Number of elements in a factor ring












3












$begingroup$


I am presented with $f(x) = 2x^3 + 3x^2 + 1$ $in mathbb{Z_5}[x]$ and need to explain why $F = frac{mathbb{Z_5}[x]}{f(x)}$ is a field and also find how many elements are in F.



So far I have shown that $f(x)$ is an irreducible polynomial and I also know that,




If $f(x)$ is an irreducible polynomial in $mathbb{Z_5}[x]$, then the factor ring $frac{mathbb{Z_5}[x]}{f(x)}$ is also a field.




Basically I am not sure how to properly find the factor ring F and also determine how many elements are in it.



Thanks in advance






abstract-algebra







share|cite|improve this question









$endgroup$

















    3












    $begingroup$


    I am presented with $f(x) = 2x^3 + 3x^2 + 1$ $in mathbb{Z_5}[x]$ and need to explain why $F = frac{mathbb{Z_5}[x]}{f(x)}$ is a field and also find how many elements are in F.



    So far I have shown that $f(x)$ is an irreducible polynomial and I also know that,




    If $f(x)$ is an irreducible polynomial in $mathbb{Z_5}[x]$, then the factor ring $frac{mathbb{Z_5}[x]}{f(x)}$ is also a field.




    Basically I am not sure how to properly find the factor ring F and also determine how many elements are in it.



    Thanks in advance






    abstract-algebra







    share|cite|improve this question









    $endgroup$















      3












      3








      3





      $begingroup$


      I am presented with $f(x) = 2x^3 + 3x^2 + 1$ $in mathbb{Z_5}[x]$ and need to explain why $F = frac{mathbb{Z_5}[x]}{f(x)}$ is a field and also find how many elements are in F.



      So far I have shown that $f(x)$ is an irreducible polynomial and I also know that,




      If $f(x)$ is an irreducible polynomial in $mathbb{Z_5}[x]$, then the factor ring $frac{mathbb{Z_5}[x]}{f(x)}$ is also a field.




      Basically I am not sure how to properly find the factor ring F and also determine how many elements are in it.



      Thanks in advance






      abstract-algebra







      share|cite|improve this question









      $endgroup$




      I am presented with $f(x) = 2x^3 + 3x^2 + 1$ $in mathbb{Z_5}[x]$ and need to explain why $F = frac{mathbb{Z_5}[x]}{f(x)}$ is a field and also find how many elements are in F.



      So far I have shown that $f(x)$ is an irreducible polynomial and I also know that,




      If $f(x)$ is an irreducible polynomial in $mathbb{Z_5}[x]$, then the factor ring $frac{mathbb{Z_5}[x]}{f(x)}$ is also a field.




      Basically I am not sure how to properly find the factor ring F and also determine how many elements are in it.



      Thanks in advance






      abstract-algebra




      abstract-algebra






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked yesterday









      MathsRookieMathsRookie

      1237




      1237






















          4 Answers
          4






          active

          oldest

          votes


















          3












          $begingroup$

          Well, in general, if a field $L$ is an extension field of some field $K$, then $L$ is also a $K$-vector space.



          If $f$ is an irreducible polynomial of degree $n$ over, say, ${Bbb Z}_p$, then
          the quotient field ${Bbb Z}_p[x]/langle frangle$ has $p^n$ elements and is a vector space over ${Bbb Z}_p$ of dimension $n$.






          share|cite|improve this answer











          $endgroup$





















            2












            $begingroup$

            Hint: Use the division algorithm to find canonical representatives for each residue class $g(x) + (f(x))$.






            share|cite|improve this answer









            $endgroup$





















              0












              $begingroup$

              Outline:



              You know $Bbb Z_5[x]/I$ where $I=langle f(x) rangle $ is a field using the mentioned result. Now, the task is to find $|Bbb Z_5[x]/I |$. Here $Bbb Z_5[x]/I$ can be considered as an entension of $Bbb Z_5$. Also $$dim_Bbb {Z_5} (Bbb Z_5[x]/I)=3=deg f$$



              Call ${v_1,v_2,v_3}$ the basis of $Bbb Z_5[x]/I$ over $Bbb Z_5$



              Now arbitrarily pick $f(x)+I in Bbb Z_5[x]/I$. Then $$f(x)+I=a_1v_1+a_2v_2+a_3v_3;;;a_i in Bbb Z_5$$



              There are $5 times 5 times 5$ choices to choose the coefficients, so $$text{number of such $f(x)+I$ }=5^3=|Bbb Z_5[x]/I|$$






              share|cite|improve this answer









              $endgroup$





















                0












                $begingroup$

                Note that two polynomials $g(x)$ and $h(x)$, both of degree 2 or less, will represent different elements in $F$. This is due to the fact that their difference
                $g(x)-h(x)$ being of degree less than 3, cannot be a multiple of the cubic polynomials $f(x)=2x^3+3x^2+1$, and so belong to different cosets.



