Justification of an isomorphism between $mathbb{Z}[t]/(t,3)$ and $mathbb{Z}/3$











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I am wondering if the following is true:



$$mathbb{Z}[t]/(t,3)congmathbb{Z}/3.$$



I can tell that the ideal $(t,3)$ is maximal. In contrast, the ideal $(t^2,2t)$ is not even irreducible. So it's less likely that we can simplify $mathbb{Z}[t]/(t^2,2t)$ further.



If the aforementioned isomorphism is true, that means $mathbb{Z}[t]/(t,3)cong(mathbb{Z}[t]/t)/(3)congmathbb{Z}/3$, where the $(3)$ is an ideal in $mathbb{Z}[t]/t$. My questions are:



1) Is this argument legitimate?



2) If it is true, is there any theorem justifying the aforementioned isomorphism?










share|cite|improve this question




























    up vote
    2
    down vote

    favorite
    1












    I am wondering if the following is true:



    $$mathbb{Z}[t]/(t,3)congmathbb{Z}/3.$$



    I can tell that the ideal $(t,3)$ is maximal. In contrast, the ideal $(t^2,2t)$ is not even irreducible. So it's less likely that we can simplify $mathbb{Z}[t]/(t^2,2t)$ further.



    If the aforementioned isomorphism is true, that means $mathbb{Z}[t]/(t,3)cong(mathbb{Z}[t]/t)/(3)congmathbb{Z}/3$, where the $(3)$ is an ideal in $mathbb{Z}[t]/t$. My questions are:



    1) Is this argument legitimate?



    2) If it is true, is there any theorem justifying the aforementioned isomorphism?










    share|cite|improve this question


























      up vote
      2
      down vote

      favorite
      1









      up vote
      2
      down vote

      favorite
      1






      1





      I am wondering if the following is true:



      $$mathbb{Z}[t]/(t,3)congmathbb{Z}/3.$$



      I can tell that the ideal $(t,3)$ is maximal. In contrast, the ideal $(t^2,2t)$ is not even irreducible. So it's less likely that we can simplify $mathbb{Z}[t]/(t^2,2t)$ further.



      If the aforementioned isomorphism is true, that means $mathbb{Z}[t]/(t,3)cong(mathbb{Z}[t]/t)/(3)congmathbb{Z}/3$, where the $(3)$ is an ideal in $mathbb{Z}[t]/t$. My questions are:



      1) Is this argument legitimate?



      2) If it is true, is there any theorem justifying the aforementioned isomorphism?










      share|cite|improve this question















      I am wondering if the following is true:



      $$mathbb{Z}[t]/(t,3)congmathbb{Z}/3.$$



      I can tell that the ideal $(t,3)$ is maximal. In contrast, the ideal $(t^2,2t)$ is not even irreducible. So it's less likely that we can simplify $mathbb{Z}[t]/(t^2,2t)$ further.



      If the aforementioned isomorphism is true, that means $mathbb{Z}[t]/(t,3)cong(mathbb{Z}[t]/t)/(3)congmathbb{Z}/3$, where the $(3)$ is an ideal in $mathbb{Z}[t]/t$. My questions are:



      1) Is this argument legitimate?



      2) If it is true, is there any theorem justifying the aforementioned isomorphism?







      abstract-algebra ring-theory






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 2 at 9:24









      Asaf Karagila

      300k32422751




      300k32422751










      asked Dec 2 at 3:23









      J. Wang

      523




      523






















          2 Answers
          2






          active

          oldest

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          up vote
          3
          down vote



          accepted










          Yes your intuition is correct and I try to explain necessary steps:




          Claim. $Bbb{Z}[x]/langle x,3ranglecongBbb{Z}/langle3rangle$




          Well, I'll prove the following steps:



          $$Bbb{Z}[x]/langle x,3ranglecong frac{Bbb{Z}[x]/langle xrangle}{langle x,3rangle/langle xrangle} congBbb{Z}/langle 3rangle$$




          • The first isomorphism is due to the 2nd Isomorphism Theorem of Ring theory.


          • I think $Bbb{Z}[x]/langle xrangle cong Bbb{Z}$ is clear. (Hint: consider the homomorphism $f(x) mapsto f(0)$ form $Bbb{Z}[x]$ to $Bbb{Z}$ and use 1st Isomorphism Theorem on Rings.


