If $Acong B$, then $Aotimes Ccong Botimes C$.











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5
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I think this is kind of true, since $squareotimes C$ is a functor, so it preserves the isomorphism.



But what if we consider the example, $mathbb{Z}otimesmathbb{Z}_2$ is not isomorphic to $2mathbb{Z}otimesmathbb{Z}_2$. Because the latter is trivial while the first is not.



We can also consider the exact sequences:
$$0tomathbb{Z}tomathbb{Z}tomathbb{Z}_2to0,$$
where the first map is simply multiple by 2,
and
$$0to2mathbb{Z}tomathbb{Z}tomathbb{Z}_2to0,$$
where the first map is the inclusion.
If we tensor by $mathbb{Z}_2$ we will get
$$mathbb{Z}_2tomathbb{Z}_2tomathbb{Z}_2to0$$
and



$$(0to)Xtomathbb{Z}_2tomathbb{Z}_2to0$$
respectively.



Apparently, the first one is the famous counterexample of non-left-exactness of tensor product. But the second one we just inject a submodule (ideal) to the whole module (ring). If $rotimes m$ is 0 in $mathbb{Z}otimesmathbb{Z}_2$, then apparently it is 0 in $2mathbb{Z}otimesmathbb{Z}_2$. So we have the left exactness of the tensor product. This is again weird, since the two exact sequences are isomorphic (by wiki), why do I have different result?










share|cite|improve this question




















  • 1




    @5xum but tensoring by $C$ is a functor in the category of $R$-modules, so it must preserve the isomorphism.
    – Xavier Yang
    Nov 22 at 14:53










  • Are ${Bbb Z}$ and ${Bbb Z}_2$ considered as $Bbb Z$ modules?
    – Wuestenfux
    Nov 22 at 14:54










  • @Wuestenfux Yes. So, you mean $2mathbb{Z}otimes_mathbb{Z}mathbb{Z}_2$ is also $mathbb{Z}_2$?
    – Xavier Yang
    Nov 22 at 14:56








  • 3




    Why is the second trivial? Shouldn’t $2 otimes 1$ be a nonzero element of $2 mathbb{Z} otimes mathbb{Z}_2$?
    – user328442
    Nov 22 at 15:01












  • You've implicitly assumed that $(-) otimes mathbb{Z}_2$ preserves inclusions, but it doesn't; in general tensoring with a module preserves monomorphisms iff it's exact iff the module is flat.
    – Qiaochu Yuan
    Nov 22 at 21:31















up vote
5
down vote

favorite












I think this is kind of true, since $squareotimes C$ is a functor, so it preserves the isomorphism.



But what if we consider the example, $mathbb{Z}otimesmathbb{Z}_2$ is not isomorphic to $2mathbb{Z}otimesmathbb{Z}_2$. Because the latter is trivial while the first is not.



We can also consider the exact sequences:
$$0tomathbb{Z}tomathbb{Z}tomathbb{Z}_2to0,$$
where the first map is simply multiple by 2,
and
$$0to2mathbb{Z}tomathbb{Z}tomathbb{Z}_2to0,$$
where the first map is the inclusion.
If we tensor by $mathbb{Z}_2$ we will get
$$mathbb{Z}_2tomathbb{Z}_2tomathbb{Z}_2to0$$
and



$$(0to)Xtomathbb{Z}_2tomathbb{Z}_2to0$$
respectively.



