Maximal ideal and invertible element
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I must be prove the following result: Let $R$ a commutative ring with unity, then
begin{equation}
bigg (ain R;text{is invertible}bigg) Longleftrightarrow bigg(ain I,text{when};Isubseteq R;text{is not maximal}bigg).
end{equation}
My attempt: ($Rightarrow$) Let $ain R$ invertible, then $R=(a)$. Indeed, if it were not $1notin(a)$, then $a$ would not be invertible. Therefore $R=(a)$, then $ain(a)$ which is not maximal. (Correct?)
$(Leftarrow)$ For this implication I know I must use the Krull-Zorn Lemma, which states that every proper ideal of $R$ is contained in a maximal ideal.
Could someone help me complete the proof? Thanks!
abstract-algebra proof-verification proof-writing proof-explanation
asked Nov 27 at 13:59
Jack J.
4601419
|
show 3 more comments
up vote
0
down vote
favorite
I must be prove the following result: Let $R$ a commutative ring with unity, then
begin{equation}
bigg (ain R;text{is invertible}bigg) Longleftrightarrow bigg(ain I,text{when};Isubseteq R;text{is not maximal}bigg).
end{equation}
My attempt: ($Rightarrow$) Let $ain R$ invertible, then $R=(a)$. Indeed, if it were not $1notin(a)$, then $a$ would not be invertible. Therefore $R=(a)$, then $ain(a)$ which is not maximal. (Correct?)
$(Leftarrow)$ For this implication I know I must use the Krull-Zorn Lemma, which states that every proper ideal of $R$ is contained in a maximal ideal.
Could someone help me complete the proof? Thanks!
abstract-algebra proof-verification proof-writing proof-explanation
asked Nov 27 at 13:59
Jack J.
4601419
For $1^{st}$ part: If $ain R$ is invertible and $ain I$ then $1in Iimplies I=R$ is not maximal.
– Yadati Kiran
Nov 27 at 14:05
Thanks for your anser, if $ain I$, then $a^{-1}in I$?
– Jack J.
Nov 27 at 14:10
Since $I$ is an ideal so we have for $ain I$ and $rin Rimplies arin I$. So take $a^{-1}=r$.
– Yadati Kiran
Nov 27 at 14:12
Ok! Sorry! Could you help me too for the second implication?
– Jack J.
Nov 27 at 14:14
1
@JackJ.: if "$a$ is not contained in any maximal ideal " then $ain I=R implies 1in Iimplies :exists:bin R ::ab=1in I$.
– Yadati Kiran
Nov 27 at 14:25
|
show 3 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I must be prove the following result: Let $R$ a commutative ring with unity, then
begin{equation}
bigg (ain R;text{is invertible}bigg) Longleftrightarrow bigg(ain I,text{when};Isubseteq R;text{is not maximal}bigg).
end{equation}
My attempt: ($Rightarrow$) Let $ain R$ invertible, then $R=(a)$. Indeed, if it were not $1notin(a)$, then $a$ would not be invertible. Therefore $R=(a)$, then $ain(a)$ which is not maximal. (Correct?)
$(Leftarrow)$ For this implication I know I must use the Krull-Zorn Lemma, which states that every proper ideal of $R$ is contained in a maximal ideal.
Could someone help me complete the proof? Thanks!
abstract-algebra proof-verification proof-writing proof-explanation
asked Nov 27 at 13:59
Jack J.
4601419
I must be prove the following result: Let $R$ a commutative ring with unity, then
begin{equation}
bigg (ain R;text{is invertible}bigg) Longleftrightarrow bigg(ain I,text{when};Isubseteq R;text{is not maximal}bigg).
end{equation}
My attempt: ($Rightarrow$) Let $ain R$ invertible, then $R=(a)$. Indeed, if it were not $1notin(a)$, then $a$ would not be invertible. Therefore $R=(a)$, then $ain(a)$ which is not maximal. (Correct?)
$(Leftarrow)$ For this implication I know I must use the Krull-Zorn Lemma, which states that every proper ideal of $R$ is contained in a maximal ideal.
Could someone help me complete the proof? Thanks!
abstract-algebra proof-verification proof-writing proof-explanation
abstract-algebra proof-verification proof-writing proof-explanation
asked Nov 27 at 13:59
Jack J.
4601419
asked Nov 27 at 13:59
Jack J.
4601419
asked Nov 27 at 13:59
Jack J.
4601419
asked Nov 27 at 13:59
Jack J.
4601419
asked Nov 27 at 13:59
Jack J.
4601419
4601419
For $1^{st}$ part: If $ain R$ is invertible and $ain I$ then $1in Iimplies I=R$ is not maximal.
– Yadati Kiran
Nov 27 at 14:05
Thanks for your anser, if $ain I$, then $a^{-1}in I$?
– Jack J.
Nov 27 at 14:10
Since $I$ is an ideal so we have for $ain I$ and $rin Rimplies arin I$. So take $a^{-1}=r$.
– Yadati Kiran
Nov 27 at 14:12
Ok! Sorry! Could you help me too for the second implication?
– Jack J.
Nov 27 at 14:14
1
@JackJ.: if "$a$ is not contained in any maximal ideal " then $ain I=R implies 1in Iimplies :exists:bin R ::ab=1in I$.
– Yadati Kiran
Nov 27 at 14:25
|
show 3 more comments
For $1^{st}$ part: If $ain R$ is invertible and $ain I$ then $1in Iimplies I=R$ is not maximal.
– Yadati Kiran
Nov 27 at 14:05
Thanks for your anser, if $ain I$, then $a^{-1}in I$?
– Jack J.
Nov 27 at 14:10
Since $I$ is an ideal so we have for $ain I$ and $rin Rimplies arin I$. So take $a^{-1}=r$.
