Proof $xin R^times wedge bin R^times Rightarrow abin R^times$ [closed]
$begingroup$
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
edited Nov 27 '18 at 23:00
greedoid
40.2k114799
asked Nov 27 '18 at 19:40
kjwemke13kjwemke13
72
$endgroup$
closed as off-topic by Adrian Keister, Saad, Brahadeesh, Lord Shark the Unknown, Leucippus Nov 28 '18 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, Saad, 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 |
$begingroup$
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
edited Nov 27 '18 at 23:00
greedoid
40.2k114799
asked Nov 27 '18 at 19:40
kjwemke13kjwemke13
72
$endgroup$
closed as off-topic by Adrian Keister, Saad, Brahadeesh, Lord Shark the Unknown, Leucippus Nov 28 '18 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, Saad, Brahadeesh, Lord Shark the Unknown, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
What do you know about $a$?
$endgroup$
– Fakemistake
Nov 27 '18 at 19:42
$begingroup$
sorry, misspelling. Edited it
$endgroup$
– kjwemke13
Nov 27 '18 at 19:43
$begingroup$
Consider the element $a^{-1}b^{-1}$
$endgroup$
– Fakemistake
Nov 27 '18 at 19:54
add a comment |
$begingroup$
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
edited Nov 27 '18 at 23:00
greedoid
40.2k114799
asked Nov 27 '18 at 19:40
kjwemke13kjwemke13
72
$endgroup$
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 '18 at 23:00
greedoid
40.2k114799
asked Nov 27 '18 at 19:40
kjwemke13kjwemke13
72
edited Nov 27 '18 at 23:00
greedoid
40.2k114799
asked Nov 27 '18 at 19:40
kjwemke13kjwemke13
72
edited Nov 27 '18 at 23:00
greedoid
40.2k114799
edited Nov 27 '18 at 23:00
greedoid
40.2k114799
edited Nov 27 '18 at 23:00
greedoid
40.2k114799
40.2k114799
asked Nov 27 '18 at 19:40
kjwemke13kjwemke13
72
asked Nov 27 '18 at 19:40
kjwemke13kjwemke13
72
asked Nov 27 '18 at 19:40
kjwemke13kjwemke13
72
72
closed as off-topic by Adrian Keister, Saad, Brahadeesh, Lord Shark the Unknown, Leucippus Nov 28 '18 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, Saad, 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, Saad, Brahadeesh, Lord Shark the Unknown, Leucippus Nov 28 '18 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, Saad, Brahadeesh, Lord Shark the Unknown, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
What do you know about $a$?
$endgroup$
– Fakemistake
Nov 27 '18 at 19:42
$begingroup$
sorry, misspelling. Edited it
$endgroup$
– kjwemke13
Nov 27 '18 at 19:43
$begingroup$
Consider the element $a^{-1}b^{-1}$
$endgroup$
– Fakemistake
Nov 27 '18 at 19:54
add a comment |
$begingroup$
What do you know about $a$?
$endgroup$
– Fakemistake
Nov 27 '18 at 19:42
$begingroup$
sorry, misspelling. Edited it
$endgroup$
– kjwemke13
Nov 27 '18 at 19:43
$begingroup$
Consider the element $a^{-1}b^{-1}$
$endgroup$
– Fakemistake
Nov 27 '18 at 19:54
$begingroup$
What do you know about $a$?
$endgroup$
– Fakemistake
Nov 27 '18 at 19:42
$begingroup$
What do you know about $a$?
$endgroup$
– Fakemistake
Nov 27 '18 at 19:42
$begingroup$
sorry, misspelling. Edited it
$endgroup$
– kjwemke13
Nov 27 '18 at 19:43
$begingroup$
sorry, misspelling. Edited it
$endgroup$
– kjwemke13
Nov 27 '18 at 19:43
$begingroup$
Consider the element $a^{-1}b^{-1}$
$endgroup$
– Fakemistake
Nov 27 '18 at 19:54
$begingroup$
Consider the element $a^{-1}b^{-1}$
$endgroup$
– Fakemistake
Nov 27 '18 at 19:54
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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 '18 at 19:44
Matthew LeingangMatthew Leingang
16.3k12244
$endgroup$
$begingroup$
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?
