Proof $xin R^times wedge bin R^times Rightarrow abin R^times$ [closed]
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Let $R$ be a communitative ring. Prove $ain R^times wedge bin R^times >Rightarrow abin R^times$ with $R^times := {xin R | x text{ >invertible}}.$
Do you have any ideas and tips on how I could prove this?
I know that this isn't really difficult but as so often in
proofs concerning algebraic groups you need the right beginning
and I can't get it...
Thanks!
algebraic-groups
closed as off-topic by Adrian Keister, user302797, Brahadeesh, Lord Shark the Unknown, Leucippus Nov 28 at 6:27
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Adrian Keister, user302797, Brahadeesh, Lord Shark the Unknown, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
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favorite
Let $R$ be a communitative ring. Prove $ain R^times wedge bin R^times >Rightarrow abin R^times$ with $R^times := {xin R | x text{ >invertible}}.$
Do you have any ideas and tips on how I could prove this?
I know that this isn't really difficult but as so often in
proofs concerning algebraic groups you need the right beginning
and I can't get it...
Thanks!
algebraic-groups
closed as off-topic by Adrian Keister, user302797, Brahadeesh, Lord Shark the Unknown, Leucippus Nov 28 at 6:27
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Adrian Keister, user302797, Brahadeesh, Lord Shark the Unknown, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
What do you know about $a$?
– Fakemistake
Nov 27 at 19:42
sorry, misspelling. Edited it
– kjwemke13
Nov 27 at 19:43
Consider the element $a^{-1}b^{-1}$
– Fakemistake
Nov 27 at 19:54
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $R$ be a communitative ring. Prove $ain R^times wedge bin R^times >Rightarrow abin R^times$ with $R^times := {xin R | x text{ >invertible}}.$
Do you have any ideas and tips on how I could prove this?
I know that this isn't really difficult but as so often in
proofs concerning algebraic groups you need the right beginning
and I can't get it...
Thanks!
algebraic-groups
Let $R$ be a communitative ring. Prove $ain R^times wedge bin R^times >Rightarrow abin R^times$ with $R^times := {xin R | x text{ >invertible}}.$
Do you have any ideas and tips on how I could prove this?
I know that this isn't really difficult but as so often in
proofs concerning algebraic groups you need the right beginning
and I can't get it...
Thanks!
algebraic-groups
algebraic-groups
edited Nov 27 at 23:00
greedoid
37k114794
37k114794
asked Nov 27 at 19:40
kjwemke13
72
72
closed as off-topic by Adrian Keister, user302797, Brahadeesh, Lord Shark the Unknown, Leucippus Nov 28 at 6:27
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Adrian Keister, user302797, Brahadeesh, Lord Shark the Unknown, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Adrian Keister, user302797, Brahadeesh, Lord Shark the Unknown, Leucippus Nov 28 at 6:27
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Adrian Keister, user302797, Brahadeesh, Lord Shark the Unknown, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
What do you know about $a$?
– Fakemistake
Nov 27 at 19:42
sorry, misspelling. Edited it
– kjwemke13
Nov 27 at 19:43
Consider the element $a^{-1}b^{-1}$
– Fakemistake
Nov 27 at 19:54
add a comment |
What do you know about $a$?
– Fakemistake
Nov 27 at 19:42
sorry, misspelling. Edited it
– kjwemke13
Nov 27 at 19:43
Consider the element $a^{-1}b^{-1}$
– Fakemistake
Nov 27 at 19:54
What do you know about $a$?
– Fakemistake
Nov 27 at 19:42
What do you know about $a$?
– Fakemistake
Nov 27 at 19:42
sorry, misspelling. Edited it
– kjwemke13
Nov 27 at 19:43
sorry, misspelling. Edited it
– kjwemke13
Nov 27 at 19:43
Consider the element $a^{-1}b^{-1}$
– Fakemistake
Nov 27 at 19:54
Consider the element $a^{-1}b^{-1}$
– Fakemistake
Nov 27 at 19:54
add a comment |
3 Answers
3
active
oldest
votes
up vote
1
down vote
Here's a beginning: “Suppose that $a,b in R^times$.”
Here's the end: “Therefore, $ab in R^times$.
The middle is up to you. For inspiration, you might try examples with $R=mathbb{Z}$ or $mathbb{Q}$. What is the inverse of a product of numbers? Is it expressible in terms of the inverses of the factors?
