphic to group of units [duplicate]
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Let $R$ be a commutative unital ring. Is it true that the group of units of $R$ is not isomorphic with the additive group of $R$?
2 answers
Is there a Ring $R$ with $(R,+) cong (R^times,cdot)$? If $R$ is finite, clearly only the trivial ring does it (for cardinality reasons). But what about infinite rings? Are there even fields as example?
abstract-algebra ring-theory commutative-algebra
asked Jul 18 at 8:18
principal-ideal-domain
2,661521
marked as duplicate by Christopher, Daniel Fischer♦ Jul 18 at 9:31
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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up vote
3
down vote
favorite
This question already has an answer here:
Let $R$ be a commutative unital ring. Is it true that the group of units of $R$ is not isomorphic with the additive group of $R$?
2 answers
Is there a Ring $R$ with $(R,+) cong (R^times,cdot)$? If $R$ is finite, clearly only the trivial ring does it (for cardinality reasons). But what about infinite rings? Are there even fields as example?
abstract-algebra ring-theory commutative-algebra
asked Jul 18 at 8:18
principal-ideal-domain
2,661521
marked as duplicate by Christopher, Daniel Fischer♦ Jul 18 at 9:31
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
This question already has an answer here:
Let $R$ be a commutative unital ring. Is it true that the group of units of $R$ is not isomorphic with the additive group of $R$?
2 answers
Is there a Ring $R$ with $(R,+) cong (R^times,cdot)$? If $R$ is finite, clearly only the trivial ring does it (for cardinality reasons). But what about infinite rings? Are there even fields as example?
abstract-algebra ring-theory commutative-algebra
asked Jul 18 at 8:18
principal-ideal-domain
2,661521
This question already has an answer here:
Let $R$ be a commutative unital ring. Is it true that the group of units of $R$ is not isomorphic with the additive group of $R$?
2 answers
Is there a Ring $R$ with $(R,+) cong (R^times,cdot)$? If $R$ is finite, clearly only the trivial ring does it (for cardinality reasons). But what about infinite rings? Are there even fields as example?
This question already has an answer here:
Let $R$ be a commutative unital ring. Is it true that the group of units of $R$ is not isomorphic with the additive group of $R$?
2 answers
abstract-algebra ring-theory commutative-algebra
asked Jul 18 at 8:18
principal-ideal-domain
2,661521
asked Jul 18 at 8:18
principal-ideal-domain
2,661521
asked Jul 18 at 8:18
principal-ideal-domain
2,661521
asked Jul 18 at 8:18
principal-ideal-domain
2,661521
2,661521
marked as duplicate by Christopher, Daniel Fischer♦ Jul 18 at 9:31
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Christopher, Daniel Fischer♦ Jul 18 at 9:31
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |Â
add a comment |Â
1 Answer
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(as pointed out in the comments, this proof works only when $R$ is a field)
By the given isomorphism, the equations $2x=0$ and $x^2=1$ have the same number of solutions. But $2x=0$ has nontrivial solutions if and only if $R$ has characteristic $2$, while $x^2=1$ has nontrivial solutions if and only if $R$ has characteristic different from $2$.
answered Jul 18 at 9:02
Slade
23.8k12463
The first paragraph doesn't hold - the initial question asked for an isomorphism between $(R, +)$ and the group of units of $R$, not the monoid of non-zero elements of $R$. But this does answer the secondary question of the case where $R$ is a field.
– Christopher
Jul 18 at 9:10
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
(as pointed out in the comments, this proof works only when $R$ is a field)
By the given isomorphism, the equations $2x=0$ and $x^2=1$ have the same number of solutions. But $2x=0$ has nontrivial solutions if and only if $R$ has characteristic $2$, while $x^2=1$ has nontrivial solutions if and only if $R$ has characteristic different from $2$.
answered Jul 18 at 9:02
Slade
23.8k12463
The first paragraph doesn't hold - the initial question asked for an isomorphism between $(R, +)$ and the group of units of $R$, not the monoid of non-zero elements of $R$. But this does answer the secondary question of the case where $R$ is a field.
– Christopher
Jul 18 at 9:10
add a comment |Â
up vote
1
down vote
(as pointed out in the comments, this proof works only when $R$ is a field)
By the given isomorphism, the equations $2x=0$ and $x^2=1$ have the same number of solutions. But $2x=0$ has nontrivial solutions if and only if $R$ has characteristic $2$, while $x^2=1$ has nontrivial solutions if and only if $R$ has characteristic different from $2$.
answered Jul 18 at 9:02
Slade
23.8k12463
The first paragraph doesn't hold - the initial question asked for an isomorphism between $(R, +)$ and the group of units of $R$, not the monoid of non-zero elements of $R$. But this does answer the secondary question of the case where $R$ is a field.
– Christopher
Jul 18 at 9:10
add a comment |Â
up vote
1
down vote
up vote
1
down vote
(as pointed out in the comments, this proof works only when $R$ is a field)
By the given isomorphism, the equations $2x=0$ and $x^2=1$ have the same number of solutions. But $2x=0$ has nontrivial solutions if and only if $R$ has characteristic $2$, while $x^2=1$ has nontrivial solutions if and only if $R$ has characteristic different from $2$.
answered Jul 18 at 9:02
Slade
23.8k12463
(as pointed out in the comments, this proof works only when $R$ is a field)
By the given isomorphism, the equations $2x=0$ and $x^2=1$ have the same number of solutions. But $2x=0$ has nontrivial solutions if and only if $R$ has characteristic $2$, while $x^2=1$ has nontrivial solutions if and only if $R$ has characteristic different from $2$.
answered Jul 18 at 9:02
Slade
23.8k12463
edited Jul 18 at 9:11
answered Jul 18 at 9:02
Slade
23.8k12463
answered Jul 18 at 9:02
Slade
23.8k12463
answered Jul 18 at 9:02
Slade
23.8k12463
23.8k12463
The first paragraph doesn't hold - the initial question asked for an isomorphism between $(R, +)$ and the group of units of $R$, not the monoid of non-zero elements of $R$. But this does answer the secondary question of the case where $R$ is a field.
– Christopher
Jul 18 at 9:10
add a comment |Â
The first paragraph doesn't hold - the initial question asked for an isomorphism between $(R, +)$ and the group of units of $R$, not the monoid of non-zero elements of $R$. But this does answer the secondary question of the case where $R$ is a field.
– Christopher
Jul 18 at 9:10
The first paragraph doesn't hold - the initial question asked for an isomorphism between $(R, +)$ and the group of units of $R$, not the monoid of non-zero elements of $R$. But this does answer the secondary question of the case where $R$ is a field.
– Christopher
Jul 18 at 9:10
The first paragraph doesn't hold - the initial question asked for an isomorphism between $(R, +)$ and the group of units of $R$, not the monoid of non-zero elements of $R$. But this does answer the secondary question of the case where $R$ is a field.
– Christopher
Jul 18 at 9:10
add a comment |Â