Apparently, the first one is the famous counterexample of non-left-exactness of tensor product. But the second one we just inject a submodule (ideal) to the whole module (ring). If $rotimes m$ is 0 in $mathbb{Z}otimesmathbb{Z}_2$, then apparently it is 0 in $2mathbb{Z}otimesmathbb{Z}_2$. So we have the left exactness of the tensor product. This is again weird, since the two exact sequences are isomorphic (by wiki), why do I have different result?










share|cite|improve this question




















  • 1




    @5xum but tensoring by $C$ is a functor in the category of $R$-modules, so it must preserve the isomorphism.
    – Xavier Yang
    Nov 22 at 14:53










  • Are ${Bbb Z}$ and ${Bbb Z}_2$ considered as $Bbb Z$ modules?
    – Wuestenfux
    Nov 22 at 14:54










  • @Wuestenfux Yes. So, you mean $2mathbb{Z}otimes_mathbb{Z}mathbb{Z}_2$ is also $mathbb{Z}_2$?
    – Xavier Yang
    Nov 22 at 14:56








  • 3




    Why is the second trivial? Shouldn’t $2 otimes 1$ be a nonzero element of $2 mathbb{Z} otimes mathbb{Z}_2$?
    – user328442
    Nov 22 at 15:01












  • You've implicitly assumed that $(-) otimes mathbb{Z}_2$ preserves inclusions, but it doesn't; in general tensoring with a module preserves monomorphisms iff it's exact iff the module is flat.
    – Qiaochu Yuan
    Nov 22 at 21:31













up vote
5
down vote

favorite









up vote
5
down vote

favorite











I think this is kind of true, since $squareotimes C$ is a functor, so it preserves the isomorphism.



But what if we consider the example, $mathbb{Z}otimesmathbb{Z}_2$ is not isomorphic to $2mathbb{Z}otimesmathbb{Z}_2$. Because the latter is trivial while the first is not.



We can also consider the exact sequences:
$$0tomathbb{Z}tomathbb{Z}tomathbb{Z}_2to0,$$
where the first map is simply multiple by 2,
and
$$0to2mathbb{Z}tomathbb{Z}tomathbb{Z}_2to0,$$
where the first map is the inclusion.
If we tensor by $mathbb{Z}_2$ we will get
$$mathbb{Z}_2tomathbb{Z}_2tomathbb{Z}_2to0$$
and



$$(0to)Xtomathbb{Z}_2tomathbb{Z}_2to0$$
respectively.



Apparently, the first one is the famous counterexample of non-left-exactness of tensor product. But the second one we just inject a submodule (ideal) to the whole module (ring). If $rotimes m$ is 0 in $mathbb{Z}otimesmathbb{Z}_2$, then apparently it is 0 in $2mathbb{Z}otimesmathbb{Z}_2$. So we have the left exactness of the tensor product. This is again weird, since the two exact sequences are isomorphic (by wiki), why do I have different result?










share|cite|improve this question















I think this is kind of true, since $squareotimes C$ is a functor, so it preserves the isomorphism.



But what if we consider the example, $mathbb{Z}otimesmathbb{Z}_2$ is not isomorphic to $2mathbb{Z}otimesmathbb{Z}_2$. Because the latter is trivial while the first is not.



We can also consider the exact sequences:
$$0tomathbb{Z}tomathbb{Z}tomathbb{Z}_2to0,$$
where the first map is simply multiple by 2,
and
$$0to2mathbb{Z}tomathbb{Z}tomathbb{Z}_2to0,$$
where the first map is the inclusion.
If we tensor by $mathbb{Z}_2$ we will get
$$mathbb{Z}_2tomathbb{Z}_2tomathbb{Z}_2to0$$
and



$$(0to)Xtomathbb{Z}_2tomathbb{Z}_2to0$$
respectively.



Apparently, the first one is the famous counterexample of non-left-exactness of tensor product. But the second one we just inject a submodule (ideal) to the whole module (ring). If $rotimes m$ is 0 in $mathbb{Z}otimesmathbb{Z}_2$, then apparently it is 0 in $2mathbb{Z}otimesmathbb{Z}_2$. So we have the left exactness of the tensor product. This is again weird, since the two exact sequences are isomorphic (by wiki), why do I have different result?







abstract-algebra modules






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago

























asked Nov 22 at 14:48









Xavier Yang

469314




469314








  • 1




    @5xum but tensoring by $C$ is a functor in the category of $R$-modules, so it must preserve the isomorphism.
    – Xavier Yang
    Nov 22 at 14:53










