Show that an element $x$ belongs to the unique maximal ideal of the commutative ring $R$ if and only if $...
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Let R be a commutative ring with $1 neq 0$. Let $M$ be the unique maximal ideal of R. I want to show that $x in M$ if and only if $1+xr$ is invertible for every $r in R$. I am a bit stuck on where to begin. Any hints are appreciated.
abstract-algebra
asked Dec 27 '18 at 0:55
user8513188
1096
add a comment |
Let R be a commutative ring with $1 neq 0$. Let $M$ be the unique maximal ideal of R. I want to show that $x in M$ if and only if $1+xr$ is invertible for every $r in R$. I am a bit stuck on where to begin. Any hints are appreciated.
abstract-algebra
asked Dec 27 '18 at 0:55
user8513188
1096
1
What happens if an element is not invertible?
– Rafael Holanda
Dec 27 '18 at 1:08
add a comment |
Let R be a commutative ring with $1 neq 0$. Let $M$ be the unique maximal ideal of R. I want to show that $x in M$ if and only if $1+xr$ is invertible for every $r in R$. I am a bit stuck on where to begin. Any hints are appreciated.
abstract-algebra
asked Dec 27 '18 at 0:55
user8513188
1096
Let R be a commutative ring with $1 neq 0$. Let $M$ be the unique maximal ideal of R. I want to show that $x in M$ if and only if $1+xr$ is invertible for every $r in R$. I am a bit stuck on where to begin. Any hints are appreciated.
abstract-algebra
abstract-algebra
asked Dec 27 '18 at 0:55
user8513188
1096
asked Dec 27 '18 at 0:55
user8513188
1096
asked Dec 27 '18 at 0:55
user8513188
1096
asked Dec 27 '18 at 0:55
user8513188
1096
asked Dec 27 '18 at 0:55
user8513188
1096
1096
1
What happens if an element is not invertible?
– Rafael Holanda
Dec 27 '18 at 1:08
add a comment |
1
What happens if an element is not invertible?
– Rafael Holanda
Dec 27 '18 at 1:08
1
1
What happens if an element is not invertible?
– Rafael Holanda
Dec 27 '18 at 1:08
What happens if an element is not invertible?
– Rafael Holanda
Dec 27 '18 at 1:08
add a comment |
2 Answers
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Suppose that $xin M, $ for every $r, xrin M$ since $M$ is an ideal. If $(1+xr)$ is not invertible, it is contained in a maximal ideal which is $M$, you deduce that $1=(1+xr)-xrin M$. Contradiction.
Suppose that $(1+xr)$ is invertible for every $r$, if $x$ is not in $M$, it is invertible, there exists $u$ such that $ux=1$, $1+(-u)x=0$ is not invertible. Contradiction.
answered Dec 27 '18 at 1:06
Tsemo Aristide
55.7k11444
add a comment |
It is the special case $rm J = (x) $ in $ (1) iff (2) $ below.
Theorem $ $ TFAE in ring $rm:R:$ with units $rm:U,:$ ideal $rm:J,:$ and Jacobson radical $rm:Jac(R)$
$rm(1)quad J subseteq Jac(R),quad $ i.e. $rm:J:$ lies in every max ideal $rm:M:$ of $rm:R$
$rm(2)quad 1+J subseteq U,quad $ i.e. $rm: 1 + j:$ is a unit for every $rm: j in J$
$rm(3)quad Ineq 1 Rightarrow I,+,J neq 1,quad $ i.e. proper ideals survive in $rm:R/J$
$rm(4)quad!! M$ max $rm,Rightarrow M+J ne 1,quad $ i.e. max ideals survive in $rm:R/J$
Proof $: $ (sketch) $ $ With $rm:i in I, j in J,:$ and max ideal $rm:M,$
$rm(1Rightarrow 2)quad j in all M Rightarrow 1+j in no M Rightarrow 1+j:$ unit.
$rm(2Rightarrow 3)quad i+j = 1 Rightarrow 1-j = i:$ unit $rm:Rightarrow: I = 1$
$rm(3Rightarrow 4) $ Let $rm:I = M:$ max.
$rm(4Rightarrow 1)quad M+J ne 1 Rightarrow J subseteq M:$ by $rm:M:$ max.
answered Dec 27 '18 at 2:01
Bill Dubuque
208k29190628
add a comment |
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2 Answers
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2 Answers
2
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oldest
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Suppose that $xin M, $ for every $r, xrin M$ since $M$ is an ideal. If $(1+xr)$ is not invertible, it is contained in a maximal ideal which is $M$, you deduce that $1=(1+xr)-xrin M$. Contradiction.
