$K[X,Y]/(XY-1)$ not isomorphic to $K[T]$ [duplicate]












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  • Show that quotient rings are not isomorphic

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Why is $K[X,Y]/(XY-1)$ not isomorphic to $K[T]?$



As a hint I should think about units. But I don't get a helpful idea.










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Feb 23 '17 at 9:14


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  • $begingroup$
    Hint: Is $T$ invertible in $K[T]$? Is $X$ invertible in $K[X,Y]/(XY-1)$? What is $XY$ equal to?
    $endgroup$
    – Michael Burr
    Feb 22 '17 at 18:21






  • 2




    $begingroup$
    $K[x,y]/(xy-1) cong K[x,frac{1}{x}] $
    $endgroup$
    – Mustafa
    Feb 22 '17 at 18:33


















3












$begingroup$



This question already has an answer here:




  • Show that quotient rings are not isomorphic

    3 answers




Why is $K[X,Y]/(XY-1)$ not isomorphic to $K[T]?$



As a hint I should think about units. But I don't get a helpful idea.










share|cite|improve this question









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marked as duplicate by user26857 abstract-algebra
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Feb 23 '17 at 9:14


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.















  • $begingroup$
    Hint: Is $T$ invertible in $K[T]$? Is $X$ invertible in $K[X,Y]/(XY-1)$? What is $XY$ equal to?
    $endgroup$
    – Michael Burr
    Feb 22 '17 at 18:21






  • 2




    $begingroup$
    $K[x,y]/(xy-1) cong K[x,frac{1}{x}] $
    $endgroup$
    – Mustafa
    Feb 22 '17 at 18:33
















3












3








3


1



$begingroup$



This question already has an answer here:




  • Show that quotient rings are not isomorphic

    3 answers




Why is $K[X,Y]/(XY-1)$ not isomorphic to $K[T]?$



As a hint I should think about units. But I don't get a helpful idea.










share|cite|improve this question









$endgroup$





This question already has an answer here:




  • Show that quotient rings are not isomorphic

    3 answers




Why is $K[X,Y]/(XY-1)$ not isomorphic to $K[T]?$



As a hint I should think about units. But I don't get a helpful idea.





This question already has an answer here:




  • Show that quotient rings are not isomorphic

    3 answers








linear-algebra abstract-algebra






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Feb 22 '17 at 18:20







user404105











marked as duplicate by user26857 abstract-algebra
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Feb 23 '17 at 9:14


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






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Feb 23 '17 at 9:14


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • $begingroup$
    Hint: Is $T$ invertible in $K[T]$? Is $X$ invertible in $K[X,Y]/(XY-1)$? What is $XY$ equal to?
    $endgroup$
    – Michael Burr
    Feb 22 '17 at 18:21






  • 2




    $begingroup$
    $K[x,y]/(xy-1) cong K[x,frac{1}{x}] $
    $endgroup$
    – Mustafa
    Feb 22 '17 at 18:33




















  • $begingroup$
    Hint: Is $T$ invertible in $K[T]$? Is $X$ invertible in $K[X,Y]/(XY-1)$? What is $XY$ equal to?
    $endgroup$
    – Michael Burr
    Feb 22 '17 at 18:21






  • 2




    $begingroup$
    $K[x,y]/(xy-1) cong K[x,frac{1}{x}] $
    $endgroup$
    – Mustafa
    Feb 22 '17 at 18:33


















$begingroup$
Hint: Is $T$ invertible in $K[T]$? Is $X$ invertible in $K[X,Y]/(XY-1)$? What is $XY$ equal to?
$endgroup$
– Michael Burr
Feb 22 '17 at 18:21




$begingroup$
Hint: Is $T$ invertible in $K[T]$? Is $X$ invertible in $K[X,Y]/(XY-1)$? What is $XY$ equal to?
$endgroup$
– Michael Burr
Feb 22 '17 at 18:21




2




2




$begingroup$
$K[x,y]/(xy-1) cong K[x,frac{1}{x}] $
$endgroup$
– Mustafa
Feb 22 '17 at 18:33






$begingroup$
$K[x,y]/(xy-1) cong K[x,frac{1}{x}] $
$endgroup$
– Mustafa
Feb 22 '17 at 18:33












2 Answers
2






active

oldest

votes


















-1












$begingroup$

The group units of $k[x]$ is $k^*$, the units of $k[x,y]/(xy-1)$ is $k^*times {x^nmid nin mathbb{Z}}$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    I don't know why this answer got 4 upvotes. If $k$ is not algebraically closed, then this argument fails, that is, the groups $k^*$ and $k^*timesmathbb Z$ can be isomorphic (for instance, when $k=mathbb Q$).
    $endgroup$
    – user26857
    Dec 3 '18 at 19:49





















5












$begingroup$

Suppose $K[x,y]/(xy-1)$ is isomorphic to $K[t]$.



Then there is a surjective homomorphism $f:K[x,y] to K[t]$ with $text{ker}(f) = (xy-1)$.



Let's assume you've previously shown that the group of units of $K[t]$ is $K^*$ (if not, you need to show this).