          • Now for the equality $langle x,3rangle/langle xranglecong langle 3rangle$ Try to define the homomorphism as follows:



          $phi :langle x,3rangle to langle 3rangle$ by mapping any element $xf(x)+3g(x)in langle x,3 rangle$ to $3g(0) in langle 3rangle$ (N.B. Here, $f,g in Bbb{Z}[x]$), i.e. $$phi: xf(x)+3g(x) mapsto 3g(0)$$



          See that this is an epimorphism. Also $ker phi=langle xrangle$. Then by 1st Isomorphism theorem $langle x,3rangle/ker phi cong langle 3rangle$. Which justifies all your steps...!!



          N.B. To find out the $ker phi$, remember for any $g(x)in Bbb{Z}[x]$ if we have $g(0)=0$ that means $x$ is a factor of $g$ and so $gin langle xrangle$.






          share|cite|improve this answer























          • $>$ is an operator; &rangle$ is a grouping symbol. They have different formatting rules, and are the reason for all the weird spaces in your answer.
            – Bob Krueger
            Dec 2 at 8:26










          • @BobKrueger...Thanks for mentioning that....I'll take care of it next time....:)
            – Indrajit Ghosh
            Dec 2 at 8:36












          • What's wrong with using the edit function to fix it this time as well? Is your time somehow more important than that of someone who might have edited this?
            – Asaf Karagila
            Dec 2 at 9:22












          • I’ll note that I didn’t edit it so that the poster would recognize the mistake. @IndrajitGhosh happy to help even if it’s not expressly math.
            – Bob Krueger
            Dec 2 at 15:28


















          up vote
          2
          down vote













          It's legitimate and a consequence of the isomorphism theorems.






          share|cite|improve this answer





















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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            3
            down vote



            accepted










            Yes your intuition is correct and I try to explain necessary steps:




            Claim. $Bbb{Z}[x]/langle x,3ranglecongBbb{Z}/langle3rangle$




            Well, I'll prove the following steps:



            $$Bbb{Z}[x]/langle x,3ranglecong frac{Bbb{Z}[x]/langle xrangle}{langle x,3rangle/langle xrangle} congBbb{Z}/langle 3rangle$$




            • The first isomorphism is due to the 2nd Isomorphism Theorem of Ring theory.


            • I think $Bbb{Z}[x]/langle xrangle cong Bbb{Z}$ is clear. (Hint: consider the homomorphism $f(x) mapsto f(0)$ form $Bbb{Z}[x]$ to $Bbb{Z}$ and use 1st Isomorphism Theorem on Rings.


            • Now for the equality $langle x,3rangle/langle xranglecong langle 3rangle$ Try to define the homomorphism as follows:



            $phi :langle x,3rangle to langle 3rangle$ by mapping any element $xf(x)+3g(x)in langle x,3 rangle$ to $3g(0) in langle 3rangle$ (N.B. Here, $f,g in Bbb{Z}[x]$), i.e. $$phi: xf(x)+3g(x) mapsto 3g(0)$$



            See that this is an epimorphism. Also $ker phi=langle xrangle$. Then by 1st Isomorphism theorem $langle x,3rangle/ker phi cong langle 3rangle$. Which justifies all your steps...!!



            N.B. To find out the $ker phi$, remember for any $g(x)in Bbb{Z}[x]$ if we have $g(0)=0$ that means $x$ is a factor of $g$ and so $gin langle xrangle$.






            share|cite|improve this answer























            • $>$ is an operator; &rangle$ is a grouping symbol. They have different formatting rules, and are the reason for all the weird spaces in your answer.
              – Bob Krueger
              Dec 2 at 8:26










            • @BobKrueger...Thanks for mentioning that....I'll take care of it next time....:)
              – Indrajit Ghosh
              Dec 2 at 8:36












            • What's wrong with using the edit function to fix it this time as well? Is your time somehow more important than that of someone who might have edited this?
              – Asaf Karagila
              Dec 2 at 9:22












            • I’ll note that I didn’t edit it so that the poster would recognize the mistake. @IndrajitGhosh happy to help even if it’s not expressly math.
              – Bob Krueger
              Dec 2 at 15:28















            up vote
            3
            down vote



            accepted










            Yes your intuition is correct and I try to explain necessary steps:




            Claim. $Bbb{Z}[x]/langle x,3ranglecongBbb{Z}/langle3rangle$




            Well, I'll prove the following steps:



            $$Bbb{Z}[x]/langle x,3ranglecong frac{Bbb{Z}[x]/langle xrangle}{langle x,3rangle/langle xrangle} congBbb{Z}/langle 3rangle$$




            • The first isomorphism is due to the 2nd Isomorphism Theorem of Ring theory.