$endgroup$
– kjwemke13
Nov 27 '18 at 19:48
$begingroup$
The inverse elements of $a,b$ exists by assumption!
$endgroup$
– Fakemistake
Nov 27 '18 at 20:03
$begingroup$
@kjwemke13: By your definition, $R^times$ is the set of invertible elements of $R$. So you have to show that $ab$ is invertible.
$endgroup$
– Matthew Leingang
Nov 27 '18 at 20:19
add a comment |
$begingroup$
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 '18 at 19:55
Robert LewisRobert Lewis
45.4k23065
$endgroup$
add a comment |
$begingroup$
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 '18 at 19:57
greedoidgreedoid
40.2k114799
$endgroup$
$begingroup$
didn't you proof that $(ab)c in R^times$ I don't see how this is a proof for $ab in R^times$
$endgroup$
– Gykonik
Nov 27 '18 at 20:38
$begingroup$
$(ab)c =e$ what does that say about $ab$? @Gykonik
$endgroup$
– greedoid
Nov 27 '18 at 20:40
$begingroup$
ehm, that $(ab)=c^{-1}$?
$endgroup$
– Gykonik
Nov 27 '18 at 20:43
$begingroup$
So...............
$endgroup$
– greedoid
Nov 27 '18 at 20:44
$begingroup$
$c^{-1}$ and $c$ has to be in $R^times$?
$endgroup$
– Gykonik
Nov 27 '18 at 20:44
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 '18 at 19:44
Matthew LeingangMatthew Leingang
16.3k12244
$endgroup$
$begingroup$
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?
$endgroup$
– kjwemke13
Nov 27 '18 at 19:48
$begingroup$
The inverse elements of $a,b$ exists by assumption!
$endgroup$
– Fakemistake
Nov 27 '18 at 20:03
$begingroup$
@kjwemke13: By your definition, $R^times$ is the set of invertible elements of $R$. So you have to show that $ab$ is invertible.
$endgroup$
– Matthew Leingang
Nov 27 '18 at 20:19
add a comment |
$begingroup$
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 '18 at 19:44
Matthew LeingangMatthew Leingang
16.3k12244
$endgroup$
$begingroup$
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?
$endgroup$
– kjwemke13
Nov 27 '18 at 19:48
$begingroup$
The inverse elements of $a,b$ exists by assumption!
$endgroup$
– Fakemistake
Nov 27 '18 at 20:03
$begingroup$
@kjwemke13: By your definition, $R^times$ is the set of invertible elements of $R$. So you have to show that $ab$ is invertible.
$endgroup$
– Matthew Leingang
Nov 27 '18 at 20:19
add a comment |
$begingroup$
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 '18 at 19:44
Matthew LeingangMatthew Leingang
16.3k12244
$endgroup$
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 '18 at 19:44
Matthew LeingangMatthew Leingang
16.3k12244
answered Nov 27 '18 at 19:44
Matthew LeingangMatthew Leingang
16.3k12244
answered Nov 27 '18 at 19:44
Matthew LeingangMatthew Leingang
16.3k12244
answered Nov 27 '18 at 19:44
Matthew LeingangMatthew Leingang
16.3k12244
16.3k12244
$begingroup$
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?
$endgroup$
– kjwemke13
Nov 27 '18 at 19:48
$begingroup$
The inverse elements of $a,b$ exists by assumption!
$endgroup$
– Fakemistake
Nov 27 '18 at 20:03
$begingroup$
@kjwemke13: By your definition, $R^times$ is the set of invertible elements of $R$. So you have to show that $ab$ is invertible.