  • Are ${Bbb Z}$ and ${Bbb Z}_2$ considered as $Bbb Z$ modules?
    – Wuestenfux
    Nov 22 at 14:54










  • @Wuestenfux Yes. So, you mean $2mathbb{Z}otimes_mathbb{Z}mathbb{Z}_2$ is also $mathbb{Z}_2$?
    – Xavier Yang
    Nov 22 at 14:56








  • 3




    Why is the second trivial? Shouldn’t $2 otimes 1$ be a nonzero element of $2 mathbb{Z} otimes mathbb{Z}_2$?
    – user328442
    Nov 22 at 15:01












  • You've implicitly assumed that $(-) otimes mathbb{Z}_2$ preserves inclusions, but it doesn't; in general tensoring with a module preserves monomorphisms iff it's exact iff the module is flat.
    – Qiaochu Yuan
    Nov 22 at 21:31














  • 1




    @5xum but tensoring by $C$ is a functor in the category of $R$-modules, so it must preserve the isomorphism.
    – Xavier Yang
    Nov 22 at 14:53










  • Are ${Bbb Z}$ and ${Bbb Z}_2$ considered as $Bbb Z$ modules?
    – Wuestenfux
    Nov 22 at 14:54










  • @Wuestenfux Yes. So, you mean $2mathbb{Z}otimes_mathbb{Z}mathbb{Z}_2$ is also $mathbb{Z}_2$?
    – Xavier Yang
    Nov 22 at 14:56








  • 3




    Why is the second trivial? Shouldn’t $2 otimes 1$ be a nonzero element of $2 mathbb{Z} otimes mathbb{Z}_2$?
    – user328442
    Nov 22 at 15:01












  • You've implicitly assumed that $(-) otimes mathbb{Z}_2$ preserves inclusions, but it doesn't; in general tensoring with a module preserves monomorphisms iff it's exact iff the module is flat.
    – Qiaochu Yuan
    Nov 22 at 21:31








1




1




@5xum but tensoring by $C$ is a functor in the category of $R$-modules, so it must preserve the isomorphism.
– Xavier Yang
Nov 22 at 14:53




@5xum but tensoring by $C$ is a functor in the category of $R$-modules, so it must preserve the isomorphism.
– Xavier Yang
Nov 22 at 14:53












Are ${Bbb Z}$ and ${Bbb Z}_2$ considered as $Bbb Z$ modules?
– Wuestenfux
Nov 22 at 14:54




Are ${Bbb Z}$ and ${Bbb Z}_2$ considered as $Bbb Z$ modules?
– Wuestenfux
Nov 22 at 14:54












@Wuestenfux Yes. So, you mean $2mathbb{Z}otimes_mathbb{Z}mathbb{Z}_2$ is also $mathbb{Z}_2$?
– Xavier Yang
Nov 22 at 14:56






@Wuestenfux Yes. So, you mean $2mathbb{Z}otimes_mathbb{Z}mathbb{Z}_2$ is also $mathbb{Z}_2$?
– Xavier Yang
Nov 22 at 14:56






3




3




Why is the second trivial? Shouldn’t $2 otimes 1$ be a nonzero element of $2 mathbb{Z} otimes mathbb{Z}_2$?
– user328442
Nov 22 at 15:01






Why is the second trivial? Shouldn’t $2 otimes 1$ be a nonzero element of $2 mathbb{Z} otimes mathbb{Z}_2$?
– user328442
Nov 22 at 15:01














You've implicitly assumed that $(-) otimes mathbb{Z}_2$ preserves inclusions, but it doesn't; in general tensoring with a module preserves monomorphisms iff it's exact iff the module is flat.
– Qiaochu Yuan
Nov 22 at 21:31




You've implicitly assumed that $(-) otimes mathbb{Z}_2$ preserves inclusions, but it doesn't; in general tensoring with a module preserves monomorphisms iff it's exact iff the module is flat.
– Qiaochu Yuan
Nov 22 at 21:31