Do you mean $(ab)^{-1}=a^{-1} cdot {b^-1}$ and therefore $a$ and $b$ has to be in $R^times$ because of the definition of a ring?
– kjwemke13
Nov 27 at 19:48
The inverse elements of $a,b$ exists by assumption!
– Fakemistake
Nov 27 at 20:03
@kjwemke13: By your definition, $R^times$ is the set of invertible elements of $R$. So you have to show that $ab$ is invertible.
– Matthew Leingang
Nov 27 at 20:19
add a comment |
up vote
0
down vote
With
$a, b in R^times, tag 1$
we have
$c, d in R^times tag 2$
with
$ac = bd = 1_R, tag 3$
where $1_R$ is the multiplicative identity of $R$; then
$(ab)(cd) = a(bc)d = a(cb)d = (ac)(bd) = 1_R 1_R = 1_R, tag 4$
that is,
$ab, cd in R^times. tag 5$
$OEDelta$.
add a comment |
up vote
0
down vote
Since $a,bin R^times $, exsist $a^{-1}$ and $b^{-1}$ in $R$. Let $c = b^{-1}a^{-1}$, then $$(ab)cdot c = aunderbrace {bcdot b^{-1}}_{=e}a^{-1} = a cdot a^{-1} = e$$
so $abin R^times $.
didn't you proof that $(ab)c in R^times$ I don't see how this is a proof for $ab in R^times$
– Gykonik
Nov 27 at 20:38
$(ab)c =e$ what does that say about $ab$? @Gykonik
– greedoid
Nov 27 at 20:40
ehm, that $(ab)=c^{-1}$?
– Gykonik
Nov 27 at 20:43
So...............
– greedoid
Nov 27 at 20:44
$c^{-1}$ and $c$ has to be in $R^times$?
– Gykonik
Nov 27 at 20:44
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Here's a beginning: “Suppose that $a,b in R^times$.”
Here's the end: “Therefore, $ab in R^times$.
The middle is up to you. For inspiration, you might try examples with $R=mathbb{Z}$ or $mathbb{Q}$. What is the inverse of a product of numbers? Is it expressible in terms of the inverses of the factors?
Do you mean $(ab)^{-1}=a^{-1} cdot {b^-1}$ and therefore $a$ and $b$ has to be in $R^times$ because of the definition of a ring?
– kjwemke13
Nov 27 at 19:48
The inverse elements of $a,b$ exists by assumption!
– Fakemistake
Nov 27 at 20:03
@kjwemke13: By your definition, $R^times$ is the set of invertible elements of $R$. So you have to show that $ab$ is invertible.
– Matthew Leingang
Nov 27 at 20:19
add a comment |
up vote
1
down vote
Here's a beginning: “Suppose that $a,b in R^times$.”
Here's the end: “Therefore, $ab in R^times$.
The middle is up to you. For inspiration, you might try examples with $R=mathbb{Z}$ or $mathbb{Q}$. What is the inverse of a product of numbers? Is it expressible in terms of the inverses of the factors?
Do you mean $(ab)^{-1}=a^{-1} cdot {b^-1}$ and therefore $a$ and $b$ has to be in $R^times$ because of the definition of a ring?
– kjwemke13
Nov 27 at 19:48
The inverse elements of $a,b$ exists by assumption!
– Fakemistake
Nov 27 at 20:03
@kjwemke13: By your definition, $R^times$ is the set of invertible elements of $R$. So you have to show that $ab$ is invertible.
– Matthew Leingang
Nov 27 at 20:19
add a comment |
up vote
1
down vote
up vote
1
down vote
Here's a beginning: “Suppose that $a,b in R^times$.”
Here's the end: “Therefore, $ab in R^times$.
The middle is up to you. For inspiration, you might try examples with $R=mathbb{Z}$ or $mathbb{Q}$. What is the inverse of a product of numbers? Is it expressible in terms of the inverses of the factors?
Here's a beginning: “Suppose that $a,b in R^times$.”
Here's the end: “Therefore, $ab in R^times$.
The middle is up to you. For inspiration, you might try examples with $R=mathbb{Z}$ or $mathbb{Q}$. What is the inverse of a product of numbers? Is it expressible in terms of the inverses of the factors?
answered Nov 27 at 19:44
Matthew Leingang
16.2k12244
16.2k12244
Do you mean $(ab)^{-1}=a^{-1} cdot {b^-1}$ and therefore $a$ and $b$ has to be in $R^times$ because of the definition of a ring?
