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







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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

















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







share|cite|improve this question






















  • 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















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







share|cite|improve this question













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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 27 at 13:59









Jack J.

5711318




5711318












  • 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










  • 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












1 Answer
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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.






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    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.






    share|cite|improve this answer

























      up vote
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      down vote



      accepted
      +50










      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.






      share|cite|improve this answer























        up vote
        2
        down vote



        accepted
        +50







        up vote
        2
        down vote



        accepted
        +50




        +50




        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.






        share|cite|improve this answer












        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 3 at 14:09









        Cornelius

        1957




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