$(a) = R$ if and only if a is a unit.
Let $⟨R;+,−,0,·,1⟩$ be a commutative ring. For $a in R$, define $(a)$ $:= {a · r | r in R }.$ How can i prove that $(a) = R$ if and only if a is a unit.
So if there exist $a' in R$ such that $a*a' = 1$ we have $ forall r in R $ $ (a*a')*r = r$ hence $(a) = R$ , right ?
If $(a) = R$ than $ forall r in R $ there exists $r' in R $ such that $ a*r' = r$ but than ...?
Is the first part correct and how can i formulate the second part?
abstract-algebra proof-verification
asked Nov 25 '18 at 10:39
Tralren
12
add a comment |
Let $⟨R;+,−,0,·,1⟩$ be a commutative ring. For $a in R$, define $(a)$ $:= {a · r | r in R }.$ How can i prove that $(a) = R$ if and only if a is a unit.
So if there exist $a' in R$ such that $a*a' = 1$ we have $ forall r in R $ $ (a*a')*r = r$ hence $(a) = R$ , right ?
If $(a) = R$ than $ forall r in R $ there exists $r' in R $ such that $ a*r' = r$ but than ...?
Is the first part correct and how can i formulate the second part?
abstract-algebra proof-verification
asked Nov 25 '18 at 10:39
Tralren
12
1
First part is correct, second isn't. You don't get $ar=r$ for all $r$. Just for all $r$ there is $r'$ such that $ar'=r$.
– Wojowu
Nov 25 '18 at 10:41
add a comment |
Let $⟨R;+,−,0,·,1⟩$ be a commutative ring. For $a in R$, define $(a)$ $:= {a · r | r in R }.$ How can i prove that $(a) = R$ if and only if a is a unit.
So if there exist $a' in R$ such that $a*a' = 1$ we have $ forall r in R $ $ (a*a')*r = r$ hence $(a) = R$ , right ?
If $(a) = R$ than $ forall r in R $ there exists $r' in R $ such that $ a*r' = r$ but than ...?
Is the first part correct and how can i formulate the second part?
abstract-algebra proof-verification
asked Nov 25 '18 at 10:39
Tralren
12
Let $⟨R;+,−,0,·,1⟩$ be a commutative ring. For $a in R$, define $(a)$ $:= {a · r | r in R }.$ How can i prove that $(a) = R$ if and only if a is a unit.
So if there exist $a' in R$ such that $a*a' = 1$ we have $ forall r in R $ $ (a*a')*r = r$ hence $(a) = R$ , right ?
If $(a) = R$ than $ forall r in R $ there exists $r' in R $ such that $ a*r' = r$ but than ...?
Is the first part correct and how can i formulate the second part?
abstract-algebra proof-verification
abstract-algebra proof-verification
asked Nov 25 '18 at 10:39
Tralren
12
asked Nov 25 '18 at 10:39
Tralren
12
edited Nov 25 '18 at 10:46
asked Nov 25 '18 at 10:39
Tralren
12
asked Nov 25 '18 at 10:39
Tralren
12
asked Nov 25 '18 at 10:39
Tralren
12
12
1
First part is correct, second isn't. You don't get $ar=r$ for all $r$. Just for all $r$ there is $r'$ such that $ar'=r$.
– Wojowu
Nov 25 '18 at 10:41
add a comment |
1
First part is correct, second isn't. You don't get $ar=r$ for all $r$. Just for all $r$ there is $r'$ such that $ar'=r$.
– Wojowu
Nov 25 '18 at 10:41
1
1
First part is correct, second isn't. You don't get $ar=r$ for all $r$. Just for all $r$ there is $r'$ such that $ar'=r$.
– Wojowu
Nov 25 '18 at 10:41
First part is correct, second isn't. You don't get $ar=r$ for all $r$. Just for all $r$ there is $r'$ such that $ar'=r$.
– Wojowu
Nov 25 '18 at 10:41
add a comment |
2 Answers
2
active
oldest
votes
The first part is correct: if $a$ is a unit, then $ab=1$ for some $bin R$; therefore, for every $rin R$,
$$
r=r1=r(ab)=a(rb)in(a)
$$
hence $Rsubseteq (a)$ and therefore $R=(a)$.
The second part is simpler: if $(a)=R$, then $1in (a)$, so there exists $bin R$ with $ab=1$.
answered Nov 25 '18 at 10:48
egreg
178k1484201
add a comment |
The first part is correct.
For the second one, your initial thought needn't be true: as an example, consider $R=mathbb{Z}_p$ and the ideal $(-1)$. Then, even though $(-1) = mathbb{Z}_p$, in general $(-1)r neq r$.