2 Answers
2






active

oldest

votes

















up vote
5
down vote













$2mathbb{Z}otimesmathbb{Z}_2$ is not trivial. It is trivial inside of $mathbb Z otimes mathbb{Z}_2$, a subtle but important difference. The "proof" you might have in mind that it is trivial would be



$$ 2n otimes 0 = 0 text{ and } 2n otimes 1 = n otimes 2 = 0 $$



but $n otimes 2$ is only an element of $2mathbb{Z}otimesmathbb{Z}_2$ if $n$ is even. Otherwise, you need the ambient module $mathbb Z otimes mathbb{Z}_2$ to make sense of it.



Secondly, for your exact sequences. Do keep in mind that the isomorphism $2mathbb{Z} to mathbb{Z}$ is a different map than the inclusion $2mathbb{Z} to mathbb{Z}$.



So if we take the exact sequence



$$ 2mathbb{Z} to mathbb{Z}tomathbb{Z}_2to 0 $$



where the leftmost map is inclusion and tensor with $mathbb{Z}_2$, we get the following commutative diagram
$require{AMScd}$
begin{CD}
2mathbb{Z} otimes mathbb{Z}_2 @>text{inclusion}>> mathbb{Z} otimes mathbb{Z}_2 @>>> mathbb{Z}_2otimes mathbb{Z}_2 @>>> 0 \
@VVV @VVV @VVV @VVV \
mathbb{Z}_2 @>times 2>> mathbb{Z}_2 @>>> mathbb{Z}_2 @>>>0
end{CD}

where the rows are exact and the vertical arrows are isomorphisms.



That map $mathbb{Z}_2 xrightarrow{times 2} mathbb{Z}_2$ on the bottom is the zero map. To see that this must be the case, let us note that the isomorphisms $to mathbb{Z}_2$ are



$$ begin{array}{ccc}
2mathbb{Z} otimes mathbb{Z}_2 & longrightarrow & mathbb{Z}_2 \
2n otimes 0 &longmapsto & 0 \
2n otimes 1 & longmapsto & n bmod 2
end{array}
qquad text{and} qquad
begin{array}{ccc}
mathbb{Z} otimes mathbb{Z}_2 & longrightarrow & mathbb{Z}_2 \
n otimes 0 &longmapsto & 0 \
n otimes 1 & longmapsto & n bmod 2
end{array} $$



So the map on the bottom we can compute by looking at the composition $mathbb{Z}_2 to 2mathbb{Z} otimes mathbb{Z}_2 to mathbb{Z} otimes mathbb{Z}_2 to mathbb{Z}_2$ and this composition takes $1 mapsto 2 otimes 1 mapsto 2 otimes 1 mapsto 0$.






share|cite|improve this answer




























    up vote
    0
    down vote













    $2mathbb{Z}$ and $mathbb{Z}_2$ are not isomorphic as $mathbb{Z}$-modules. Take a generator of each module and multiply it by 2.






    share|cite|improve this answer





















    • Yes, this is clear.
      – Xavier Yang
      Nov 22 at 17:16











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    2 Answers
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    2 Answers
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    active

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    up vote
    5
    down vote













    $2mathbb{Z}otimesmathbb{Z}_2$ is not trivial. It is trivial inside of $mathbb Z otimes mathbb{Z}_2$, a subtle but important difference. The "proof" you might have in mind that it is trivial would be



    $$ 2n otimes 0 = 0 text{ and } 2n otimes 1 = n otimes 2 = 0 $$



    but $n otimes 2$ is only an element of $2mathbb{Z}otimesmathbb{Z}_2$ if $n$ is even. Otherwise, you need the ambient module $mathbb Z otimes mathbb{Z}_2$ to make sense of it.



    Secondly, for your exact sequences. Do keep in mind that the isomorphism $2mathbb{Z} to mathbb{Z}$ is a different map than the inclusion $2mathbb{Z} to mathbb{Z}$.