            • I think $Bbb{Z}[x]/langle xrangle cong Bbb{Z}$ is clear. (Hint: consider the homomorphism $f(x) mapsto f(0)$ form $Bbb{Z}[x]$ to $Bbb{Z}$ and use 1st Isomorphism Theorem on Rings.


            • Now for the equality $langle x,3rangle/langle xranglecong langle 3rangle$ Try to define the homomorphism as follows:



            $phi :langle x,3rangle to langle 3rangle$ by mapping any element $xf(x)+3g(x)in langle x,3 rangle$ to $3g(0) in langle 3rangle$ (N.B. Here, $f,g in Bbb{Z}[x]$), i.e. $$phi: xf(x)+3g(x) mapsto 3g(0)$$



            See that this is an epimorphism. Also $ker phi=langle xrangle$. Then by 1st Isomorphism theorem $langle x,3rangle/ker phi cong langle 3rangle$. Which justifies all your steps...!!



            N.B. To find out the $ker phi$, remember for any $g(x)in Bbb{Z}[x]$ if we have $g(0)=0$ that means $x$ is a factor of $g$ and so $gin langle xrangle$.






            share|cite|improve this answer























            • $>$ is an operator; &rangle$ is a grouping symbol. They have different formatting rules, and are the reason for all the weird spaces in your answer.
              – Bob Krueger
              Dec 2 at 8:26










            • @BobKrueger...Thanks for mentioning that....I'll take care of it next time....:)
              – Indrajit Ghosh
              Dec 2 at 8:36












            • What's wrong with using the edit function to fix it this time as well? Is your time somehow more important than that of someone who might have edited this?
              – Asaf Karagila
              Dec 2 at 9:22












            • I’ll note that I didn’t edit it so that the poster would recognize the mistake. @IndrajitGhosh happy to help even if it’s not expressly math.
              – Bob Krueger
              Dec 2 at 15:28













            up vote
            3
            down vote



            accepted







            up vote
            3
            down vote



            accepted






            Yes your intuition is correct and I try to explain necessary steps:




            Claim. $Bbb{Z}[x]/langle x,3ranglecongBbb{Z}/langle3rangle$




            Well, I'll prove the following steps:



            $$Bbb{Z}[x]/langle x,3ranglecong frac{Bbb{Z}[x]/langle xrangle}{langle x,3rangle/langle xrangle} congBbb{Z}/langle 3rangle$$




            • The first isomorphism is due to the 2nd Isomorphism Theorem of Ring theory.


            • I think $Bbb{Z}[x]/langle xrangle cong Bbb{Z}$ is clear. (Hint: consider the homomorphism $f(x) mapsto f(0)$ form $Bbb{Z}[x]$ to $Bbb{Z}$ and use 1st Isomorphism Theorem on Rings.


            • Now for the equality $langle x,3rangle/langle xranglecong langle 3rangle$ Try to define the homomorphism as follows:



            $phi :langle x,3rangle to langle 3rangle$ by mapping any element $xf(x)+3g(x)in langle x,3 rangle$ to $3g(0) in langle 3rangle$ (N.B. Here, $f,g in Bbb{Z}[x]$), i.e. $$phi: xf(x)+3g(x) mapsto 3g(0)$$



            See that this is an epimorphism. Also $ker phi=langle xrangle$. Then by 1st Isomorphism theorem $langle x,3rangle/ker phi cong langle 3rangle$. Which justifies all your steps...!!



            N.B. To find out the $ker phi$, remember for any $g(x)in Bbb{Z}[x]$ if we have $g(0)=0$ that means $x$ is a factor of $g$ and so $gin langle xrangle$.






            share|cite|improve this answer














            Yes your intuition is correct and I try to explain necessary steps:




            Claim. $Bbb{Z}[x]/langle x,3ranglecongBbb{Z}/langle3rangle$




            Well, I'll prove the following steps:



            $$Bbb{Z}[x]/langle x,3ranglecong frac{Bbb{Z}[x]/langle xrangle}{langle x,3rangle/langle xrangle} congBbb{Z}/langle 3rangle$$




            • The first isomorphism is due to the 2nd Isomorphism Theorem of Ring theory.


            • I think $Bbb{Z}[x]/langle xrangle cong Bbb{Z}$ is clear. (Hint: consider the homomorphism $f(x) mapsto f(0)$ form $Bbb{Z}[x]$ to $Bbb{Z}$ and use 1st Isomorphism Theorem on Rings.