$endgroup$
– Matthew Leingang
Nov 27 '18 at 20:19
add a comment |
$begingroup$
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?
$endgroup$
– kjwemke13
Nov 27 '18 at 19:48
$begingroup$
The inverse elements of $a,b$ exists by assumption!
$endgroup$
– Fakemistake
Nov 27 '18 at 20:03
$begingroup$
@kjwemke13: By your definition, $R^times$ is the set of invertible elements of $R$. So you have to show that $ab$ is invertible.
$endgroup$
– Matthew Leingang
Nov 27 '18 at 20:19
$begingroup$
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?
$endgroup$
– kjwemke13
Nov 27 '18 at 19:48
$begingroup$
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?
$endgroup$
– kjwemke13
Nov 27 '18 at 19:48
$begingroup$
The inverse elements of $a,b$ exists by assumption!
$endgroup$
– Fakemistake
Nov 27 '18 at 20:03
$begingroup$
The inverse elements of $a,b$ exists by assumption!
$endgroup$
– Fakemistake
Nov 27 '18 at 20:03
$begingroup$
@kjwemke13: By your definition, $R^times$ is the set of invertible elements of $R$. So you have to show that $ab$ is invertible.
$endgroup$
– Matthew Leingang
Nov 27 '18 at 20:19
$begingroup$
@kjwemke13: By your definition, $R^times$ is the set of invertible elements of $R$. So you have to show that $ab$ is invertible.
$endgroup$
– Matthew Leingang
Nov 27 '18 at 20:19
add a comment |
$begingroup$
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 '18 at 19:55
Robert LewisRobert Lewis
45.4k23065
$endgroup$
add a comment |
$begingroup$
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 '18 at 19:55
Robert LewisRobert Lewis
45.4k23065
$endgroup$
add a comment |
$begingroup$
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 '18 at 19:55
Robert LewisRobert Lewis
45.4k23065
$endgroup$
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 '18 at 19:55
Robert LewisRobert Lewis
45.4k23065
answered Nov 27 '18 at 19:55
Robert LewisRobert Lewis
45.4k23065
answered Nov 27 '18 at 19:55
Robert LewisRobert Lewis
45.4k23065
answered Nov 27 '18 at 19:55
Robert LewisRobert Lewis
45.4k23065
45.4k23065
add a comment |
add a comment |
$begingroup$
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 '18 at 19:57
greedoidgreedoid
40.2k114799
$endgroup$
$begingroup$
didn't you proof that $(ab)c in R^times$ I don't see how this is a proof for $ab in R^times$
$endgroup$
– Gykonik
Nov 27 '18 at 20:38
$begingroup$
$(ab)c =e$ what does that say about $ab$? @Gykonik
$endgroup$
– greedoid
Nov 27 '18 at 20:40
$begingroup$
ehm, that $(ab)=c^{-1}$?
$endgroup$
– Gykonik
Nov 27 '18 at 20:43
$begingroup$
So...............
$endgroup$
– greedoid
Nov 27 '18 at 20:44
$begingroup$
$c^{-1}$ and $c$ has to be in $R^times$?
$endgroup$
– Gykonik
Nov 27 '18 at 20:44
add a comment |
$begingroup$
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 '18 at 19:57
greedoidgreedoid
40.2k114799
$endgroup$
$begingroup$
didn't you proof that $(ab)c in R^times$ I don't see how this is a proof for $ab in R^times$
$endgroup$
– Gykonik
Nov 27 '18 at 20:38
$begingroup$
$(ab)c =e$ what does that say about $ab$? @Gykonik
$endgroup$
– greedoid
Nov 27 '18 at 20:40
$begingroup$
ehm, that $(ab)=c^{-1}$?
$endgroup$
– Gykonik
Nov 27 '18 at 20:43
$begingroup$
So...............
$endgroup$
– greedoid
Nov 27 '18 at 20:44
$begingroup$
$c^{-1}$ and $c$ has to be in $R^times$?