The only implication is that $forall r in R quadexists r'$ such that $ar=r'$.From this, putting $r'=1$, you see that $a$ must be a unit.
answered Nov 25 '18 at 10:49
LucaRand
2615
add a comment |
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2 Answers
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2 Answers
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The first part is correct: if $a$ is a unit, then $ab=1$ for some $bin R$; therefore, for every $rin R$,
$$
r=r1=r(ab)=a(rb)in(a)
$$
hence $Rsubseteq (a)$ and therefore $R=(a)$.
The second part is simpler: if $(a)=R$, then $1in (a)$, so there exists $bin R$ with $ab=1$.
answered Nov 25 '18 at 10:48
egreg
178k1484201
add a comment |
The first part is correct: if $a$ is a unit, then $ab=1$ for some $bin R$; therefore, for every $rin R$,
$$
r=r1=r(ab)=a(rb)in(a)
$$
hence $Rsubseteq (a)$ and therefore $R=(a)$.
The second part is simpler: if $(a)=R$, then $1in (a)$, so there exists $bin R$ with $ab=1$.
answered Nov 25 '18 at 10:48
egreg
178k1484201
add a comment |
The first part is correct: if $a$ is a unit, then $ab=1$ for some $bin R$; therefore, for every $rin R$,
$$
r=r1=r(ab)=a(rb)in(a)
$$
hence $Rsubseteq (a)$ and therefore $R=(a)$.
The second part is simpler: if $(a)=R$, then $1in (a)$, so there exists $bin R$ with $ab=1$.
answered Nov 25 '18 at 10:48
egreg
178k1484201
The first part is correct: if $a$ is a unit, then $ab=1$ for some $bin R$; therefore, for every $rin R$,
$$
r=r1=r(ab)=a(rb)in(a)
$$
hence $Rsubseteq (a)$ and therefore $R=(a)$.
The second part is simpler: if $(a)=R$, then $1in (a)$, so there exists $bin R$ with $ab=1$.
answered Nov 25 '18 at 10:48
egreg
178k1484201
answered Nov 25 '18 at 10:48
egreg
178k1484201
answered Nov 25 '18 at 10:48
egreg
178k1484201
answered Nov 25 '18 at 10:48
egreg
178k1484201
178k1484201
add a comment |
add a comment |
The first part is correct.
For the second one, your initial thought needn't be true: as an example, consider $R=mathbb{Z}_p$ and the ideal $(-1)$. Then, even though $(-1) = mathbb{Z}_p$, in general $(-1)r neq r$.
The only implication is that $forall r in R quadexists r'$ such that $ar=r'$.From this, putting $r'=1$, you see that $a$ must be a unit.
answered Nov 25 '18 at 10:49
LucaRand
2615
add a comment |
The first part is correct.
For the second one, your initial thought needn't be true: as an example, consider $R=mathbb{Z}_p$ and the ideal $(-1)$. Then, even though $(-1) = mathbb{Z}_p$, in general $(-1)r neq r$.
The only implication is that $forall r in R quadexists r'$ such that $ar=r'$.From this, putting $r'=1$, you see that $a$ must be a unit.
answered Nov 25 '18 at 10:49
LucaRand
2615
add a comment |
The first part is correct.
For the second one, your initial thought needn't be true: as an example, consider $R=mathbb{Z}_p$ and the ideal $(-1)$. Then, even though $(-1) = mathbb{Z}_p$, in general $(-1)r neq r$.
The only implication is that $forall r in R quadexists r'$ such that $ar=r'$.From this, putting $r'=1$, you see that $a$ must be a unit.
answered Nov 25 '18 at 10:49
LucaRand
2615
The first part is correct.
For the second one, your initial thought needn't be true: as an example, consider $R=mathbb{Z}_p$ and the ideal $(-1)$. Then, even though $(-1) = mathbb{Z}_p$, in general $(-1)r neq r$.
The only implication is that $forall r in R quadexists r'$ such that $ar=r'$.From this, putting $r'=1$, you see that $a$ must be a unit.
answered Nov 25 '18 at 10:49
LucaRand
2615
edited Nov 26 '18 at 17:16
answered Nov 25 '18 at 10:49
LucaRand
2615
answered Nov 25 '18 at 10:49
LucaRand
2615
answered Nov 25 '18 at 10:49
LucaRand
2615
2615
add a comment |
add a comment |
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First part is correct, second isn't. You don't get $ar=r$ for all $r$. Just for all $r$ there is $r'$ such that $ar'=r$.
– Wojowu
Nov 25 '18 at 10:41