What is wrong with this argument? $R/I approx R$












0












$begingroup$


Let $I subset R$ be an ideal of $R$. Define $pi(a + I) = a$ with $a in R$, so $pi: R/I to R$



Then $ker pi = I = bar{0} in R/I$ and this is clearly onto so it is an isomorphism. So this map says $R$ is always isomorphic to its quotient. Isn't there something not right about this argument? Or is this just a simple trivialization of the correspondence property?



Suppose we have the containment of ideals $I subset A subset R$



Then the correspondence theorem say $I subset A subset R stackrel{pi}leftrightsquigarrow I subset A/I subset R/I$



So considering this, set $A = R$, then $R subset R$ is an ideal of itself and so is $R/I subset R/I$, we get what I wrote previously?



Why am I asking this? It is because I was working several problems in Dummit chapter 7.4 and at least half of them could be answered by this without writing much more, but I feel like I m cheating my way here.



Here is one that made me wrote this question




If $R$ is commutative (with $1$) then if $P$ is a prime ideal of $R$ with no zero divisors, then $R$ is integral domain.




$P$ is prime so $R/P$ is integral domain. By the correspondence $R/P approx R$. So $P$ having no zero divisors seem to play no role.










share|cite|improve this question









$endgroup$








  • 4




    $begingroup$
    Your mapping $pi$ isn't well defined.
    $endgroup$
    – Nathanael Skrepek
    Dec 21 '18 at 11:08
















0












$begingroup$


Let $I subset R$ be an ideal of $R$. Define $pi(a + I) = a$ with $a in R$, so $pi: R/I to R$



Then $ker pi = I = bar{0} in R/I$ and this is clearly onto so it is an isomorphism. So this map says $R$ is always isomorphic to its quotient. Isn't there something not right about this argument? Or is this just a simple trivialization of the correspondence property?



Suppose we have the containment of ideals $I subset A subset R$



Then the correspondence theorem say $I subset A subset R stackrel{pi}leftrightsquigarrow I subset A/I subset R/I$



So considering this, set $A = R$, then $R subset R$ is an ideal of itself and so is $R/I subset R/I$, we get what I wrote previously?



Why am I asking this? It is because I was working several problems in Dummit chapter 7.4 and at least half of them could be answered by this without writing much more, but I feel like I m cheating my way here.



Here is one that made me wrote this question




If $R$ is commutative (with $1$) then if $P$ is a prime ideal of $R$ with no zero divisors, then $R$ is integral domain.




$P$ is prime so $R/P$ is integral domain. By the correspondence $R/P approx R$. So $P$ having no zero divisors seem to play no role.










share|cite|improve this question









$endgroup$








  • 4




    $begingroup$
    Your mapping $pi$ isn't well defined.
    $endgroup$
    – Nathanael Skrepek
    Dec 21 '18 at 11:08














0












0








0





$begingroup$


Let $I subset R$ be an ideal of $R$. Define $pi(a + I) = a$ with $a in R$, so $pi: R/I to R$



Then $ker pi = I = bar{0} in R/I$ and this is clearly onto so it is an isomorphism. So this map says $R$ is always isomorphic to its quotient. Isn't there something not right about this argument? Or is this just a simple trivialization of the correspondence property?



Suppose we have the containment of ideals $I subset A subset R$



Then the correspondence theorem say $I subset A subset R stackrel{pi}leftrightsquigarrow I subset A/I subset R/I$



So considering this, set $A = R$, then $R subset R$ is an ideal of itself and so is $R/I subset R/I$, we get what I wrote previously?



Why am I asking this? It is because I was working several problems in Dummit chapter 7.4 and at least half of them could be answered by this without writing much more, but I feel like I m cheating my way here.



Here is one that made me wrote this question




If $R$ is commutative (with $1$) then if $P$ is a prime ideal of $R$ with no zero divisors, then $R$ is integral domain.




$P$ is prime so $R/P$ is integral domain. By the correspondence $R/P approx R$. So $P$ having no zero divisors seem to play no role.










share|cite|improve this question









$endgroup$




Let $I subset R$ be an ideal of $R$. Define $pi(a + I) = a$ with $a in R$, so $pi: R/I to R$



Then $ker pi = I = bar{0} in R/I$ and this is clearly onto so it is an isomorphism. So this map says $R$ is always isomorphic to its quotient. Isn't there something not right about this argument? Or is this just a simple trivialization of the correspondence property?