– kjwemke13
Nov 27 at 19:48
The inverse elements of $a,b$ exists by assumption!
– Fakemistake
Nov 27 at 20:03
@kjwemke13: By your definition, $R^times$ is the set of invertible elements of $R$. So you have to show that $ab$ is invertible.
– Matthew Leingang
Nov 27 at 20:19
add a comment |
Do you mean $(ab)^{-1}=a^{-1} cdot {b^-1}$ and therefore $a$ and $b$ has to be in $R^times$ because of the definition of a ring?
– kjwemke13
Nov 27 at 19:48
The inverse elements of $a,b$ exists by assumption!
– Fakemistake
Nov 27 at 20:03
@kjwemke13: By your definition, $R^times$ is the set of invertible elements of $R$. So you have to show that $ab$ is invertible.
– Matthew Leingang
Nov 27 at 20:19
Do you mean $(ab)^{-1}=a^{-1} cdot {b^-1}$ and therefore $a$ and $b$ has to be in $R^times$ because of the definition of a ring?
– kjwemke13
Nov 27 at 19:48
Do you mean $(ab)^{-1}=a^{-1} cdot {b^-1}$ and therefore $a$ and $b$ has to be in $R^times$ because of the definition of a ring?
– kjwemke13
Nov 27 at 19:48
The inverse elements of $a,b$ exists by assumption!
– Fakemistake
Nov 27 at 20:03
The inverse elements of $a,b$ exists by assumption!
– Fakemistake
Nov 27 at 20:03
@kjwemke13: By your definition, $R^times$ is the set of invertible elements of $R$. So you have to show that $ab$ is invertible.
– Matthew Leingang
Nov 27 at 20:19
@kjwemke13: By your definition, $R^times$ is the set of invertible elements of $R$. So you have to show that $ab$ is invertible.
– Matthew Leingang
Nov 27 at 20:19
add a comment |
up vote
0
down vote
With
$a, b in R^times, tag 1$
we have
$c, d in R^times tag 2$
with
$ac = bd = 1_R, tag 3$
where $1_R$ is the multiplicative identity of $R$; then
$(ab)(cd) = a(bc)d = a(cb)d = (ac)(bd) = 1_R 1_R = 1_R, tag 4$
that is,
$ab, cd in R^times. tag 5$
$OEDelta$.
add a comment |
up vote
0
down vote
With
$a, b in R^times, tag 1$
we have
$c, d in R^times tag 2$
with
$ac = bd = 1_R, tag 3$
where $1_R$ is the multiplicative identity of $R$; then
$(ab)(cd) = a(bc)d = a(cb)d = (ac)(bd) = 1_R 1_R = 1_R, tag 4$
that is,
$ab, cd in R^times. tag 5$
$OEDelta$.
add a comment |
up vote
0
down vote
up vote
0
down vote
With
$a, b in R^times, tag 1$
we have
$c, d in R^times tag 2$
with
$ac = bd = 1_R, tag 3$
where $1_R$ is the multiplicative identity of $R$; then
$(ab)(cd) = a(bc)d = a(cb)d = (ac)(bd) = 1_R 1_R = 1_R, tag 4$
that is,
$ab, cd in R^times. tag 5$
$OEDelta$.
With
$a, b in R^times, tag 1$
we have
$c, d in R^times tag 2$
with
$ac = bd = 1_R, tag 3$
where $1_R$ is the multiplicative identity of $R$; then
$(ab)(cd) = a(bc)d = a(cb)d = (ac)(bd) = 1_R 1_R = 1_R, tag 4$
that is,
$ab, cd in R^times. tag 5$
$OEDelta$.
answered Nov 27 at 19:55
Robert Lewis
43k22863
43k22863
add a comment |
add a comment |
up vote
0
down vote
Since $a,bin R^times $, exsist $a^{-1}$ and $b^{-1}$ in $R$. Let $c = b^{-1}a^{-1}$, then $$(ab)cdot c = aunderbrace {bcdot b^{-1}}_{=e}a^{-1} = a cdot a^{-1} = e$$
so $abin R^times $.
didn't you proof that $(ab)c in R^times$ I don't see how this is a proof for $ab in R^times$
– Gykonik
Nov 27 at 20:38
$(ab)c =e$ what does that say about $ab$? @Gykonik
– greedoid
Nov 27 at 20:40
ehm, that $(ab)=c^{-1}$?