                Now a polynomial $h(x)$ of degree 3 or more can be divided by $f(x)$ to get quotient $q(x)$ and remainder $r(x)$
                such that $h(x)=q(x)f(x) + r(x)$



                Clearly $r(x)$ and $h(x)$ will represent the same coset. This shows that the field $F$ in your question simply consists of polynomials of degree less than $3$. This is easy to count as each coefficient has only a finite number of choices.






                share|cite|improve this answer









                $endgroup$














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                  4 Answers
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                  4 Answers
                  4






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                  active

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                  active

                  oldest

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                  3












                  $begingroup$

                  Well, in general, if a field $L$ is an extension field of some field $K$, then $L$ is also a $K$-vector space.



                  If $f$ is an irreducible polynomial of degree $n$ over, say, ${Bbb Z}_p$, then
                  the quotient field ${Bbb Z}_p[x]/langle frangle$ has $p^n$ elements and is a vector space over ${Bbb Z}_p$ of dimension $n$.






                  share|cite|improve this answer











                  $endgroup$


















                    3












                    $begingroup$

                    Well, in general, if a field $L$ is an extension field of some field $K$, then $L$ is also a $K$-vector space.



                    If $f$ is an irreducible polynomial of degree $n$ over, say, ${Bbb Z}_p$, then
                    the quotient field ${Bbb Z}_p[x]/langle frangle$ has $p^n$ elements and is a vector space over ${Bbb Z}_p$ of dimension $n$.






                    share|cite|improve this answer











                    $endgroup$
















                      3












                      3








                      3





                      $begingroup$

                      Well, in general, if a field $L$ is an extension field of some field $K$, then $L$ is also a $K$-vector space.



                      If $f$ is an irreducible polynomial of degree $n$ over, say, ${Bbb Z}_p$, then
                      the quotient field ${Bbb Z}_p[x]/langle frangle$ has $p^n$ elements and is a vector space over ${Bbb Z}_p$ of dimension $n$.






                      share|cite|improve this answer











                      $endgroup$



                      Well, in general, if a field $L$ is an extension field of some field $K$, then $L$ is also a $K$-vector space.



                      If $f$ is an irreducible polynomial of degree $n$ over, say, ${Bbb Z}_p$, then
                      the quotient field ${Bbb Z}_p[x]/langle frangle$ has $p^n$ elements and is a vector space over ${Bbb Z}_p$ of dimension $n$.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited yesterday

























                      answered yesterday









                      WuestenfuxWuestenfux

                      5,4841513




                      5,4841513























                          2












                          $begingroup$

                          Hint: Use the division algorithm to find canonical representatives for each residue class $g(x) + (f(x))$.






                          share|cite|improve this answer









                          $endgroup$


















                            2












                            $begingroup$

                            Hint: Use the division algorithm to find canonical representatives for each residue class $g(x) + (f(x))$.






                            share|cite|improve this answer









                            $endgroup$
















                              2












                              2








                              2





                              $begingroup$

                              Hint: Use the division algorithm to find canonical representatives for each residue class $g(x) + (f(x))$.






                              share|cite|improve this answer









                              $endgroup$



                              Hint: Use the division algorithm to find canonical representatives for each residue class $g(x) + (f(x))$.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered yesterday









                              lhflhf

                              167k11172404




                              167k11172404























                                  0












                                  $begingroup$

                                  Outline:



                                  You know $Bbb Z_5[x]/I$ where $I=langle f(x) rangle $ is a field using the mentioned result. Now, the task is to find $|Bbb Z_5[x]/I |$. Here $Bbb Z_5[x]/I$ can be considered as an entension of $Bbb Z_5$. Also $$dim_Bbb {Z_5} (Bbb Z_5[x]/I)=3=deg f$$



                                  Call ${v_1,v_2,v_3}$ the basis of $Bbb Z_5[x]/I$ over $Bbb Z_5$



                                  Now arbitrarily pick $f(x)+I in Bbb Z_5[x]/I$. Then $$f(x)+I=a_1v_1+a_2v_2+a_3v_3;;;a_i in Bbb Z_5$$



                                  There are $5 times 5 times 5$ choices to choose the coefficients, so $$text{number of such $f(x)+I$ }=5^3=|Bbb Z_5[x]/I|$$






                                  share|cite|improve this answer









                                  $endgroup$


















                                    0












                                    $begingroup$

                                    Outline:



