Finding units and zero divisors in a polynomial quotient ring












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I am trying to study for an exam and I am not sure of this solution my professor posted to an exercise. I am given the polynomial quotient ring $mathbb{Z}_6/(x^2+2x)$ and have to find all units and zero divisors.



My initial idea was to do a multiplication table, but with a group this big I am sure there must be another way. For units, I was thinking all polynomials that are coprime to $x$ and $x+2$, but this isn’t what the solutions have.



He lists the units as $1, 5, x+1, 2x+5, 3x+1, 3x+5, 4x+1, 5x+5$, but gives no reason why. I understand these are all coprime to the ideal, but why isn’t $x+3$ an ideal for example?



He then says the zero divisors are everything that isn’t a unit or zero. Why is this? I thought it was possible for an element to be neither a unit nor a zero divisor.



Thank you very much.






abstract-algebra polynomial-rings







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    $begingroup$


    I am trying to study for an exam and I am not sure of this solution my professor posted to an exercise. I am given the polynomial quotient ring $mathbb{Z}_6/(x^2+2x)$ and have to find all units and zero divisors.



    My initial idea was to do a multiplication table, but with a group this big I am sure there must be another way. For units, I was thinking all polynomials that are coprime to $x$ and $x+2$, but this isn’t what the solutions have.



    He lists the units as $1, 5, x+1, 2x+5, 3x+1, 3x+5, 4x+1, 5x+5$, but gives no reason why. I understand these are all coprime to the ideal, but why isn’t $x+3$ an ideal for example?



    He then says the zero divisors are everything that isn’t a unit or zero. Why is this? I thought it was possible for an element to be neither a unit nor a zero divisor.



    Thank you very much.






    abstract-algebra polynomial-rings







    share|cite|improve this question









    $endgroup$















      0












      0








      0


      1



      $begingroup$


      I am trying to study for an exam and I am not sure of this solution my professor posted to an exercise. I am given the polynomial quotient ring $mathbb{Z}_6/(x^2+2x)$ and have to find all units and zero divisors.



      My initial idea was to do a multiplication table, but with a group this big I am sure there must be another way. For units, I was thinking all polynomials that are coprime to $x$ and $x+2$, but this isn’t what the solutions have.



      He lists the units as $1, 5, x+1, 2x+5, 3x+1, 3x+5, 4x+1, 5x+5$, but gives no reason why. I understand these are all coprime to the ideal, but why isn’t $x+3$ an ideal for example?



      He then says the zero divisors are everything that isn’t a unit or zero. Why is this? I thought it was possible for an element to be neither a unit nor a zero divisor.



      Thank you very much.






      abstract-algebra polynomial-rings







      share|cite|improve this question









      $endgroup$




      I am trying to study for an exam and I am not sure of this solution my professor posted to an exercise. I am given the polynomial quotient ring $mathbb{Z}_6/(x^2+2x)$ and have to find all units and zero divisors.



      My initial idea was to do a multiplication table, but with a group this big I am sure there must be another way. For units, I was thinking all polynomials that are coprime to $x$ and $x+2$, but this isn’t what the solutions have.



      He lists the units as $1, 5, x+1, 2x+5, 3x+1, 3x+5, 4x+1, 5x+5$, but gives no reason why. I understand these are all coprime to the ideal, but why isn’t $x+3$ an ideal for example?



      He then says the zero divisors are everything that isn’t a unit or zero. Why is this? I thought it was possible for an element to be neither a unit nor a zero divisor.



      Thank you very much.






      abstract-algebra polynomial-rings




      abstract-algebra polynomial-rings






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      asked Dec 12 '18 at 1:34









      Tyler6Tyler6

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          2 Answers
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          $begingroup$

          Hint: first we can think of elements of $mathbb{Z}_6[x]/(x^2+2x)$ is of the form $a+bx$ where $a,bin mathbb{Z}_6$.