Since $f$ is a homomorphism, $f$ must map units to units, hence
$f(K^*) subseteq K^*$. It follows that $f(K) subseteq K$.



Let $p = f(x)$, $q = f(y)$.



Then $0 = f(xy - 1) = f(x)f(y) - f(1) = pq - 1$, hence $pq = 1$.



Thus, $p,q$ are units in $K[t]$, so $p,q in K^*$.



Let $m in K[x,y]$ be a monomial. Write $m = ax^iy^j$ where $a in K$ and $i,j$ are nonnegative integers.



Then $f(m) = f(ax^iy^j) = f(a)f(x^i)f(x^j) = f(a)f(x)^if(y)^j = f(a)p^iq^j$, hence $f(m) in K$.



Let $s in K[x,y]$.



Then $s$ is a sum of finitely many monomials, hence since $f$ maps each monomial to an element of $K$, it follows that $f(s) in K$.



Thus, $f$ maps every element of $K[x,y]$ to an element of $K$, so $f$ is not surjective, contradiction.



It follows that $K[x,y]$ is not isomorphic to $K[t]$.






share|cite|improve this answer









$endgroup$



















    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    -1












    $begingroup$

    The group units of $k[x]$ is $k^*$, the units of $k[x,y]/(xy-1)$ is $k^*times {x^nmid nin mathbb{Z}}$.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      I don't know why this answer got 4 upvotes. If $k$ is not algebraically closed, then this argument fails, that is, the groups $k^*$ and $k^*timesmathbb Z$ can be isomorphic (for instance, when $k=mathbb Q$).
      $endgroup$
      – user26857
      Dec 3 '18 at 19:49


















    -1












    $begingroup$

    The group units of $k[x]$ is $k^*$, the units of $k[x,y]/(xy-1)$ is $k^*times {x^nmid nin mathbb{Z}}$.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      I don't know why this answer got 4 upvotes. If $k$ is not algebraically closed, then this argument fails, that is, the groups $k^*$ and $k^*timesmathbb Z$ can be isomorphic (for instance, when $k=mathbb Q$).
      $endgroup$
      – user26857
      Dec 3 '18 at 19:49
















    -1












    -1








    -1





    $begingroup$

    The group units of $k[x]$ is $k^*$, the units of $k[x,y]/(xy-1)$ is $k^*times {x^nmid nin mathbb{Z}}$.






    share|cite|improve this answer











    $endgroup$



    The group units of $k[x]$ is $k^*$, the units of $k[x,y]/(xy-1)$ is $k^*times {x^nmid nin mathbb{Z}}$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 3 '18 at 19:15









    user26857

    39.3k124183




    39.3k124183










    answered Feb 22 '17 at 18:25









    Rene SchipperusRene Schipperus

    32.2k11960




    32.2k11960








    • 1




      $begingroup$
      I don't know why this answer got 4 upvotes. If $k$ is not algebraically closed, then this argument fails, that is, the groups $k^*$ and $k^*timesmathbb Z$ can be isomorphic (for instance, when $k=mathbb Q$).
      $endgroup$
      – user26857
      Dec 3 '18 at 19:49
















    • 1




      $begingroup$
      I don't know why this answer got 4 upvotes. If $k$ is not algebraically closed, then this argument fails, that is, the groups $k^*$ and $k^*timesmathbb Z$ can be isomorphic (for instance, when $k=mathbb Q$).
      $endgroup$
      – user26857
      Dec 3 '18 at 19:49










    1




    1




    $begingroup$
    I don't know why this answer got 4 upvotes. If $k$ is not algebraically closed, then this argument fails, that is, the groups $k^*$ and $k^*timesmathbb Z$ can be isomorphic (for instance, when $k=mathbb Q$).
    $endgroup$
    – user26857
    Dec 3 '18 at 19:49






    $begingroup$
    I don't know why this answer got 4 upvotes. If $k$ is not algebraically closed, then this argument fails, that is, the groups $k^*$ and $k^*timesmathbb Z$ can be isomorphic (for instance, when $k=mathbb Q$).
    $endgroup$
    – user26857
    Dec 3 '18 at 19:49













    5












    $begingroup$

    Suppose $K[x,y]/(xy-1)$ is isomorphic to $K[t]$.



    Then there is a surjective homomorphism $f:K[x,y] to K[t]$ with $text{ker}(f) = (xy-1)$.



    Let's assume you've previously shown that the group of units of $K[t]$ is $K^*$ (if not, you need to show this).



    Since $f$ is a homomorphism, $f$ must map units to units, hence
    $f(K^*) subseteq K^*$. It follows that $f(K) subseteq K$.



    Let $p = f(x)$, $q = f(y)$.



    Then $0 = f(xy - 1) = f(x)f(y) - f(1) = pq - 1$, hence $pq = 1$.



    Thus, $p,q$ are units in $K[t]$, so $p,q in K^*$.



    Let $m in K[x,y]$ be a monomial. Write $m = ax^iy^j$ where $a in K$ and $i,j$ are nonnegative integers.



    Then $f(m) = f(ax^iy^j) = f(a)f(x^i)f(x^j) = f(a)f(x)^if(y)^j = f(a)p^iq^j$, hence $f(m) in K$.