Suppose that $(1+xr)$ is invertible for every $r$, if $x$ is not in $M$, it is invertible, there exists $u$ such that $ux=1$, $1+(-u)x=0$ is not invertible. Contradiction.
answered Dec 27 '18 at 1:06
Tsemo Aristide
55.7k11444
add a comment |
Suppose that $xin M, $ for every $r, xrin M$ since $M$ is an ideal. If $(1+xr)$ is not invertible, it is contained in a maximal ideal which is $M$, you deduce that $1=(1+xr)-xrin M$. Contradiction.
Suppose that $(1+xr)$ is invertible for every $r$, if $x$ is not in $M$, it is invertible, there exists $u$ such that $ux=1$, $1+(-u)x=0$ is not invertible. Contradiction.
answered Dec 27 '18 at 1:06
Tsemo Aristide
55.7k11444
add a comment |
Suppose that $xin M, $ for every $r, xrin M$ since $M$ is an ideal. If $(1+xr)$ is not invertible, it is contained in a maximal ideal which is $M$, you deduce that $1=(1+xr)-xrin M$. Contradiction.
Suppose that $(1+xr)$ is invertible for every $r$, if $x$ is not in $M$, it is invertible, there exists $u$ such that $ux=1$, $1+(-u)x=0$ is not invertible. Contradiction.
answered Dec 27 '18 at 1:06
Tsemo Aristide
55.7k11444
Suppose that $xin M, $ for every $r, xrin M$ since $M$ is an ideal. If $(1+xr)$ is not invertible, it is contained in a maximal ideal which is $M$, you deduce that $1=(1+xr)-xrin M$. Contradiction.
Suppose that $(1+xr)$ is invertible for every $r$, if $x$ is not in $M$, it is invertible, there exists $u$ such that $ux=1$, $1+(-u)x=0$ is not invertible. Contradiction.
answered Dec 27 '18 at 1:06
Tsemo Aristide
55.7k11444
answered Dec 27 '18 at 1:06
Tsemo Aristide
55.7k11444
answered Dec 27 '18 at 1:06
Tsemo Aristide
55.7k11444
answered Dec 27 '18 at 1:06
Tsemo Aristide
55.7k11444
55.7k11444
add a comment |
add a comment |
It is the special case $rm J = (x) $ in $ (1) iff (2) $ below.
Theorem $ $ TFAE in ring $rm:R:$ with units $rm:U,:$ ideal $rm:J,:$ and Jacobson radical $rm:Jac(R)$
$rm(1)quad J subseteq Jac(R),quad $ i.e. $rm:J:$ lies in every max ideal $rm:M:$ of $rm:R$
$rm(2)quad 1+J subseteq U,quad $ i.e. $rm: 1 + j:$ is a unit for every $rm: j in J$
$rm(3)quad Ineq 1 Rightarrow I,+,J neq 1,quad $ i.e. proper ideals survive in $rm:R/J$
$rm(4)quad!! M$ max $rm,Rightarrow M+J ne 1,quad $ i.e. max ideals survive in $rm:R/J$
Proof $: $ (sketch) $ $ With $rm:i in I, j in J,:$ and max ideal $rm:M,$
$rm(1Rightarrow 2)quad j in all M Rightarrow 1+j in no M Rightarrow 1+j:$ unit.
$rm(2Rightarrow 3)quad i+j = 1 Rightarrow 1-j = i:$ unit $rm:Rightarrow: I = 1$
$rm(3Rightarrow 4) $ Let $rm:I = M:$ max.
$rm(4Rightarrow 1)quad M+J ne 1 Rightarrow J subseteq M:$ by $rm:M:$ max.
answered Dec 27 '18 at 2:01
Bill Dubuque
208k29190628
add a comment |
It is the special case $rm J = (x) $ in $ (1) iff (2) $ below.