Suppose we have the containment of ideals $I subset A subset R$



Then the correspondence theorem say $I subset A subset R stackrel{pi}leftrightsquigarrow I subset A/I subset R/I$



So considering this, set $A = R$, then $R subset R$ is an ideal of itself and so is $R/I subset R/I$, we get what I wrote previously?



Why am I asking this? It is because I was working several problems in Dummit chapter 7.4 and at least half of them could be answered by this without writing much more, but I feel like I m cheating my way here.



Here is one that made me wrote this question




If $R$ is commutative (with $1$) then if $P$ is a prime ideal of $R$ with no zero divisors, then $R$ is integral domain.




$P$ is prime so $R/P$ is integral domain. By the correspondence $R/P approx R$. So $P$ having no zero divisors seem to play no role.







abstract-algebra proof-verification






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 21 '18 at 11:05









HawkHawk

5,5801140110




5,5801140110








  • 4




    $begingroup$
    Your mapping $pi$ isn't well defined.
    $endgroup$
    – Nathanael Skrepek
    Dec 21 '18 at 11:08














  • 4




    $begingroup$
    Your mapping $pi$ isn't well defined.
    $endgroup$
    – Nathanael Skrepek
    Dec 21 '18 at 11:08








4




4




$begingroup$
Your mapping $pi$ isn't well defined.
$endgroup$
– Nathanael Skrepek
Dec 21 '18 at 11:08




$begingroup$
Your mapping $pi$ isn't well defined.
$endgroup$
– Nathanael Skrepek
Dec 21 '18 at 11:08










3 Answers
3






active

oldest

votes


















2












$begingroup$

Your $pi$ is not a function in general. For example consider $R=mathbb{Z}$ and $I=2mathbb{Z}$. Then $0+2mathbb{Z}=2+2mathbb{Z}$. So what is $pi(0+2mathbb{Z})=pi(2+2mathbb{Z})$? Is it $0$ or $2$?



Of course you can arbitrarly choose for example $pi(0+2mathbb{Z}):= 0$ (and therefore $pi(2+2mathbb{Z})= 0$) and so on for every coset but then you will find out that $pi$ is no longer a homomorphism. Indeed, let's simplify notation, I will use $[x]$ instead of $x+I$. In our case $R/I={[0],[1]}$. Note that we have $[1]+[1]=[0]$. Put for example $pi([0])=0$ and $pi([1])=1$. Then



$$pi([1]+[1])=pi([0])=0$$
$$pi([1])+pi([1])=1+1=2$$



and so $pi$ is not a homomorphism. And in this particular case there is no way to fix that. The only choice of values making $phi$ a homomorphism is $pi([x])=0$. Not to mention that $R/I$ has two elements while $R$ is infinite so there's no chance for $pi$ to be surjective.



So as you can see there are no shortcuts. Also the existence of an isomorphism $R/Isimeq R$ for $Ineq 0$ is a very rare situation. First of all if $R$ is nontrivial then $R/R$ is trivial. If you are looking for an example of proper ideal then consider $R$ which is not simple (so it has a proper, nontrivial ideal). Then there's always a proper quotient $R/I$ not isomorphic to $R$, namely for any maximal ideal $I$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So after writing out a lot of stuff, I just came to the conclusion that maps going from the quotient $R/I$ to $R$ usually come with a lot of problems as opposed to defining $R to R/I$.
    $endgroup$
    – Hawk
    Dec 21 '18 at 12:30










  • $begingroup$
    @Hawk yes, the situation is not symmetric at all.
    $endgroup$
    – freakish
    Dec 21 '18 at 12:31










  • $begingroup$
    No what i mean is that if I had defined it $pi: R to R/I$, it would be easier to prove it is an isomorphism and to identify the problems, which in this case is well-definedness.
    $endgroup$
    – Hawk
    Dec 21 '18 at 12:33



















1












$begingroup$

Your function $pi$ is only well-defined if $I$ is the ideal ${0}$. E.g., with $R = Bbb{Z}$ and $I = 2Bbb{Z}$, we have $1 + I = 3 + I$, so if $pi(a + I) = a $, we would have to have $1 = 3$, which holds in $R/I$, but does not hold in $R$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    For each $ain R$ there are several $bin R$ such that $a+I=b+I$ (unless $I={0}$ in the first place). So it is unclear what $a$ you are referring to.






    share|cite|improve this answer











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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Your $pi$ is not a function in general. For example consider $R=mathbb{Z}$ and $I=2mathbb{Z}$. Then $0+2mathbb{Z}=2+2mathbb{Z}$. So what is $pi(0+2mathbb{Z})=pi(2+2mathbb{Z})$? Is it $0$ or $2$?