    So if we take the exact sequence



    $$ 2mathbb{Z} to mathbb{Z}tomathbb{Z}_2to 0 $$



    where the leftmost map is inclusion and tensor with $mathbb{Z}_2$, we get the following commutative diagram
    $require{AMScd}$
    begin{CD}
    2mathbb{Z} otimes mathbb{Z}_2 @>text{inclusion}>> mathbb{Z} otimes mathbb{Z}_2 @>>> mathbb{Z}_2otimes mathbb{Z}_2 @>>> 0 \
    @VVV @VVV @VVV @VVV \
    mathbb{Z}_2 @>times 2>> mathbb{Z}_2 @>>> mathbb{Z}_2 @>>>0
    end{CD}

    where the rows are exact and the vertical arrows are isomorphisms.



    That map $mathbb{Z}_2 xrightarrow{times 2} mathbb{Z}_2$ on the bottom is the zero map. To see that this must be the case, let us note that the isomorphisms $to mathbb{Z}_2$ are



    $$ begin{array}{ccc}
    2mathbb{Z} otimes mathbb{Z}_2 & longrightarrow & mathbb{Z}_2 \
    2n otimes 0 &longmapsto & 0 \
    2n otimes 1 & longmapsto & n bmod 2
    end{array}
    qquad text{and} qquad
    begin{array}{ccc}
    mathbb{Z} otimes mathbb{Z}_2 & longrightarrow & mathbb{Z}_2 \
    n otimes 0 &longmapsto & 0 \
    n otimes 1 & longmapsto & n bmod 2
    end{array} $$



    So the map on the bottom we can compute by looking at the composition $mathbb{Z}_2 to 2mathbb{Z} otimes mathbb{Z}_2 to mathbb{Z} otimes mathbb{Z}_2 to mathbb{Z}_2$ and this composition takes $1 mapsto 2 otimes 1 mapsto 2 otimes 1 mapsto 0$.






    share|cite|improve this answer

























      up vote
      5
      down vote













      $2mathbb{Z}otimesmathbb{Z}_2$ is not trivial. It is trivial inside of $mathbb Z otimes mathbb{Z}_2$, a subtle but important difference. The "proof" you might have in mind that it is trivial would be



      $$ 2n otimes 0 = 0 text{ and } 2n otimes 1 = n otimes 2 = 0 $$



      but $n otimes 2$ is only an element of $2mathbb{Z}otimesmathbb{Z}_2$ if $n$ is even. Otherwise, you need the ambient module $mathbb Z otimes mathbb{Z}_2$ to make sense of it.



      Secondly, for your exact sequences. Do keep in mind that the isomorphism $2mathbb{Z} to mathbb{Z}$ is a different map than the inclusion $2mathbb{Z} to mathbb{Z}$.



      So if we take the exact sequence



      $$ 2mathbb{Z} to mathbb{Z}tomathbb{Z}_2to 0 $$



      where the leftmost map is inclusion and tensor with $mathbb{Z}_2$, we get the following commutative diagram
      $require{AMScd}$
      begin{CD}
      2mathbb{Z} otimes mathbb{Z}_2 @>text{inclusion}>> mathbb{Z} otimes mathbb{Z}_2 @>>> mathbb{Z}_2otimes mathbb{Z}_2 @>>> 0 \
      @VVV @VVV @VVV @VVV \
      mathbb{Z}_2 @>times 2>> mathbb{Z}_2 @>>> mathbb{Z}_2 @>>>0
      end{CD}

      where the rows are exact and the vertical arrows are isomorphisms.