                                    You know $Bbb Z_5[x]/I$ where $I=langle f(x) rangle $ is a field using the mentioned result. Now, the task is to find $|Bbb Z_5[x]/I |$. Here $Bbb Z_5[x]/I$ can be considered as an entension of $Bbb Z_5$. Also $$dim_Bbb {Z_5} (Bbb Z_5[x]/I)=3=deg f$$



                                    Call ${v_1,v_2,v_3}$ the basis of $Bbb Z_5[x]/I$ over $Bbb Z_5$



                                    Now arbitrarily pick $f(x)+I in Bbb Z_5[x]/I$. Then $$f(x)+I=a_1v_1+a_2v_2+a_3v_3;;;a_i in Bbb Z_5$$



                                    There are $5 times 5 times 5$ choices to choose the coefficients, so $$text{number of such $f(x)+I$ }=5^3=|Bbb Z_5[x]/I|$$






                                    share|cite|improve this answer









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      Outline:



                                      You know $Bbb Z_5[x]/I$ where $I=langle f(x) rangle $ is a field using the mentioned result. Now, the task is to find $|Bbb Z_5[x]/I |$. Here $Bbb Z_5[x]/I$ can be considered as an entension of $Bbb Z_5$. Also $$dim_Bbb {Z_5} (Bbb Z_5[x]/I)=3=deg f$$



                                      Call ${v_1,v_2,v_3}$ the basis of $Bbb Z_5[x]/I$ over $Bbb Z_5$



                                      Now arbitrarily pick $f(x)+I in Bbb Z_5[x]/I$. Then $$f(x)+I=a_1v_1+a_2v_2+a_3v_3;;;a_i in Bbb Z_5$$



                                      There are $5 times 5 times 5$ choices to choose the coefficients, so $$text{number of such $f(x)+I$ }=5^3=|Bbb Z_5[x]/I|$$






                                      share|cite|improve this answer









                                      $endgroup$



                                      Outline:



                                      You know $Bbb Z_5[x]/I$ where $I=langle f(x) rangle $ is a field using the mentioned result. Now, the task is to find $|Bbb Z_5[x]/I |$. Here $Bbb Z_5[x]/I$ can be considered as an entension of $Bbb Z_5$. Also $$dim_Bbb {Z_5} (Bbb Z_5[x]/I)=3=deg f$$



                                      Call ${v_1,v_2,v_3}$ the basis of $Bbb Z_5[x]/I$ over $Bbb Z_5$



                                      Now arbitrarily pick $f(x)+I in Bbb Z_5[x]/I$. Then $$f(x)+I=a_1v_1+a_2v_2+a_3v_3;;;a_i in Bbb Z_5$$



                                      There are $5 times 5 times 5$ choices to choose the coefficients, so $$text{number of such $f(x)+I$ }=5^3=|Bbb Z_5[x]/I|$$







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered yesterday









                                      Chinnapparaj RChinnapparaj R

                                      6,1612929




                                      6,1612929























                                          0












                                          $begingroup$

                                          Note that two polynomials $g(x)$ and $h(x)$, both of degree 2 or less, will represent different elements in $F$. This is due to the fact that their difference
                                          $g(x)-h(x)$ being of degree less than 3, cannot be a multiple of the cubic polynomials $f(x)=2x^3+3x^2+1$, and so belong to different cosets.



                                          Now a polynomial $h(x)$ of degree 3 or more can be divided by $f(x)$ to get quotient $q(x)$ and remainder $r(x)$
                                          such that $h(x)=q(x)f(x) + r(x)$



                                          Clearly $r(x)$ and $h(x)$ will represent the same coset. This shows that the field $F$ in your question simply consists of polynomials of degree less than $3$. This is easy to count as each coefficient has only a finite number of choices.






                                          share|cite|improve this answer









                                          $endgroup$


















                                            0












                                            $begingroup$

                                            Note that two polynomials $g(x)$ and $h(x)$, both of degree 2 or less, will represent different elements in $F$. This is due to the fact that their difference
                                            $g(x)-h(x)$ being of degree less than 3, cannot be a multiple of the cubic polynomials $f(x)=2x^3+3x^2+1$, and so belong to different cosets.



                                            Now a polynomial $h(x)$ of degree 3 or more can be divided by $f(x)$ to get quotient $q(x)$ and remainder $r(x)$
                                            such that $h(x)=q(x)f(x) + r(x)$



                                            Clearly $r(x)$ and $h(x)$ will represent the same coset. This shows that the field $F$ in your question simply consists of polynomials of degree less than $3$. This is easy to count as each coefficient has only a finite number of choices.






