Show that $mathbb{Z}[x] to mathbb{Z}[x]/(2) times mathbb{Z}[x]/(x)$ is not surjective.












3












$begingroup$


In Aluffi's book it says that $mathbb{Z}[x] to mathbb{Z}[x]/(2) times mathbb{Z}[x]/(x)$ is not surjective. To prove that I need to know how that means at all! Why it is not surjective if for any polynomial $q(x)$ we can write $(q(x)+2n_1, q(x)+n_2x)$?






abstract-algebra







share|cite|improve this question









$endgroup$












  • $begingroup$
    How is the map defined?
    $endgroup$
    – Servaes
    Dec 3 '18 at 7:47






  • 1




    $begingroup$
    @Servaes, natural projection. It's an example to a theorem for $phi$ a natural projection.
    $endgroup$
    – 72D
    Dec 3 '18 at 7:48


















3












$begingroup$


In Aluffi's book it says that $mathbb{Z}[x] to mathbb{Z}[x]/(2) times mathbb{Z}[x]/(x)$ is not surjective. To prove that I need to know how that means at all! Why it is not surjective if for any polynomial $q(x)$ we can write $(q(x)+2n_1, q(x)+n_2x)$?






abstract-algebra







share|cite|improve this question









$endgroup$












  • $begingroup$
    How is the map defined?
    $endgroup$
    – Servaes
    Dec 3 '18 at 7:47






  • 1




    $begingroup$
    @Servaes, natural projection. It's an example to a theorem for $phi$ a natural projection.
    $endgroup$
    – 72D
    Dec 3 '18 at 7:48
















3












3








3





$begingroup$


In Aluffi's book it says that $mathbb{Z}[x] to mathbb{Z}[x]/(2) times mathbb{Z}[x]/(x)$ is not surjective. To prove that I need to know how that means at all! Why it is not surjective if for any polynomial $q(x)$ we can write $(q(x)+2n_1, q(x)+n_2x)$?






abstract-algebra







share|cite|improve this question









$endgroup$




In Aluffi's book it says that $mathbb{Z}[x] to mathbb{Z}[x]/(2) times mathbb{Z}[x]/(x)$ is not surjective. To prove that I need to know how that means at all! Why it is not surjective if for any polynomial $q(x)$ we can write $(q(x)+2n_1, q(x)+n_2x)$?






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 Dec 3 '18 at 7:44









72D72D

512116




512116












  • $begingroup$
    How is the map defined?
    $endgroup$
    – Servaes
    Dec 3 '18 at 7:47






  • 1




    $begingroup$
    @Servaes, natural projection. It's an example to a theorem for $phi$ a natural projection.
    $endgroup$
    – 72D
    Dec 3 '18 at 7:48




















  • $begingroup$
    How is the map defined?
    $endgroup$
    – Servaes
    Dec 3 '18 at 7:47






  • 1




    $begingroup$
    @Servaes, natural projection. It's an example to a theorem for $phi$ a natural projection.
    $endgroup$
    – 72D
    Dec 3 '18 at 7:48


















$begingroup$
How is the map defined?
$endgroup$
– Servaes
Dec 3 '18 at 7:47




$begingroup$
How is the map defined?
$endgroup$
– Servaes
Dec 3 '18 at 7:47




1




1




$begingroup$
@Servaes, natural projection. It's an example to a theorem for $phi$ a natural projection.
$endgroup$
– 72D
Dec 3 '18 at 7:48






$begingroup$
@Servaes, natural projection. It's an example to a theorem for $phi$ a natural projection.
$endgroup$
– 72D
Dec 3 '18 at 7:48












2 Answers
2






active

oldest

votes


















3












$begingroup$

The product on the right hand side is generated by $(1,0)$ and $(0,1)$. So the map is surjective if and only if these are in the image of the map, i.e. if there exist polynomials $f,ginBbb{Z}[x]$ such that
$$f longmapsto (1,0)qquadtext{ and }qquad g longmapsto (0,1).$$
Show that this is impossible.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    How are the typical/general elements of $mathbb{Z}[x]/(2)$ and $mathbb{Z}[x]/(x)$?
    $endgroup$
    – 72D
    Dec 3 '18 at 7:50