      Of course you can arbitrarly choose for example $pi(0+2mathbb{Z}):= 0$ (and therefore $pi(2+2mathbb{Z})= 0$) and so on for every coset but then you will find out that $pi$ is no longer a homomorphism. Indeed, let's simplify notation, I will use $[x]$ instead of $x+I$. In our case $R/I={[0],[1]}$. Note that we have $[1]+[1]=[0]$. Put for example $pi([0])=0$ and $pi([1])=1$. Then



      $$pi([1]+[1])=pi([0])=0$$
      $$pi([1])+pi([1])=1+1=2$$



      and so $pi$ is not a homomorphism. And in this particular case there is no way to fix that. The only choice of values making $phi$ a homomorphism is $pi([x])=0$. Not to mention that $R/I$ has two elements while $R$ is infinite so there's no chance for $pi$ to be surjective.



      So as you can see there are no shortcuts. Also the existence of an isomorphism $R/Isimeq R$ for $Ineq 0$ is a very rare situation. First of all if $R$ is nontrivial then $R/R$ is trivial. If you are looking for an example of proper ideal then consider $R$ which is not simple (so it has a proper, nontrivial ideal). Then there's always a proper quotient $R/I$ not isomorphic to $R$, namely for any maximal ideal $I$.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        So after writing out a lot of stuff, I just came to the conclusion that maps going from the quotient $R/I$ to $R$ usually come with a lot of problems as opposed to defining $R to R/I$.
        $endgroup$
        – Hawk
        Dec 21 '18 at 12:30










      • $begingroup$
        @Hawk yes, the situation is not symmetric at all.
        $endgroup$
        – freakish
        Dec 21 '18 at 12:31










      • $begingroup$
        No what i mean is that if I had defined it $pi: R to R/I$, it would be easier to prove it is an isomorphism and to identify the problems, which in this case is well-definedness.
        $endgroup$
        – Hawk
        Dec 21 '18 at 12:33
















      2












      $begingroup$

      Your $pi$ is not a function in general. For example consider $R=mathbb{Z}$ and $I=2mathbb{Z}$. Then $0+2mathbb{Z}=2+2mathbb{Z}$. So what is $pi(0+2mathbb{Z})=pi(2+2mathbb{Z})$? Is it $0$ or $2$?



      Of course you can arbitrarly choose for example $pi(0+2mathbb{Z}):= 0$ (and therefore $pi(2+2mathbb{Z})= 0$) and so on for every coset but then you will find out that $pi$ is no longer a homomorphism. Indeed, let's simplify notation, I will use $[x]$ instead of $x+I$. In our case $R/I={[0],[1]}$. Note that we have $[1]+[1]=[0]$. Put for example $pi([0])=0$ and $pi([1])=1$. Then



      $$pi([1]+[1])=pi([0])=0$$
      $$pi([1])+pi([1])=1+1=2$$



      and so $pi$ is not a homomorphism. And in this particular case there is no way to fix that. The only choice of values making $phi$ a homomorphism is $pi([x])=0$. Not to mention that $R/I$ has two elements while $R$ is infinite so there's no chance for $pi$ to be surjective.



      So as you can see there are no shortcuts. Also the existence of an isomorphism $R/Isimeq R$ for $Ineq 0$ is a very rare situation. First of all if $R$ is nontrivial then $R/R$ is trivial. If you are looking for an example of proper ideal then consider $R$ which is not simple (so it has a proper, nontrivial ideal). Then there's always a proper quotient $R/I$ not isomorphic to $R$, namely for any maximal ideal $I$.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        So after writing out a lot of stuff, I just came to the conclusion that maps going from the quotient $R/I$ to $R$ usually come with a lot of problems as opposed to defining $R to R/I$.
        $endgroup$
        – Hawk
        Dec 21 '18 at 12:30










      • $begingroup$
        @Hawk yes, the situation is not symmetric at all.
        $endgroup$
        – freakish
        Dec 21 '18 at 12:31










      • $begingroup$
        No what i mean is that if I had defined it $pi: R to R/I$, it would be easier to prove it is an isomorphism and to identify the problems, which in this case is well-definedness.
        $endgroup$
        – Hawk
        Dec 21 '18 at 12:33














      2












      2








      2





      $begingroup$

      Your $pi$ is not a function in general. For example consider $R=mathbb{Z}$ and $I=2mathbb{Z}$. Then $0+2mathbb{Z}=2+2mathbb{Z}$. So what is $pi(0+2mathbb{Z})=pi(2+2mathbb{Z})$? Is it $0$ or $2$?