      That map $mathbb{Z}_2 xrightarrow{times 2} mathbb{Z}_2$ on the bottom is the zero map. To see that this must be the case, let us note that the isomorphisms $to mathbb{Z}_2$ are



      $$ begin{array}{ccc}
      2mathbb{Z} otimes mathbb{Z}_2 & longrightarrow & mathbb{Z}_2 \
      2n otimes 0 &longmapsto & 0 \
      2n otimes 1 & longmapsto & n bmod 2
      end{array}
      qquad text{and} qquad
      begin{array}{ccc}
      mathbb{Z} otimes mathbb{Z}_2 & longrightarrow & mathbb{Z}_2 \
      n otimes 0 &longmapsto & 0 \
      n otimes 1 & longmapsto & n bmod 2
      end{array} $$



      So the map on the bottom we can compute by looking at the composition $mathbb{Z}_2 to 2mathbb{Z} otimes mathbb{Z}_2 to mathbb{Z} otimes mathbb{Z}_2 to mathbb{Z}_2$ and this composition takes $1 mapsto 2 otimes 1 mapsto 2 otimes 1 mapsto 0$.






      share|cite|improve this answer























        up vote
        5
        down vote










        up vote
        5
        down vote









        $2mathbb{Z}otimesmathbb{Z}_2$ is not trivial. It is trivial inside of $mathbb Z otimes mathbb{Z}_2$, a subtle but important difference. The "proof" you might have in mind that it is trivial would be



        $$ 2n otimes 0 = 0 text{ and } 2n otimes 1 = n otimes 2 = 0 $$



        but $n otimes 2$ is only an element of $2mathbb{Z}otimesmathbb{Z}_2$ if $n$ is even. Otherwise, you need the ambient module $mathbb Z otimes mathbb{Z}_2$ to make sense of it.



        Secondly, for your exact sequences. Do keep in mind that the isomorphism $2mathbb{Z} to mathbb{Z}$ is a different map than the inclusion $2mathbb{Z} to mathbb{Z}$.



        So if we take the exact sequence



        $$ 2mathbb{Z} to mathbb{Z}tomathbb{Z}_2to 0 $$



        where the leftmost map is inclusion and tensor with $mathbb{Z}_2$, we get the following commutative diagram
        $require{AMScd}$
        begin{CD}
        2mathbb{Z} otimes mathbb{Z}_2 @>text{inclusion}>> mathbb{Z} otimes mathbb{Z}_2 @>>> mathbb{Z}_2otimes mathbb{Z}_2 @>>> 0 \
        @VVV @VVV @VVV @VVV \
        mathbb{Z}_2 @>times 2>> mathbb{Z}_2 @>>> mathbb{Z}_2 @>>>0
        end{CD}

        where the rows are exact and the vertical arrows are isomorphisms.



        That map $mathbb{Z}_2 xrightarrow{times 2} mathbb{Z}_2$ on the bottom is the zero map. To see that this must be the case, let us note that the isomorphisms $to mathbb{Z}_2$ are



        $$ begin{array}{ccc}
        2mathbb{Z} otimes mathbb{Z}_2 & longrightarrow & mathbb{Z}_2 \
        2n otimes 0 &longmapsto & 0 \
        2n otimes 1 & longmapsto & n bmod 2
        end{array}
        qquad text{and} qquad
        begin{array}{ccc}
        mathbb{Z} otimes mathbb{Z}_2 & longrightarrow & mathbb{Z}_2 \
        n otimes 0 &longmapsto & 0 \
        n otimes 1 & longmapsto & n bmod 2
        end{array} $$



        So the map on the bottom we can compute by looking at the composition $mathbb{Z}_2 to 2mathbb{Z} otimes mathbb{Z}_2 to mathbb{Z} otimes mathbb{Z}_2 to mathbb{Z}_2$ and this composition takes $1 mapsto 2 otimes 1 mapsto 2 otimes 1 mapsto 0$.






        share|cite|improve this answer












        $2mathbb{Z}otimesmathbb{Z}_2$ is not trivial. It is trivial inside of $mathbb Z otimes mathbb{Z}_2$, a subtle but important difference. The "proof" you might have in mind that it is trivial would be



        $$ 2n otimes 0 = 0 text{ and } 2n otimes 1 = n otimes 2 = 0 $$



        but $n otimes 2$ is only an element of $2mathbb{Z}otimesmathbb{Z}_2$ if $n$ is even. Otherwise, you need the ambient module $mathbb Z otimes mathbb{Z}_2$ to make sense of it.