  • 1




    $begingroup$
    They are polynomials with integer coefficients mod $2$, and simply integers.
    $endgroup$
    – Servaes
    Dec 3 '18 at 7:51












  • $begingroup$
    I could understand now their meanings but still don't understand the meaning for surjection esp because the rhs is 'double'-element. To try to show that (first) it is surjective (then fail), so we try to find a polynomial such that simultaneously (??) is an integer and a polynomial with integer coefficients mod 2? But it is not intersection it is product.
    $endgroup$
    – 72D
    Dec 3 '18 at 7:59










  • $begingroup$
    I've expanded my hint into a sketch of an answer.
    $endgroup$
    – Servaes
    Dec 3 '18 at 8:05






  • 1




    $begingroup$
    "Which elements of Z[x] map to (1,0)" : so is this a counterexample? : for any polynomial q(x) in Z[x], 2q(x)+1 is the only element of Z[x] such that it is 1 in Z[x]/(2) but 2q(x)+1 is never 0 in Z[x]/(x); am I right?
    $endgroup$
    – 72D
    Dec 3 '18 at 8:05



















0












$begingroup$

Hint $ R to R/I times R/J $ surjective $Rightarrow I + J = 1$



since $,rto (1,0),Rightarrow, 1 = (1!-!r) + rin I + J$






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

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    3












    $begingroup$

    The product on the right hand side is generated by $(1,0)$ and $(0,1)$. So the map is surjective if and only if these are in the image of the map, i.e. if there exist polynomials $f,ginBbb{Z}[x]$ such that
    $$f longmapsto (1,0)qquadtext{ and }qquad g longmapsto (0,1).$$
    Show that this is impossible.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      How are the typical/general elements of $mathbb{Z}[x]/(2)$ and $mathbb{Z}[x]/(x)$?
      $endgroup$
      – 72D
      Dec 3 '18 at 7:50






    • 1




      $begingroup$
      They are polynomials with integer coefficients mod $2$, and simply integers.
      $endgroup$
      – Servaes
      Dec 3 '18 at 7:51












    • $begingroup$
      I could understand now their meanings but still don't understand the meaning for surjection esp because the rhs is 'double'-element. To try to show that (first) it is surjective (then fail), so we try to find a polynomial such that simultaneously (??) is an integer and a polynomial with integer coefficients mod 2? But it is not intersection it is product.
      $endgroup$
      – 72D
      Dec 3 '18 at 7:59










    • $begingroup$
      I've expanded my hint into a sketch of an answer.
      $endgroup$
      – Servaes
      Dec 3 '18 at 8:05






    • 1




      $begingroup$
      "Which elements of Z[x] map to (1,0)" : so is this a counterexample? : for any polynomial q(x) in Z[x], 2q(x)+1 is the only element of Z[x] such that it is 1 in Z[x]/(2) but 2q(x)+1 is never 0 in Z[x]/(x); am I right?
      $endgroup$
      – 72D
      Dec 3 '18 at 8:05
















    3












    $begingroup$

    The product on the right hand side is generated by $(1,0)$ and $(0,1)$. So the map is surjective if and only if these are in the image of the map, i.e. if there exist polynomials $f,ginBbb{Z}[x]$ such that
    $$f longmapsto (1,0)qquadtext{ and }qquad g longmapsto (0,1).$$
    Show that this is impossible.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      How are the typical/general elements of $mathbb{Z}[x]/(2)$ and $mathbb{Z}[x]/(x)$?
      $endgroup$
      – 72D
      Dec 3 '18 at 7:50






    • 1




      $begingroup$
      They are polynomials with integer coefficients mod $2$, and simply integers.
      $endgroup$
      – Servaes
      Dec 3 '18 at 7:51