      Of course you can arbitrarly choose for example $pi(0+2mathbb{Z}):= 0$ (and therefore $pi(2+2mathbb{Z})= 0$) and so on for every coset but then you will find out that $pi$ is no longer a homomorphism. Indeed, let's simplify notation, I will use $[x]$ instead of $x+I$. In our case $R/I={[0],[1]}$. Note that we have $[1]+[1]=[0]$. Put for example $pi([0])=0$ and $pi([1])=1$. Then



      $$pi([1]+[1])=pi([0])=0$$
      $$pi([1])+pi([1])=1+1=2$$



      and so $pi$ is not a homomorphism. And in this particular case there is no way to fix that. The only choice of values making $phi$ a homomorphism is $pi([x])=0$. Not to mention that $R/I$ has two elements while $R$ is infinite so there's no chance for $pi$ to be surjective.



      So as you can see there are no shortcuts. Also the existence of an isomorphism $R/Isimeq R$ for $Ineq 0$ is a very rare situation. First of all if $R$ is nontrivial then $R/R$ is trivial. If you are looking for an example of proper ideal then consider $R$ which is not simple (so it has a proper, nontrivial ideal). Then there's always a proper quotient $R/I$ not isomorphic to $R$, namely for any maximal ideal $I$.






      share|cite|improve this answer











      $endgroup$



      Your $pi$ is not a function in general. For example consider $R=mathbb{Z}$ and $I=2mathbb{Z}$. Then $0+2mathbb{Z}=2+2mathbb{Z}$. So what is $pi(0+2mathbb{Z})=pi(2+2mathbb{Z})$? Is it $0$ or $2$?



      Of course you can arbitrarly choose for example $pi(0+2mathbb{Z}):= 0$ (and therefore $pi(2+2mathbb{Z})= 0$) and so on for every coset but then you will find out that $pi$ is no longer a homomorphism. Indeed, let's simplify notation, I will use $[x]$ instead of $x+I$. In our case $R/I={[0],[1]}$. Note that we have $[1]+[1]=[0]$. Put for example $pi([0])=0$ and $pi([1])=1$. Then



      $$pi([1]+[1])=pi([0])=0$$
      $$pi([1])+pi([1])=1+1=2$$



      and so $pi$ is not a homomorphism. And in this particular case there is no way to fix that. The only choice of values making $phi$ a homomorphism is $pi([x])=0$. Not to mention that $R/I$ has two elements while $R$ is infinite so there's no chance for $pi$ to be surjective.



      So as you can see there are no shortcuts. Also the existence of an isomorphism $R/Isimeq R$ for $Ineq 0$ is a very rare situation. First of all if $R$ is nontrivial then $R/R$ is trivial. If you are looking for an example of proper ideal then consider $R$ which is not simple (so it has a proper, nontrivial ideal). Then there's always a proper quotient $R/I$ not isomorphic to $R$, namely for any maximal ideal $I$.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Dec 21 '18 at 16:03

























      answered Dec 21 '18 at 11:09









      freakishfreakish

      12.9k1630




      12.9k1630












      • $begingroup$
        So after writing out a lot of stuff, I just came to the conclusion that maps going from the quotient $R/I$ to $R$ usually come with a lot of problems as opposed to defining $R to R/I$.
        $endgroup$
        – Hawk
        Dec 21 '18 at 12:30










      • $begingroup$
        @Hawk yes, the situation is not symmetric at all.
        $endgroup$
        – freakish
        Dec 21 '18 at 12:31










      • $begingroup$
        No what i mean is that if I had defined it $pi: R to R/I$, it would be easier to prove it is an isomorphism and to identify the problems, which in this case is well-definedness.
        $endgroup$
        – Hawk
        Dec 21 '18 at 12:33


