$endgroup$
– Gykonik
Nov 27 '18 at 20:44
add a comment |
$begingroup$
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 '18 at 19:57
greedoidgreedoid
40.2k114799
$endgroup$
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 '18 at 19:57
greedoidgreedoid
40.2k114799
answered Nov 27 '18 at 19:57
greedoidgreedoid
40.2k114799
answered Nov 27 '18 at 19:57
greedoidgreedoid
40.2k114799
answered Nov 27 '18 at 19:57
greedoidgreedoid
40.2k114799
40.2k114799
$begingroup$
didn't you proof that $(ab)c in R^times$ I don't see how this is a proof for $ab in R^times$
$endgroup$
– Gykonik
Nov 27 '18 at 20:38
$begingroup$
$(ab)c =e$ what does that say about $ab$? @Gykonik
$endgroup$
– greedoid
Nov 27 '18 at 20:40
$begingroup$
ehm, that $(ab)=c^{-1}$?
$endgroup$
– Gykonik
Nov 27 '18 at 20:43
$begingroup$
So...............
$endgroup$
– greedoid
Nov 27 '18 at 20:44
$begingroup$
$c^{-1}$ and $c$ has to be in $R^times$?
$endgroup$
– Gykonik
Nov 27 '18 at 20:44
add a comment |
$begingroup$
didn't you proof that $(ab)c in R^times$ I don't see how this is a proof for $ab in R^times$
$endgroup$
– Gykonik
Nov 27 '18 at 20:38
$begingroup$
$(ab)c =e$ what does that say about $ab$? @Gykonik
$endgroup$
– greedoid
Nov 27 '18 at 20:40
$begingroup$
ehm, that $(ab)=c^{-1}$?
$endgroup$
– Gykonik
Nov 27 '18 at 20:43
$begingroup$
So...............
$endgroup$
– greedoid
Nov 27 '18 at 20:44
$begingroup$
$c^{-1}$ and $c$ has to be in $R^times$?
$endgroup$
– Gykonik
Nov 27 '18 at 20:44
$begingroup$
didn't you proof that $(ab)c in R^times$ I don't see how this is a proof for $ab in R^times$
$endgroup$
– Gykonik
Nov 27 '18 at 20:38
$begingroup$
didn't you proof that $(ab)c in R^times$ I don't see how this is a proof for $ab in R^times$
$endgroup$
– Gykonik
Nov 27 '18 at 20:38
$begingroup$
$(ab)c =e$ what does that say about $ab$? @Gykonik
$endgroup$
– greedoid
Nov 27 '18 at 20:40
$begingroup$
$(ab)c =e$ what does that say about $ab$? @Gykonik
$endgroup$
– greedoid
Nov 27 '18 at 20:40
$begingroup$
ehm, that $(ab)=c^{-1}$?
$endgroup$
– Gykonik
Nov 27 '18 at 20:43
$begingroup$
ehm, that $(ab)=c^{-1}$?
$endgroup$
– Gykonik
Nov 27 '18 at 20:43
$begingroup$
So...............
$endgroup$
– greedoid
Nov 27 '18 at 20:44
$begingroup$
So...............
$endgroup$
– greedoid
Nov 27 '18 at 20:44
$begingroup$
$c^{-1}$ and $c$ has to be in $R^times$?
$endgroup$
– Gykonik
Nov 27 '18 at 20:44
$begingroup$
$c^{-1}$ and $c$ has to be in $R^times$?
$endgroup$
– Gykonik
Nov 27 '18 at 20:44
add a comment |
$begingroup$
What do you know about $a$?
$endgroup$
– Fakemistake
Nov 27 '18 at 19:42
$begingroup$
sorry, misspelling. Edited it
$endgroup$
– kjwemke13
Nov 27 '18 at 19:43
$begingroup$
Consider the element $a^{-1}b^{-1}$
$endgroup$
– Fakemistake
Nov 27 '18 at 19:54