Hom of the direct product of $mathbb{Z}_{n}$ to the rationals is nonzero.












3












$begingroup$


Why is $mathrm{Hom}_{mathbb{Z}}left(prod_{n geq 2}mathbb{Z}_{n},mathbb{Q}right)$ nonzero?



Context: This is problem $2.25 (iii)$ of page $69$ Rotman's Introduction to Homological Algebra:




Prove that
$$mathrm{Hom}_{mathbb{Z}}left(prod_{n geq 2}mathbb{Z}_{n},mathbb{Q}right) ncong prod_{n geq 2}mathrm{Hom}_{mathbb{Z}}(mathbb{Z}_{n},mathbb{Q}).$$




The right hand side is $0$ because $mathbb{Z}_{n}$ is torsion and $mathbb{Q}$ is not.






abstract-algebra ring-theory modules ring-homomorphism ring-isomorphism







share|cite|improve this question











$endgroup$












  • $begingroup$
    Haven't worked it out completely (so I'm not sure that this actually works) but maybe sounds like a job for Zorn's Lemma? Show that the set of partial homomorphisms ${rm Hom}(S, mathbb{Q})$ where $S$ ranges over all subgroups of $prod_{n ge 2} mathbb{Z}_n$, satisfies the hypotheses of Zorn's Lemma, and then show that any partially defined map can be extended to a larger subgroup.
    $endgroup$
    – Ted
    Mar 19 '12 at 4:36












  • $begingroup$
    @Ted: Thanks, just added a comment.
    $endgroup$
    – user6495
    Mar 19 '12 at 4:48










  • $begingroup$
    sorry, my suggestion obviously doesn't work... I was confused and thinking $mathbb{Q}/mathbb{Z}$ instead of $mathbb{Q}$.
    $endgroup$
    – Ted
    Mar 19 '12 at 7:50
















3












$begingroup$


Why is $mathrm{Hom}_{mathbb{Z}}left(prod_{n geq 2}mathbb{Z}_{n},mathbb{Q}right)$ nonzero?



Context: This is problem $2.25 (iii)$ of page $69$ Rotman's Introduction to Homological Algebra:




Prove that
$$mathrm{Hom}_{mathbb{Z}}left(prod_{n geq 2}mathbb{Z}_{n},mathbb{Q}right) ncong prod_{n geq 2}mathrm{Hom}_{mathbb{Z}}(mathbb{Z}_{n},mathbb{Q}).$$




The right hand side is $0$ because $mathbb{Z}_{n}$ is torsion and $mathbb{Q}$ is not.






abstract-algebra ring-theory modules ring-homomorphism ring-isomorphism







share|cite|improve this question











$endgroup$












  • $begingroup$
    Haven't worked it out completely (so I'm not sure that this actually works) but maybe sounds like a job for Zorn's Lemma? Show that the set of partial homomorphisms ${rm Hom}(S, mathbb{Q})$ where $S$ ranges over all subgroups of $prod_{n ge 2} mathbb{Z}_n$, satisfies the hypotheses of Zorn's Lemma, and then show that any partially defined map can be extended to a larger subgroup.
    $endgroup$
    – Ted
    Mar 19 '12 at 4:36












  • $begingroup$
    @Ted: Thanks, just added a comment.
    $endgroup$
    – user6495
    Mar 19 '12 at 4:48










  • $begingroup$
    sorry, my suggestion obviously doesn't work... I was confused and thinking $mathbb{Q}/mathbb{Z}$ instead of $mathbb{Q}$.
    $endgroup$
    – Ted
    Mar 19 '12 at 7:50














3












3








3


1



$begingroup$


Why is $mathrm{Hom}_{mathbb{Z}}left(prod_{n geq 2}mathbb{Z}_{n},mathbb{Q}right)$ nonzero?



Context: This is problem $2.25 (iii)$ of page $69$ Rotman's Introduction to Homological Algebra:




Prove that
$$mathrm{Hom}_{mathbb{Z}}left(prod_{n geq 2}mathbb{Z}_{n},mathbb{Q}right) ncong prod_{n geq 2}mathrm{Hom}_{mathbb{Z}}(mathbb{Z}_{n},mathbb{Q}).$$




The right hand side is $0$ because $mathbb{Z}_{n}$ is torsion and $mathbb{Q}$ is not.






abstract-algebra ring-theory modules ring-homomorphism ring-isomorphism







share|cite|improve this question











$endgroup$




Why is $mathrm{Hom}_{mathbb{Z}}left(prod_{n geq 2}mathbb{Z}_{n},mathbb{Q}right)$ nonzero?