      • $begingroup$
        So after writing out a lot of stuff, I just came to the conclusion that maps going from the quotient $R/I$ to $R$ usually come with a lot of problems as opposed to defining $R to R/I$.
        $endgroup$
        – Hawk
        Dec 21 '18 at 12:30










      • $begingroup$
        @Hawk yes, the situation is not symmetric at all.
        $endgroup$
        – freakish
        Dec 21 '18 at 12:31










      • $begingroup$
        No what i mean is that if I had defined it $pi: R to R/I$, it would be easier to prove it is an isomorphism and to identify the problems, which in this case is well-definedness.
        $endgroup$
        – Hawk
        Dec 21 '18 at 12:33
















      $begingroup$
      So after writing out a lot of stuff, I just came to the conclusion that maps going from the quotient $R/I$ to $R$ usually come with a lot of problems as opposed to defining $R to R/I$.
      $endgroup$
      – Hawk
      Dec 21 '18 at 12:30




      $begingroup$
      So after writing out a lot of stuff, I just came to the conclusion that maps going from the quotient $R/I$ to $R$ usually come with a lot of problems as opposed to defining $R to R/I$.
      $endgroup$
      – Hawk
      Dec 21 '18 at 12:30












      $begingroup$
      @Hawk yes, the situation is not symmetric at all.
      $endgroup$
      – freakish
      Dec 21 '18 at 12:31




      $begingroup$
      @Hawk yes, the situation is not symmetric at all.
      $endgroup$
      – freakish
      Dec 21 '18 at 12:31












      $begingroup$
      No what i mean is that if I had defined it $pi: R to R/I$, it would be easier to prove it is an isomorphism and to identify the problems, which in this case is well-definedness.
      $endgroup$
      – Hawk
      Dec 21 '18 at 12:33




      $begingroup$
      No what i mean is that if I had defined it $pi: R to R/I$, it would be easier to prove it is an isomorphism and to identify the problems, which in this case is well-definedness.
      $endgroup$
      – Hawk
      Dec 21 '18 at 12:33











      1












      $begingroup$

      Your function $pi$ is only well-defined if $I$ is the ideal ${0}$. E.g., with $R = Bbb{Z}$ and $I = 2Bbb{Z}$, we have $1 + I = 3 + I$, so if $pi(a + I) = a $, we would have to have $1 = 3$, which holds in $R/I$, but does not hold in $R$.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Your function $pi$ is only well-defined if $I$ is the ideal ${0}$. E.g., with $R = Bbb{Z}$ and $I = 2Bbb{Z}$, we have $1 + I = 3 + I$, so if $pi(a + I) = a $, we would have to have $1 = 3$, which holds in $R/I$, but does not hold in $R$.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Your function $pi$ is only well-defined if $I$ is the ideal ${0}$. E.g., with $R = Bbb{Z}$ and $I = 2Bbb{Z}$, we have $1 + I = 3 + I$, so if $pi(a + I) = a $, we would have to have $1 = 3$, which holds in $R/I$, but does not hold in $R$.






          share|cite|improve this answer









          $endgroup$



          Your function $pi$ is only well-defined if $I$ is the ideal ${0}$. E.g., with $R = Bbb{Z}$ and $I = 2Bbb{Z}$, we have $1 + I = 3 + I$, so if $pi(a + I) = a $, we would have to have $1 = 3$, which holds in $R/I$, but does not hold in $R$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 21 '18 at 11:10









          Rob ArthanRob Arthan

          29.6k42967




          29.6k42967























              0












              $begingroup$

              For each $ain R$ there are several $bin R$ such that $a+I=b+I$ (unless $I={0}$ in the first place). So it is unclear what $a$ you are referring to.






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                For each $ain R$ there are several $bin R$ such that $a+I=b+I$ (unless $I={0}$ in the first place). So it is unclear what $a$ you are referring to.






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  For each $ain R$ there are several $bin R$ such that $a+I=b+I$ (unless $I={0}$ in the first place). So it is unclear what $a$ you are referring to.






                  share|cite|improve this answer











                  $endgroup$



                  For each $ain R$ there are several $bin R$ such that $a+I=b+I$ (unless $I={0}$ in the first place). So it is unclear what $a$ you are referring to.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 21 '18 at 11:14

























                  answered Dec 21 '18 at 11:09









                  Saucy O'PathSaucy O'Path

                  6,5751627




                  6,5751627






























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