        Secondly, for your exact sequences. Do keep in mind that the isomorphism $2mathbb{Z} to mathbb{Z}$ is a different map than the inclusion $2mathbb{Z} to mathbb{Z}$.



        So if we take the exact sequence



        $$ 2mathbb{Z} to mathbb{Z}tomathbb{Z}_2to 0 $$



        where the leftmost map is inclusion and tensor with $mathbb{Z}_2$, we get the following commutative diagram
        $require{AMScd}$
        begin{CD}
        2mathbb{Z} otimes mathbb{Z}_2 @>text{inclusion}>> mathbb{Z} otimes mathbb{Z}_2 @>>> mathbb{Z}_2otimes mathbb{Z}_2 @>>> 0 \
        @VVV @VVV @VVV @VVV \
        mathbb{Z}_2 @>times 2>> mathbb{Z}_2 @>>> mathbb{Z}_2 @>>>0
        end{CD}

        where the rows are exact and the vertical arrows are isomorphisms.



        That map $mathbb{Z}_2 xrightarrow{times 2} mathbb{Z}_2$ on the bottom is the zero map. To see that this must be the case, let us note that the isomorphisms $to mathbb{Z}_2$ are



        $$ begin{array}{ccc}
        2mathbb{Z} otimes mathbb{Z}_2 & longrightarrow & mathbb{Z}_2 \
        2n otimes 0 &longmapsto & 0 \
        2n otimes 1 & longmapsto & n bmod 2
        end{array}
        qquad text{and} qquad
        begin{array}{ccc}
        mathbb{Z} otimes mathbb{Z}_2 & longrightarrow & mathbb{Z}_2 \
        n otimes 0 &longmapsto & 0 \
        n otimes 1 & longmapsto & n bmod 2
        end{array} $$



        So the map on the bottom we can compute by looking at the composition $mathbb{Z}_2 to 2mathbb{Z} otimes mathbb{Z}_2 to mathbb{Z} otimes mathbb{Z}_2 to mathbb{Z}_2$ and this composition takes $1 mapsto 2 otimes 1 mapsto 2 otimes 1 mapsto 0$.







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        answered Nov 22 at 16:20









        Trevor Gunn

        13.9k32045




        13.9k32045






















            up vote
            0
            down vote













            $2mathbb{Z}$ and $mathbb{Z}_2$ are not isomorphic as $mathbb{Z}$-modules. Take a generator of each module and multiply it by 2.






            share|cite|improve this answer





















            • Yes, this is clear.
              – Xavier Yang
              Nov 22 at 17:16















            up vote
            0
            down vote













            $2mathbb{Z}$ and $mathbb{Z}_2$ are not isomorphic as $mathbb{Z}$-modules. Take a generator of each module and multiply it by 2.






            share|cite|improve this answer





















            • Yes, this is clear.
              – Xavier Yang
              Nov 22 at 17:16













            up vote
            0
            down vote










            up vote
            0
            down vote









            $2mathbb{Z}$ and $mathbb{Z}_2$ are not isomorphic as $mathbb{Z}$-modules. Take a generator of each module and multiply it by 2.






            share|cite|improve this answer












            $2mathbb{Z}$ and $mathbb{Z}_2$ are not isomorphic as $mathbb{Z}$-modules. Take a generator of each module and multiply it by 2.







            share|cite|improve this answer












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            answered Nov 22 at 15:54









            Joel Pereira

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            • Yes, this is clear.
              – Xavier Yang
              Nov 22 at 17:16


















            • Yes, this is clear.
              – Xavier Yang
              Nov 22 at 17:16
















            Yes, this is clear.
            – Xavier Yang
            Nov 22 at 17:16




            Yes, this is clear.
            – Xavier Yang
            Nov 22 at 17:16


















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