    • $begingroup$
      I could understand now their meanings but still don't understand the meaning for surjection esp because the rhs is 'double'-element. To try to show that (first) it is surjective (then fail), so we try to find a polynomial such that simultaneously (??) is an integer and a polynomial with integer coefficients mod 2? But it is not intersection it is product.
      $endgroup$
      – 72D
      Dec 3 '18 at 7:59










    • $begingroup$
      I've expanded my hint into a sketch of an answer.
      $endgroup$
      – Servaes
      Dec 3 '18 at 8:05






    • 1




      $begingroup$
      "Which elements of Z[x] map to (1,0)" : so is this a counterexample? : for any polynomial q(x) in Z[x], 2q(x)+1 is the only element of Z[x] such that it is 1 in Z[x]/(2) but 2q(x)+1 is never 0 in Z[x]/(x); am I right?
      $endgroup$
      – 72D
      Dec 3 '18 at 8:05














    3












    3








    3





    $begingroup$

    The product on the right hand side is generated by $(1,0)$ and $(0,1)$. So the map is surjective if and only if these are in the image of the map, i.e. if there exist polynomials $f,ginBbb{Z}[x]$ such that
    $$f longmapsto (1,0)qquadtext{ and }qquad g longmapsto (0,1).$$
    Show that this is impossible.






    share|cite|improve this answer











    $endgroup$



    The product on the right hand side is generated by $(1,0)$ and $(0,1)$. So the map is surjective if and only if these are in the image of the map, i.e. if there exist polynomials $f,ginBbb{Z}[x]$ such that
    $$f longmapsto (1,0)qquadtext{ and }qquad g longmapsto (0,1).$$
    Show that this is impossible.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 3 '18 at 8:05

























    answered Dec 3 '18 at 7:48









    ServaesServaes

    23.4k33893




    23.4k33893












    • $begingroup$
      How are the typical/general elements of $mathbb{Z}[x]/(2)$ and $mathbb{Z}[x]/(x)$?
      $endgroup$
      – 72D
      Dec 3 '18 at 7:50






    • 1




      $begingroup$
      They are polynomials with integer coefficients mod $2$, and simply integers.
      $endgroup$
      – Servaes
      Dec 3 '18 at 7:51












    • $begingroup$
      I could understand now their meanings but still don't understand the meaning for surjection esp because the rhs is 'double'-element. To try to show that (first) it is surjective (then fail), so we try to find a polynomial such that simultaneously (??) is an integer and a polynomial with integer coefficients mod 2? But it is not intersection it is product.
      $endgroup$
      – 72D
      Dec 3 '18 at 7:59










    • $begingroup$
      I've expanded my hint into a sketch of an answer.
      $endgroup$
      – Servaes
      Dec 3 '18 at 8:05






    • 1




      $begingroup$
      "Which elements of Z[x] map to (1,0)" : so is this a counterexample? : for any polynomial q(x) in Z[x], 2q(x)+1 is the only element of Z[x] such that it is 1 in Z[x]/(2) but 2q(x)+1 is never 0 in Z[x]/(x); am I right?
      $endgroup$
      – 72D
      Dec 3 '18 at 8:05


















    • $begingroup$
      How are the typical/general elements of $mathbb{Z}[x]/(2)$ and $mathbb{Z}[x]/(x)$?
      $endgroup$
      – 72D
      Dec 3 '18 at 7:50






    • 1




      $begingroup$
      They are polynomials with integer coefficients mod $2$, and simply integers.
      $endgroup$
      – Servaes
      Dec 3 '18 at 7:51












    • $begingroup$
      I could understand now their meanings but still don't understand the meaning for surjection esp because the rhs is 'double'-element. To try to show that (first) it is surjective (then fail), so we try to find a polynomial such that simultaneously (??) is an integer and a polynomial with integer coefficients mod 2? But it is not intersection it is product.
      $endgroup$
      – 72D
      Dec 3 '18 at 7:59










    • $begingroup$
      I've expanded my hint into a sketch of an answer.
      $endgroup$
      – Servaes
      Dec 3 '18 at 8:05






