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Question on Closure Property of Rings
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This is perhaps a very trivial question, but I just want to be sure.
Suppose that, by assumption, some element of a ring is invertible. Rings need not be closed under multiplicative inverses, surely, but is it reasonable or necessary to conclude that this inverse is also an element of the ring?
I've been trying to piece together an example wherein this is guaranteed by closure under multiplication or the fact that, if $y$ is the inverse of some $x in R$, then clearly $x$ is also the inverse of $y$, but this doesn't seem to guarantee anything, unless I'm missing something obvious.
Thanks in advance.
ring-theory
asked Jul 28 at 13:38
Matt.P
666313
add a comment |Â
up vote
1
down vote
favorite
This is perhaps a very trivial question, but I just want to be sure.
Suppose that, by assumption, some element of a ring is invertible. Rings need not be closed under multiplicative inverses, surely, but is it reasonable or necessary to conclude that this inverse is also an element of the ring?
I've been trying to piece together an example wherein this is guaranteed by closure under multiplication or the fact that, if $y$ is the inverse of some $x in R$, then clearly $x$ is also the inverse of $y$, but this doesn't seem to guarantee anything, unless I'm missing something obvious.
Thanks in advance.
ring-theory
asked Jul 28 at 13:38
Matt.P
666313
3
If an element of a ring is invertible, it means that it has an inverse in that ring surely?
– Lord Shark the Unknown
Jul 28 at 13:39
1
Sometimes, an element is a zero divisor. In this case, it is either zero or cannot be invertible in any super-ring.
– AlgebraicsAnonymous
Jul 28 at 13:40
1
I don't think you have to conclude anything that is already given by definition. . I don't follow the thought in the last paragraph either. The definition of invertibility is totally symmetric so that $x$ is an invertible element iff its inverse is an invertible element.
– rschwieb
Jul 30 at 13:37
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
This is perhaps a very trivial question, but I just want to be sure.
Suppose that, by assumption, some element of a ring is invertible. Rings need not be closed under multiplicative inverses, surely, but is it reasonable or necessary to conclude that this inverse is also an element of the ring?
I've been trying to piece together an example wherein this is guaranteed by closure under multiplication or the fact that, if $y$ is the inverse of some $x in R$, then clearly $x$ is also the inverse of $y$, but this doesn't seem to guarantee anything, unless I'm missing something obvious.
Thanks in advance.
ring-theory
asked Jul 28 at 13:38
Matt.P
666313
This is perhaps a very trivial question, but I just want to be sure.
Suppose that, by assumption, some element of a ring is invertible. Rings need not be closed under multiplicative inverses, surely, but is it reasonable or necessary to conclude that this inverse is also an element of the ring?
I've been trying to piece together an example wherein this is guaranteed by closure under multiplication or the fact that, if $y$ is the inverse of some $x in R$, then clearly $x$ is also the inverse of $y$, but this doesn't seem to guarantee anything, unless I'm missing something obvious.
Thanks in advance.
ring-theory
asked Jul 28 at 13:38
Matt.P
666313
asked Jul 28 at 13:38
Matt.P
666313
asked Jul 28 at 13:38
Matt.P
666313
asked Jul 28 at 13:38
Matt.P
666313
666313
3
If an element of a ring is invertible, it means that it has an inverse in that ring surely?
– Lord Shark the Unknown
Jul 28 at 13:39
1
Sometimes, an element is a zero divisor. In this case, it is either zero or cannot be invertible in any super-ring.
– AlgebraicsAnonymous
Jul 28 at 13:40
1
I don't think you have to conclude anything that is already given by definition. . I don't follow the thought in the last paragraph either. The definition of invertibility is totally symmetric so that $x$ is an invertible element iff its inverse is an invertible element.
– rschwieb
Jul 30 at 13:37
add a comment |Â
3
If an element of a ring is invertible, it means that it has an inverse in that ring surely?
– Lord Shark the Unknown
Jul 28 at 13:39
1
Sometimes, an element is a zero divisor. In this case, it is either zero or cannot be invertible in any super-ring.
– AlgebraicsAnonymous
Jul 28 at 13:40
1
I don't think you have to conclude anything that is already given by definition. . I don't follow the thought in the last paragraph either. The definition of invertibility is totally symmetric so that $x$ is an invertible element iff its inverse is an invertible element.
– rschwieb
Jul 30 at 13:37
3
3
If an element of a ring is invertible, it means that it has an inverse in that ring surely?
– Lord Shark the Unknown
Jul 28 at 13:39
If an element of a ring is invertible, it means that it has an inverse in that ring surely?
– Lord Shark the Unknown
Jul 28 at 13:39
1
1
Sometimes, an element is a zero divisor. In this case, it is either zero or cannot be invertible in any super-ring.
– AlgebraicsAnonymous
Jul 28 at 13:40
Sometimes, an element is a zero divisor. In this case, it is either zero or cannot be invertible in any super-ring.
– AlgebraicsAnonymous
Jul 28 at 13:40
1
1
I don't think you have to conclude anything that is already given by definition. . I don't follow the thought in the last paragraph either. The definition of invertibility is totally symmetric so that $x$ is an invertible element iff its inverse is an invertible element.
– rschwieb
Jul 30 at 13:37
I don't think you have to conclude anything that is already given by definition. . I don't follow the thought in the last paragraph either. The definition of invertibility is totally symmetric so that $x$ is an invertible element iff its inverse is an invertible element.
– rschwieb
Jul 30 at 13:37
add a comment |Â
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3
If an element of a ring is invertible, it means that it has an inverse in that ring surely?
– Lord Shark the Unknown
Jul 28 at 13:39
1
Sometimes, an element is a zero divisor. In this case, it is either zero or cannot be invertible in any super-ring.
– AlgebraicsAnonymous
Jul 28 at 13:40
1
I don't think you have to conclude anything that is already given by definition. . I don't follow the thought in the last paragraph either. The definition of invertibility is totally symmetric so that $x$ is an invertible element iff its inverse is an invertible element.
– rschwieb
Jul 30 at 13:37