    Let $s in K[x,y]$.



    Then $s$ is a sum of finitely many monomials, hence since $f$ maps each monomial to an element of $K$, it follows that $f(s) in K$.



    Thus, $f$ maps every element of $K[x,y]$ to an element of $K$, so $f$ is not surjective, contradiction.



    It follows that $K[x,y]$ is not isomorphic to $K[t]$.






    share|cite|improve this answer









    $endgroup$


















      5












      $begingroup$

      Suppose $K[x,y]/(xy-1)$ is isomorphic to $K[t]$.



      Then there is a surjective homomorphism $f:K[x,y] to K[t]$ with $text{ker}(f) = (xy-1)$.



      Let's assume you've previously shown that the group of units of $K[t]$ is $K^*$ (if not, you need to show this).



      Since $f$ is a homomorphism, $f$ must map units to units, hence
      $f(K^*) subseteq K^*$. It follows that $f(K) subseteq K$.



      Let $p = f(x)$, $q = f(y)$.



      Then $0 = f(xy - 1) = f(x)f(y) - f(1) = pq - 1$, hence $pq = 1$.



      Thus, $p,q$ are units in $K[t]$, so $p,q in K^*$.



      Let $m in K[x,y]$ be a monomial. Write $m = ax^iy^j$ where $a in K$ and $i,j$ are nonnegative integers.



      Then $f(m) = f(ax^iy^j) = f(a)f(x^i)f(x^j) = f(a)f(x)^if(y)^j = f(a)p^iq^j$, hence $f(m) in K$.



      Let $s in K[x,y]$.



      Then $s$ is a sum of finitely many monomials, hence since $f$ maps each monomial to an element of $K$, it follows that $f(s) in K$.



      Thus, $f$ maps every element of $K[x,y]$ to an element of $K$, so $f$ is not surjective, contradiction.



      It follows that $K[x,y]$ is not isomorphic to $K[t]$.






      share|cite|improve this answer









      $endgroup$
















        5












        5








        5





        $begingroup$

        Suppose $K[x,y]/(xy-1)$ is isomorphic to $K[t]$.



        Then there is a surjective homomorphism $f:K[x,y] to K[t]$ with $text{ker}(f) = (xy-1)$.



        Let's assume you've previously shown that the group of units of $K[t]$ is $K^*$ (if not, you need to show this).



        Since $f$ is a homomorphism, $f$ must map units to units, hence
        $f(K^*) subseteq K^*$. It follows that $f(K) subseteq K$.



        Let $p = f(x)$, $q = f(y)$.



        Then $0 = f(xy - 1) = f(x)f(y) - f(1) = pq - 1$, hence $pq = 1$.



        Thus, $p,q$ are units in $K[t]$, so $p,q in K^*$.



        Let $m in K[x,y]$ be a monomial. Write $m = ax^iy^j$ where $a in K$ and $i,j$ are nonnegative integers.



        Then $f(m) = f(ax^iy^j) = f(a)f(x^i)f(x^j) = f(a)f(x)^if(y)^j = f(a)p^iq^j$, hence $f(m) in K$.



        Let $s in K[x,y]$.



        Then $s$ is a sum of finitely many monomials, hence since $f$ maps each monomial to an element of $K$, it follows that $f(s) in K$.



        Thus, $f$ maps every element of $K[x,y]$ to an element of $K$, so $f$ is not surjective, contradiction.



        It follows that $K[x,y]$ is not isomorphic to $K[t]$.






        share|cite|improve this answer









        $endgroup$



        Suppose $K[x,y]/(xy-1)$ is isomorphic to $K[t]$.



        Then there is a surjective homomorphism $f:K[x,y] to K[t]$ with $text{ker}(f) = (xy-1)$.



        Let's assume you've previously shown that the group of units of $K[t]$ is $K^*$ (if not, you need to show this).



        Since $f$ is a homomorphism, $f$ must map units to units, hence
        $f(K^*) subseteq K^*$. It follows that $f(K) subseteq K$.



        Let $p = f(x)$, $q = f(y)$.



        Then $0 = f(xy - 1) = f(x)f(y) - f(1) = pq - 1$, hence $pq = 1$.



        Thus, $p,q$ are units in $K[t]$, so $p,q in K^*$.



        Let $m in K[x,y]$ be a monomial. Write $m = ax^iy^j$ where $a in K$ and $i,j$ are nonnegative integers.



        Then $f(m) = f(ax^iy^j) = f(a)f(x^i)f(x^j) = f(a)f(x)^if(y)^j = f(a)p^iq^j$, hence $f(m) in K$.



        Let $s in K[x,y]$.



        Then $s$ is a sum of finitely many monomials, hence since $f$ maps each monomial to an element of $K$, it follows that $f(s) in K$.



        Thus, $f$ maps every element of $K[x,y]$ to an element of $K$, so $f$ is not surjective, contradiction.



        It follows that $K[x,y]$ is not isomorphic to $K[t]$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Feb 22 '17 at 20:31









        quasiquasi

        36k22663




        36k22663















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