    • 1




      $begingroup$
      "Which elements of Z[x] map to (1,0)" : so is this a counterexample? : for any polynomial q(x) in Z[x], 2q(x)+1 is the only element of Z[x] such that it is 1 in Z[x]/(2) but 2q(x)+1 is never 0 in Z[x]/(x); am I right?
      $endgroup$
      – 72D
      Dec 3 '18 at 8:05
















    $begingroup$
    How are the typical/general elements of $mathbb{Z}[x]/(2)$ and $mathbb{Z}[x]/(x)$?
    $endgroup$
    – 72D
    Dec 3 '18 at 7:50




    $begingroup$
    How are the typical/general elements of $mathbb{Z}[x]/(2)$ and $mathbb{Z}[x]/(x)$?
    $endgroup$
    – 72D
    Dec 3 '18 at 7:50




    1




    1




    $begingroup$
    They are polynomials with integer coefficients mod $2$, and simply integers.
    $endgroup$
    – Servaes
    Dec 3 '18 at 7:51






    $begingroup$
    They are polynomials with integer coefficients mod $2$, and simply integers.
    $endgroup$
    – Servaes
    Dec 3 '18 at 7:51














    $begingroup$
    I could understand now their meanings but still don't understand the meaning for surjection esp because the rhs is 'double'-element. To try to show that (first) it is surjective (then fail), so we try to find a polynomial such that simultaneously (??) is an integer and a polynomial with integer coefficients mod 2? But it is not intersection it is product.
    $endgroup$
    – 72D
    Dec 3 '18 at 7:59




    $begingroup$
    I could understand now their meanings but still don't understand the meaning for surjection esp because the rhs is 'double'-element. To try to show that (first) it is surjective (then fail), so we try to find a polynomial such that simultaneously (??) is an integer and a polynomial with integer coefficients mod 2? But it is not intersection it is product.
    $endgroup$
    – 72D
    Dec 3 '18 at 7:59












    $begingroup$
    I've expanded my hint into a sketch of an answer.
    $endgroup$
    – Servaes
    Dec 3 '18 at 8:05




    $begingroup$
    I've expanded my hint into a sketch of an answer.
    $endgroup$
    – Servaes
    Dec 3 '18 at 8:05




    1




    1




    $begingroup$
    "Which elements of Z[x] map to (1,0)" : so is this a counterexample? : for any polynomial q(x) in Z[x], 2q(x)+1 is the only element of Z[x] such that it is 1 in Z[x]/(2) but 2q(x)+1 is never 0 in Z[x]/(x); am I right?
    $endgroup$
    – 72D
    Dec 3 '18 at 8:05




    $begingroup$
    "Which elements of Z[x] map to (1,0)" : so is this a counterexample? : for any polynomial q(x) in Z[x], 2q(x)+1 is the only element of Z[x] such that it is 1 in Z[x]/(2) but 2q(x)+1 is never 0 in Z[x]/(x); am I right?
    $endgroup$
    – 72D
    Dec 3 '18 at 8:05











    0












    $begingroup$

    Hint $ R to R/I times R/J $ surjective $Rightarrow I + J = 1$



    since $,rto (1,0),Rightarrow, 1 = (1!-!r) + rin I + J$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Hint $ R to R/I times R/J $ surjective $Rightarrow I + J = 1$



      since $,rto (1,0),Rightarrow, 1 = (1!-!r) + rin I + J$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Hint $ R to R/I times R/J $ surjective $Rightarrow I + J = 1$



        since $,rto (1,0),Rightarrow, 1 = (1!-!r) + rin I + J$






        share|cite|improve this answer









        $endgroup$



        Hint $ R to R/I times R/J $ surjective $Rightarrow I + J = 1$



        since $,rto (1,0),Rightarrow, 1 = (1!-!r) + rin I + J$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 3 '18 at 23:03









        Bill DubuqueBill Dubuque

        210k29191639




        210k29191639






























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