            • Now for the equality $langle x,3rangle/langle xranglecong langle 3rangle$ Try to define the homomorphism as follows:



            $phi :langle x,3rangle to langle 3rangle$ by mapping any element $xf(x)+3g(x)in langle x,3 rangle$ to $3g(0) in langle 3rangle$ (N.B. Here, $f,g in Bbb{Z}[x]$), i.e. $$phi: xf(x)+3g(x) mapsto 3g(0)$$



            See that this is an epimorphism. Also $ker phi=langle xrangle$. Then by 1st Isomorphism theorem $langle x,3rangle/ker phi cong langle 3rangle$. Which justifies all your steps...!!



            N.B. To find out the $ker phi$, remember for any $g(x)in Bbb{Z}[x]$ if we have $g(0)=0$ that means $x$ is a factor of $g$ and so $gin langle xrangle$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 3 at 3:57

























            answered Dec 2 at 4:23









            Indrajit Ghosh

            1,0431717




            1,0431717












            • $>$ is an operator; &rangle$ is a grouping symbol. They have different formatting rules, and are the reason for all the weird spaces in your answer.
              – Bob Krueger
              Dec 2 at 8:26










            • @BobKrueger...Thanks for mentioning that....I'll take care of it next time....:)
              – Indrajit Ghosh
              Dec 2 at 8:36












            • What's wrong with using the edit function to fix it this time as well? Is your time somehow more important than that of someone who might have edited this?
              – Asaf Karagila
              Dec 2 at 9:22












            • I’ll note that I didn’t edit it so that the poster would recognize the mistake. @IndrajitGhosh happy to help even if it’s not expressly math.
              – Bob Krueger
              Dec 2 at 15:28


















            • $>$ is an operator; &rangle$ is a grouping symbol. They have different formatting rules, and are the reason for all the weird spaces in your answer.
              – Bob Krueger
              Dec 2 at 8:26










            • @BobKrueger...Thanks for mentioning that....I'll take care of it next time....:)
              – Indrajit Ghosh
              Dec 2 at 8:36












            • What's wrong with using the edit function to fix it this time as well? Is your time somehow more important than that of someone who might have edited this?
              – Asaf Karagila
              Dec 2 at 9:22












            • I’ll note that I didn’t edit it so that the poster would recognize the mistake. @IndrajitGhosh happy to help even if it’s not expressly math.
              – Bob Krueger
              Dec 2 at 15:28
















            $>$ is an operator; &rangle$ is a grouping symbol. They have different formatting rules, and are the reason for all the weird spaces in your answer.
            – Bob Krueger
            Dec 2 at 8:26




            $>$ is an operator; &rangle$ is a grouping symbol. They have different formatting rules, and are the reason for all the weird spaces in your answer.
            – Bob Krueger
            Dec 2 at 8:26












            @BobKrueger...Thanks for mentioning that....I'll take care of it next time....:)
            – Indrajit Ghosh
            Dec 2 at 8:36






            @BobKrueger...Thanks for mentioning that....I'll take care of it next time....:)
            – Indrajit Ghosh
            Dec 2 at 8:36














            What's wrong with using the edit function to fix it this time as well? Is your time somehow more important than that of someone who might have edited this?
            – Asaf Karagila
            Dec 2 at 9:22






            What's wrong with using the edit function to fix it this time as well? Is your time somehow more important than that of someone who might have edited this?
            – Asaf Karagila
            Dec 2 at 9:22














            I’ll note that I didn’t edit it so that the poster would recognize the mistake. @IndrajitGhosh happy to help even if it’s not expressly math.
            – Bob Krueger
            Dec 2 at 15:28




            I’ll note that I didn’t edit it so that the poster would recognize the mistake. @IndrajitGhosh happy to help even if it’s not expressly math.
            – Bob Krueger
            Dec 2 at 15:28










            up vote
            2
            down vote













            It's legitimate and a consequence of the isomorphism theorems.






            share|cite|improve this answer

























              up vote
              2
              down vote













              It's legitimate and a consequence of the isomorphism theorems.






              share|cite|improve this answer























                up vote
                2
                down vote










                up vote
                2
                down vote









                It's legitimate and a consequence of the isomorphism theorems.






                share|cite|improve this answer












                It's legitimate and a consequence of the isomorphism theorems.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 2 at 4:24









                CyclotomicField

                2,1541312




                2,1541312






























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