          Now $a+bx$ is a unit if and only if there is $c+dx$ such that $(a+bx)(c+dx)=1$.
          Note that since $x^2=-2x$ in the ring, we have
          begin{align}1=(a+bx)(c+dx)&=ac+(ad+bc)x+bdx^2\&=ac+(ad+bc)x+bd(-2x)\
          &=ac+(ad+bc-2bd)x
          end{align}



          Then you need to solve system of equation $ac=1$ and $ad+bc-2bd=0$.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            First, note that this quotient ring is finite (since $x^2+2x$ is monic). In fact it must have $36$ elements.



            Now in a finite (commutative) ring $R$, every element is either a unit, zero or a zero-divisor.



            Why? Well, for any nonzero element $x$ in $R$, consider the sequence $x^n$ for $nin Bbb{N}$. By pigeonhole $x^n=x^m$ for some $n,m$ with $n<m$. Then $x^nx^{m-n}=x^n$, or $x^n(x^{m-n}-1)=0$. There are two possibilities. Either $x^{m-n}-1=0$ and $x^{m-n}=1$, so $x$ is a unit, or $x^{m-n}-1ne 0$. In this case, let $a$ be the least integer such that $x^a(x^{m-n}-1)=0$. Then $c=x^{a-1}(x^{m-n}-1)ne 0$ but $xc=0$.



            For the specifics of finding the actual units, user9077 has already given a good answer, so I won't duplicate that.






            share|cite|improve this answer









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              2 Answers
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              $begingroup$

              Hint: first we can think of elements of $mathbb{Z}_6[x]/(x^2+2x)$ is of the form $a+bx$ where $a,bin mathbb{Z}_6$.



              Now $a+bx$ is a unit if and only if there is $c+dx$ such that $(a+bx)(c+dx)=1$.
              Note that since $x^2=-2x$ in the ring, we have
              begin{align}1=(a+bx)(c+dx)&=ac+(ad+bc)x+bdx^2\&=ac+(ad+bc)x+bd(-2x)\
              &=ac+(ad+bc-2bd)x
              end{align}



              Then you need to solve system of equation $ac=1$ and $ad+bc-2bd=0$.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Hint: first we can think of elements of $mathbb{Z}_6[x]/(x^2+2x)$ is of the form $a+bx$ where $a,bin mathbb{Z}_6$.



                Now $a+bx$ is a unit if and only if there is $c+dx$ such that $(a+bx)(c+dx)=1$.
                Note that since $x^2=-2x$ in the ring, we have
                begin{align}1=(a+bx)(c+dx)&=ac+(ad+bc)x+bdx^2\&=ac+(ad+bc)x+bd(-2x)\
                &=ac+(ad+bc-2bd)x
                end{align}



                Then you need to solve system of equation $ac=1$ and $ad+bc-2bd=0$.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Hint: first we can think of elements of $mathbb{Z}_6[x]/(x^2+2x)$ is of the form $a+bx$ where $a,bin mathbb{Z}_6$.



                  Now $a+bx$ is a unit if and only if there is $c+dx$ such that $(a+bx)(c+dx)=1$.
                  Note that since $x^2=-2x$ in the ring, we have
                  begin{align}1=(a+bx)(c+dx)&=ac+(ad+bc)x+bdx^2\&=ac+(ad+bc)x+bd(-2x)\
                  &=ac+(ad+bc-2bd)x
                  end{align}



                  Then you need to solve system of equation $ac=1$ and $ad+bc-2bd=0$.






                  share|cite|improve this answer









                  $endgroup$



                  Hint: first we can think of elements of $mathbb{Z}_6[x]/(x^2+2x)$ is of the form $a+bx$ where $a,bin mathbb{Z}_6$.



                  Now $a+bx$ is a unit if and only if there is $c+dx$ such that $(a+bx)(c+dx)=1$.
                  Note that since $x^2=-2x$ in the ring, we have
                  begin{align}1=(a+bx)(c+dx)&=ac+(ad+bc)x+bdx^2\&=ac+(ad+bc)x+bd(-2x)\
                  &=ac+(ad+bc-2bd)x
                  end{align}



                  Then you need to solve system of equation $ac=1$ and $ad+bc-2bd=0$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 12 '18 at 1:41









                  user9077user9077

                  1,249612




                  1,249612























                      1












                      $begingroup$

                      First, note that this quotient ring is finite (since $x^2+2x$ is monic). In fact it must have $36$ elements.