                                            share|cite|improve this answer









                                            $endgroup$
















                                              0












                                              0








                                              0





                                              $begingroup$

                                              Note that two polynomials $g(x)$ and $h(x)$, both of degree 2 or less, will represent different elements in $F$. This is due to the fact that their difference
                                              $g(x)-h(x)$ being of degree less than 3, cannot be a multiple of the cubic polynomials $f(x)=2x^3+3x^2+1$, and so belong to different cosets.



                                              Now a polynomial $h(x)$ of degree 3 or more can be divided by $f(x)$ to get quotient $q(x)$ and remainder $r(x)$
                                              such that $h(x)=q(x)f(x) + r(x)$



                                              Clearly $r(x)$ and $h(x)$ will represent the same coset. This shows that the field $F$ in your question simply consists of polynomials of degree less than $3$. This is easy to count as each coefficient has only a finite number of choices.






                                              share|cite|improve this answer









                                              $endgroup$



                                              Note that two polynomials $g(x)$ and $h(x)$, both of degree 2 or less, will represent different elements in $F$. This is due to the fact that their difference
                                              $g(x)-h(x)$ being of degree less than 3, cannot be a multiple of the cubic polynomials $f(x)=2x^3+3x^2+1$, and so belong to different cosets.



                                              Now a polynomial $h(x)$ of degree 3 or more can be divided by $f(x)$ to get quotient $q(x)$ and remainder $r(x)$
                                              such that $h(x)=q(x)f(x) + r(x)$



                                              Clearly $r(x)$ and $h(x)$ will represent the same coset. This shows that the field $F$ in your question simply consists of polynomials of degree less than $3$. This is easy to count as each coefficient has only a finite number of choices.







                                              share|cite|improve this answer












                                              share|cite|improve this answer



                                              share|cite|improve this answer










                                              answered yesterday









                                              P VanchinathanP Vanchinathan

                                              15.5k12136




                                              15.5k12136






























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Interview With Slayer Guitarist Kerry King””2008-02-23: Wiltern, Los Angeles, CA, USA””Slayer's Kerry King To Perform With Megadeth Tonight! - Oct. 21, 2010”originalet”Dave Lombardo - Biography”Slayer Case DismissedArkiveradUltimate Classic Rock: Slayer guitarist Jeff Hanneman dead at 49.”Slayer: "We could never do any thing like Some Kind Of Monster..."””Cannibal Corpse'S Pat O'Brien Will Step In As Slayer'S Guest Guitarist | The Official Slayer Site”originalet”Slayer Wins 'Best Metal' Grammy Award””Slayer Guitarist Jeff Hanneman Dies””Kerrang! Awards 2006 Blog: Kerrang! Hall Of Fame””Kerrang! Awards 2013: Kerrang! Legend”originalet”Metallica, Slayer, Iron Maien Among Winners At Metal Hammer Awards””Metal Hammer Golden Gods Awards””Bullet For My Valentine Booed At Metal Hammer Golden Gods Awards””Metal Storm Awards 2006””Metal Storm Awards 2015””Slayer's Concert History””Slayer - Relationships””Slayer - Releases”Slayers officiella webbplatsSlayer på MusicBrainzOfficiell webbplatsSlayerSlayerr1373445760000 0001 1540 47353068615-5086262726cb13906545x(data)6033143kn20030215029

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                                                  SlayerMusikgrupper bildade 1981Amerikanska thrash metal-grupper amerikansktthrash metalHuntington Park1981gitarristernaJeff HannemanKerry KingbasistenTom ArayatrummisenDave LombardoGary HoltPaul Bostaphcertifierats med guldMetallicaAnthraxMegadeththrash metal1986extrema metalgenrerdeath metal22 januari2018USAKanadaEuropaHuntington ParkKalifornien1981gitarristernaJeff HannemanKerry KingbassångDave Lombardoheavy metalbrittiskaIron MaidenJudas PriestthrashMetallicaRattSteeler1982demoundergroundMetal Blade1983high schoolGene HoglanDeathStrapping Young Ladkassettvinyl1984LondonMarquee Club1984Eldorado StudiosTrack Record StudiosLos Angelesmastrade1985CDCombat Recordsextreme metalVenomExodusStudio 54 club1985VHSCombat RecordsBelgienUFORick RubinDef Jam Recordship hoprockakterThe Black CrowesRun–D.M.C.LL Cool JBeastie BoysthrashbpmAngel of DeathRaining BloodAndy WallaceHowie Weinberg1986bildskivaBillboard 200RIAAMetal HammerKerrang!Angela Gossowdeath metalArch Enemythrash metalOverk...
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