– Yadati Kiran
Nov 27 at 14:12
Ok! Sorry! Could you help me too for the second implication?
– Jack J.
Nov 27 at 14:14
1
@JackJ.: if "$a$ is not contained in any maximal ideal " then $ain I=R implies 1in Iimplies :exists:bin R ::ab=1in I$.
– Yadati Kiran
Nov 27 at 14:25
For $1^{st}$ part: If $ain R$ is invertible and $ain I$ then $1in Iimplies I=R$ is not maximal.
– Yadati Kiran
Nov 27 at 14:05
For $1^{st}$ part: If $ain R$ is invertible and $ain I$ then $1in Iimplies I=R$ is not maximal.
– Yadati Kiran
Nov 27 at 14:05
Thanks for your anser, if $ain I$, then $a^{-1}in I$?
– Jack J.
Nov 27 at 14:10
Thanks for your anser, if $ain I$, then $a^{-1}in I$?
– Jack J.
Nov 27 at 14:10
Since $I$ is an ideal so we have for $ain I$ and $rin Rimplies arin I$. So take $a^{-1}=r$.
– Yadati Kiran
Nov 27 at 14:12
Since $I$ is an ideal so we have for $ain I$ and $rin Rimplies arin I$. So take $a^{-1}=r$.
– Yadati Kiran
Nov 27 at 14:12
Ok! Sorry! Could you help me too for the second implication?
– Jack J.
Nov 27 at 14:14
Ok! Sorry! Could you help me too for the second implication?
– Jack J.
Nov 27 at 14:14
1
1
@JackJ.: if "$a$ is not contained in any maximal ideal " then $ain I=R implies 1in Iimplies :exists:bin R ::ab=1in I$.
– Yadati Kiran
Nov 27 at 14:25
@JackJ.: if "$a$ is not contained in any maximal ideal " then $ain I=R implies 1in Iimplies :exists:bin R ::ab=1in I$.
– Yadati Kiran
Nov 27 at 14:25
|
show 3 more comments
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
The proof for $(Rightarrow)$ is correct. For the implication $(Leftarrow)$, suppose that for every ideal $I$ of $R$, if $a in I$ then $I$ is not maximal. Now, if $a$ is not invertible then the ideal generated by $a$ is a proper ideal of $R$ and hence, by Krull-Zorn Lemma, we have that $(a) subseteq mathfrak{m} $ for some maximal ideal $mathfrak{m}$ of $R$ which contradicts our hypothesis thus, $a$ is invetrible.
answered Dec 3 at 14:09
Cornelius
1957
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The proof for $(Rightarrow)$ is correct. For the implication $(Leftarrow)$, suppose that for every ideal $I$ of $R$, if $a in I$ then $I$ is not maximal. Now, if $a$ is not invertible then the ideal generated by $a$ is a proper ideal of $R$ and hence, by Krull-Zorn Lemma, we have that $(a) subseteq mathfrak{m} $ for some maximal ideal $mathfrak{m}$ of $R$ which contradicts our hypothesis thus, $a$ is invetrible.
answered Dec 3 at 14:09
Cornelius
1957
add a comment |
up vote
2
down vote
accepted
The proof for $(Rightarrow)$ is correct. For the implication $(Leftarrow)$, suppose that for every ideal $I$ of $R$, if $a in I$ then $I$ is not maximal. Now, if $a$ is not invertible then the ideal generated by $a$ is a proper ideal of $R$ and hence, by Krull-Zorn Lemma, we have that $(a) subseteq mathfrak{m} $ for some maximal ideal $mathfrak{m}$ of $R$ which contradicts our hypothesis thus, $a$ is invetrible.
answered Dec 3 at 14:09
Cornelius
1957
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The proof for $(Rightarrow)$ is correct. For the implication $(Leftarrow)$, suppose that for every ideal $I$ of $R$, if $a in I$ then $I$ is not maximal. Now, if $a$ is not invertible then the ideal generated by $a$ is a proper ideal of $R$ and hence, by Krull-Zorn Lemma, we have that $(a) subseteq mathfrak{m} $ for some maximal ideal $mathfrak{m}$ of $R$ which contradicts our hypothesis thus, $a$ is invetrible.
answered Dec 3 at 14:09
Cornelius
1957
The proof for $(Rightarrow)$ is correct. For the implication $(Leftarrow)$, suppose that for every ideal $I$ of $R$, if $a in I$ then $I$ is not maximal. Now, if $a$ is not invertible then the ideal generated by $a$ is a proper ideal of $R$ and hence, by Krull-Zorn Lemma, we have that $(a) subseteq mathfrak{m} $ for some maximal ideal $mathfrak{m}$ of $R$ which contradicts our hypothesis thus, $a$ is invetrible.
answered Dec 3 at 14:09
Cornelius
1957
answered Dec 3 at 14:09
Cornelius
1957
answered Dec 3 at 14:09
Cornelius
1957
answered Dec 3 at 14:09
Cornelius
1957
1957
add a comment |
add a comment |
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For $1^{st}$ part: If $ain R$ is invertible and $ain I$ then $1in Iimplies I=R$ is not maximal.
– Yadati Kiran
Nov 27 at 14:05
Thanks for your anser, if $ain I$, then $a^{-1}in I$?
– Jack J.
Nov 27 at 14:10
Since $I$ is an ideal so we have for $ain I$ and $rin Rimplies arin I$. So take $a^{-1}=r$.
– Yadati Kiran
Nov 27 at 14:12
Ok! Sorry! Could you help me too for the second implication?
– Jack J.
Nov 27 at 14:14
1
@JackJ.: if "$a$ is not contained in any maximal ideal " then $ain I=R implies 1in Iimplies :exists:bin R ::ab=1in I$.
– Yadati Kiran
Nov 27 at 14:25