                      Now in a finite (commutative) ring $R$, every element is either a unit, zero or a zero-divisor.



                      Why? Well, for any nonzero element $x$ in $R$, consider the sequence $x^n$ for $nin Bbb{N}$. By pigeonhole $x^n=x^m$ for some $n,m$ with $n<m$. Then $x^nx^{m-n}=x^n$, or $x^n(x^{m-n}-1)=0$. There are two possibilities. Either $x^{m-n}-1=0$ and $x^{m-n}=1$, so $x$ is a unit, or $x^{m-n}-1ne 0$. In this case, let $a$ be the least integer such that $x^a(x^{m-n}-1)=0$. Then $c=x^{a-1}(x^{m-n}-1)ne 0$ but $xc=0$.



                      For the specifics of finding the actual units, user9077 has already given a good answer, so I won't duplicate that.






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        First, note that this quotient ring is finite (since $x^2+2x$ is monic). In fact it must have $36$ elements.



                        Now in a finite (commutative) ring $R$, every element is either a unit, zero or a zero-divisor.



                        Why? Well, for any nonzero element $x$ in $R$, consider the sequence $x^n$ for $nin Bbb{N}$. By pigeonhole $x^n=x^m$ for some $n,m$ with $n<m$. Then $x^nx^{m-n}=x^n$, or $x^n(x^{m-n}-1)=0$. There are two possibilities. Either $x^{m-n}-1=0$ and $x^{m-n}=1$, so $x$ is a unit, or $x^{m-n}-1ne 0$. In this case, let $a$ be the least integer such that $x^a(x^{m-n}-1)=0$. Then $c=x^{a-1}(x^{m-n}-1)ne 0$ but $xc=0$.



                        For the specifics of finding the actual units, user9077 has already given a good answer, so I won't duplicate that.






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          First, note that this quotient ring is finite (since $x^2+2x$ is monic). In fact it must have $36$ elements.



                          Now in a finite (commutative) ring $R$, every element is either a unit, zero or a zero-divisor.



                          Why? Well, for any nonzero element $x$ in $R$, consider the sequence $x^n$ for $nin Bbb{N}$. By pigeonhole $x^n=x^m$ for some $n,m$ with $n<m$. Then $x^nx^{m-n}=x^n$, or $x^n(x^{m-n}-1)=0$. There are two possibilities. Either $x^{m-n}-1=0$ and $x^{m-n}=1$, so $x$ is a unit, or $x^{m-n}-1ne 0$. In this case, let $a$ be the least integer such that $x^a(x^{m-n}-1)=0$. Then $c=x^{a-1}(x^{m-n}-1)ne 0$ but $xc=0$.



                          For the specifics of finding the actual units, user9077 has already given a good answer, so I won't duplicate that.






                          share|cite|improve this answer









                          $endgroup$



                          First, note that this quotient ring is finite (since $x^2+2x$ is monic). In fact it must have $36$ elements.



                          Now in a finite (commutative) ring $R$, every element is either a unit, zero or a zero-divisor.



                          Why? Well, for any nonzero element $x$ in $R$, consider the sequence $x^n$ for $nin Bbb{N}$. By pigeonhole $x^n=x^m$ for some $n,m$ with $n<m$. Then $x^nx^{m-n}=x^n$, or $x^n(x^{m-n}-1)=0$. There are two possibilities. Either $x^{m-n}-1=0$ and $x^{m-n}=1$, so $x$ is a unit, or $x^{m-n}-1ne 0$. In this case, let $a$ be the least integer such that $x^a(x^{m-n}-1)=0$. Then $c=x^{a-1}(x^{m-n}-1)ne 0$ but $xc=0$.



                          For the specifics of finding the actual units, user9077 has already given a good answer, so I won't duplicate that.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 12 '18 at 1:46









                          jgonjgon

                          16.3k32143




                          16.3k32143






























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