– Gykonik
Nov 27 at 20:43
So...............
– greedoid
Nov 27 at 20:44
$c^{-1}$ and $c$ has to be in $R^times$?
– Gykonik
Nov 27 at 20:44
add a comment |
up vote
0
down vote
Since $a,bin R^times $, exsist $a^{-1}$ and $b^{-1}$ in $R$. Let $c = b^{-1}a^{-1}$, then $$(ab)cdot c = aunderbrace {bcdot b^{-1}}_{=e}a^{-1} = a cdot a^{-1} = e$$
so $abin R^times $.
didn't you proof that $(ab)c in R^times$ I don't see how this is a proof for $ab in R^times$
– Gykonik
Nov 27 at 20:38
$(ab)c =e$ what does that say about $ab$? @Gykonik
– greedoid
Nov 27 at 20:40
ehm, that $(ab)=c^{-1}$?
– Gykonik
Nov 27 at 20:43
So...............
– greedoid
Nov 27 at 20:44
$c^{-1}$ and $c$ has to be in $R^times$?
– Gykonik
Nov 27 at 20:44
add a comment |
up vote
0
down vote
up vote
0
down vote
Since $a,bin R^times $, exsist $a^{-1}$ and $b^{-1}$ in $R$. Let $c = b^{-1}a^{-1}$, then $$(ab)cdot c = aunderbrace {bcdot b^{-1}}_{=e}a^{-1} = a cdot a^{-1} = e$$
so $abin R^times $.
Since $a,bin R^times $, exsist $a^{-1}$ and $b^{-1}$ in $R$. Let $c = b^{-1}a^{-1}$, then $$(ab)cdot c = aunderbrace {bcdot b^{-1}}_{=e}a^{-1} = a cdot a^{-1} = e$$
so $abin R^times $.
answered Nov 27 at 19:57
greedoid
37k114794
37k114794
didn't you proof that $(ab)c in R^times$ I don't see how this is a proof for $ab in R^times$
– Gykonik
Nov 27 at 20:38
$(ab)c =e$ what does that say about $ab$? @Gykonik
– greedoid
Nov 27 at 20:40
ehm, that $(ab)=c^{-1}$?
– Gykonik
Nov 27 at 20:43
So...............
– greedoid
Nov 27 at 20:44
$c^{-1}$ and $c$ has to be in $R^times$?
– Gykonik
Nov 27 at 20:44
add a comment |
didn't you proof that $(ab)c in R^times$ I don't see how this is a proof for $ab in R^times$
– Gykonik
Nov 27 at 20:38
$(ab)c =e$ what does that say about $ab$? @Gykonik
– greedoid
Nov 27 at 20:40
ehm, that $(ab)=c^{-1}$?
– Gykonik
Nov 27 at 20:43
So...............
– greedoid
Nov 27 at 20:44
$c^{-1}$ and $c$ has to be in $R^times$?
– Gykonik
Nov 27 at 20:44
didn't you proof that $(ab)c in R^times$ I don't see how this is a proof for $ab in R^times$
– Gykonik
Nov 27 at 20:38
didn't you proof that $(ab)c in R^times$ I don't see how this is a proof for $ab in R^times$
– Gykonik
Nov 27 at 20:38
$(ab)c =e$ what does that say about $ab$? @Gykonik
– greedoid
Nov 27 at 20:40
$(ab)c =e$ what does that say about $ab$? @Gykonik
– greedoid
Nov 27 at 20:40
ehm, that $(ab)=c^{-1}$?
– Gykonik
Nov 27 at 20:43
ehm, that $(ab)=c^{-1}$?
– Gykonik
Nov 27 at 20:43
So...............
– greedoid
Nov 27 at 20:44
So...............
– greedoid
Nov 27 at 20:44
$c^{-1}$ and $c$ has to be in $R^times$?
– Gykonik
Nov 27 at 20:44
$c^{-1}$ and $c$ has to be in $R^times$?
– Gykonik
Nov 27 at 20:44
add a comment |
What do you know about $a$?
– Fakemistake
Nov 27 at 19:42
sorry, misspelling. Edited it
– kjwemke13
Nov 27 at 19:43
Consider the element $a^{-1}b^{-1}$
– Fakemistake
Nov 27 at 19:54