Theorem $ $ TFAE in ring $rm:R:$ with units $rm:U,:$ ideal $rm:J,:$ and Jacobson radical $rm:Jac(R)$
$rm(1)quad J subseteq Jac(R),quad $ i.e. $rm:J:$ lies in every max ideal $rm:M:$ of $rm:R$
$rm(2)quad 1+J subseteq U,quad $ i.e. $rm: 1 + j:$ is a unit for every $rm: j in J$
$rm(3)quad Ineq 1 Rightarrow I,+,J neq 1,quad $ i.e. proper ideals survive in $rm:R/J$
$rm(4)quad!! M$ max $rm,Rightarrow M+J ne 1,quad $ i.e. max ideals survive in $rm:R/J$
Proof $: $ (sketch) $ $ With $rm:i in I, j in J,:$ and max ideal $rm:M,$
$rm(1Rightarrow 2)quad j in all M Rightarrow 1+j in no M Rightarrow 1+j:$ unit.
$rm(2Rightarrow 3)quad i+j = 1 Rightarrow 1-j = i:$ unit $rm:Rightarrow: I = 1$
$rm(3Rightarrow 4) $ Let $rm:I = M:$ max.
$rm(4Rightarrow 1)quad M+J ne 1 Rightarrow J subseteq M:$ by $rm:M:$ max.
answered Dec 27 '18 at 2:01
Bill Dubuque
208k29190628
add a comment |
It is the special case $rm J = (x) $ in $ (1) iff (2) $ below.
Theorem $ $ TFAE in ring $rm:R:$ with units $rm:U,:$ ideal $rm:J,:$ and Jacobson radical $rm:Jac(R)$
$rm(1)quad J subseteq Jac(R),quad $ i.e. $rm:J:$ lies in every max ideal $rm:M:$ of $rm:R$
$rm(2)quad 1+J subseteq U,quad $ i.e. $rm: 1 + j:$ is a unit for every $rm: j in J$
$rm(3)quad Ineq 1 Rightarrow I,+,J neq 1,quad $ i.e. proper ideals survive in $rm:R/J$
$rm(4)quad!! M$ max $rm,Rightarrow M+J ne 1,quad $ i.e. max ideals survive in $rm:R/J$
Proof $: $ (sketch) $ $ With $rm:i in I, j in J,:$ and max ideal $rm:M,$
$rm(1Rightarrow 2)quad j in all M Rightarrow 1+j in no M Rightarrow 1+j:$ unit.
$rm(2Rightarrow 3)quad i+j = 1 Rightarrow 1-j = i:$ unit $rm:Rightarrow: I = 1$
$rm(3Rightarrow 4) $ Let $rm:I = M:$ max.
$rm(4Rightarrow 1)quad M+J ne 1 Rightarrow J subseteq M:$ by $rm:M:$ max.
answered Dec 27 '18 at 2:01
Bill Dubuque
208k29190628
It is the special case $rm J = (x) $ in $ (1) iff (2) $ below.
Theorem $ $ TFAE in ring $rm:R:$ with units $rm:U,:$ ideal $rm:J,:$ and Jacobson radical $rm:Jac(R)$
$rm(1)quad J subseteq Jac(R),quad $ i.e. $rm:J:$ lies in every max ideal $rm:M:$ of $rm:R$
$rm(2)quad 1+J subseteq U,quad $ i.e. $rm: 1 + j:$ is a unit for every $rm: j in J$
$rm(3)quad Ineq 1 Rightarrow I,+,J neq 1,quad $ i.e. proper ideals survive in $rm:R/J$
$rm(4)quad!! M$ max $rm,Rightarrow M+J ne 1,quad $ i.e. max ideals survive in $rm:R/J$
Proof $: $ (sketch) $ $ With $rm:i in I, j in J,:$ and max ideal $rm:M,$
$rm(1Rightarrow 2)quad j in all M Rightarrow 1+j in no M Rightarrow 1+j:$ unit.
$rm(2Rightarrow 3)quad i+j = 1 Rightarrow 1-j = i:$ unit $rm:Rightarrow: I = 1$
$rm(3Rightarrow 4) $ Let $rm:I = M:$ max.
$rm(4Rightarrow 1)quad M+J ne 1 Rightarrow J subseteq M:$ by $rm:M:$ max.
answered Dec 27 '18 at 2:01
Bill Dubuque
208k29190628
edited Dec 27 '18 at 2:09
answered Dec 27 '18 at 2:01
Bill Dubuque
208k29190628
answered Dec 27 '18 at 2:01
Bill Dubuque
208k29190628
answered Dec 27 '18 at 2:01
Bill Dubuque
208k29190628
208k29190628
add a comment |
add a comment |
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1
What happens if an element is not invertible?
– Rafael Holanda
Dec 27 '18 at 1:08