Context: This is problem $2.25 (iii)$ of page $69$ Rotman's Introduction to Homological Algebra:




Prove that
$$mathrm{Hom}_{mathbb{Z}}left(prod_{n geq 2}mathbb{Z}_{n},mathbb{Q}right) ncong prod_{n geq 2}mathrm{Hom}_{mathbb{Z}}(mathbb{Z}_{n},mathbb{Q}).$$




The right hand side is $0$ because $mathbb{Z}_{n}$ is torsion and $mathbb{Q}$ is not.






abstract-algebra ring-theory modules ring-homomorphism ring-isomorphism




abstract-algebra ring-theory modules ring-homomorphism ring-isomorphism






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 18 '18 at 15:40







user593746

















asked Mar 19 '12 at 1:58









user6495user6495

1,7341518




1,7341518












  • $begingroup$
    Haven't worked it out completely (so I'm not sure that this actually works) but maybe sounds like a job for Zorn's Lemma? Show that the set of partial homomorphisms ${rm Hom}(S, mathbb{Q})$ where $S$ ranges over all subgroups of $prod_{n ge 2} mathbb{Z}_n$, satisfies the hypotheses of Zorn's Lemma, and then show that any partially defined map can be extended to a larger subgroup.
    $endgroup$
    – Ted
    Mar 19 '12 at 4:36












  • $begingroup$
    @Ted: Thanks, just added a comment.
    $endgroup$
    – user6495
    Mar 19 '12 at 4:48










  • $begingroup$
    sorry, my suggestion obviously doesn't work... I was confused and thinking $mathbb{Q}/mathbb{Z}$ instead of $mathbb{Q}$.
    $endgroup$
    – Ted
    Mar 19 '12 at 7:50


















  • $begingroup$
    Haven't worked it out completely (so I'm not sure that this actually works) but maybe sounds like a job for Zorn's Lemma? Show that the set of partial homomorphisms ${rm Hom}(S, mathbb{Q})$ where $S$ ranges over all subgroups of $prod_{n ge 2} mathbb{Z}_n$, satisfies the hypotheses of Zorn's Lemma, and then show that any partially defined map can be extended to a larger subgroup.
    $endgroup$
    – Ted
    Mar 19 '12 at 4:36












  • $begingroup$
    @Ted: Thanks, just added a comment.
    $endgroup$
    – user6495
    Mar 19 '12 at 4:48










  • $begingroup$
    sorry, my suggestion obviously doesn't work... I was confused and thinking $mathbb{Q}/mathbb{Z}$ instead of $mathbb{Q}$.
    $endgroup$
    – Ted
    Mar 19 '12 at 7:50
















$begingroup$
Haven't worked it out completely (so I'm not sure that this actually works) but maybe sounds like a job for Zorn's Lemma? Show that the set of partial homomorphisms ${rm Hom}(S, mathbb{Q})$ where $S$ ranges over all subgroups of $prod_{n ge 2} mathbb{Z}_n$, satisfies the hypotheses of Zorn's Lemma, and then show that any partially defined map can be extended to a larger subgroup.
$endgroup$
– Ted
Mar 19 '12 at 4:36






$begingroup$
Haven't worked it out completely (so I'm not sure that this actually works) but maybe sounds like a job for Zorn's Lemma? Show that the set of partial homomorphisms ${rm Hom}(S, mathbb{Q})$ where $S$ ranges over all subgroups of $prod_{n ge 2} mathbb{Z}_n$, satisfies the hypotheses of Zorn's Lemma, and then show that any partially defined map can be extended to a larger subgroup.
$endgroup$
– Ted
Mar 19 '12 at 4:36














$begingroup$
@Ted: Thanks, just added a comment.
$endgroup$
– user6495
Mar 19 '12 at 4:48




$begingroup$
@Ted: Thanks, just added a comment.
$endgroup$
– user6495
Mar 19 '12 at 4:48












$begingroup$
sorry, my suggestion obviously doesn't work... I was confused and thinking $mathbb{Q}/mathbb{Z}$ instead of $mathbb{Q}$.
$endgroup$
– Ted
Mar 19 '12 at 7:50




$begingroup$
sorry, my suggestion obviously doesn't work... I was confused and thinking $mathbb{Q}/mathbb{Z}$ instead of $mathbb{Q}$.
$endgroup$
– Ted
Mar 19 '12 at 7:50










1 Answer
1






active

oldest

votes


















4












$begingroup$

Let $G=prod_{ngeq2}mathbb Z_n$ and let $t(G)$ be the torsion subgroup, which is properly contained in $G$ (the element $(1,1,1,dots)$ is not in $t(G)$, for example) Then $G/t(G)$ is a torsion-free abelian group, which therefore embeds into its localization $(G/t(G))otimes_{mathbb Z}mathbb Q$, which is a non-zero rational vector space, and in fact generates it as a vector space. There is a non-zero $mathbb Q$-linear map $(G/t(G))otimes_{mathbb Z}mathbb Qtomathbb Q$ (here the Choice Police will observe that we are using the axiom of choice, of course...). Composing, we get a non-zero morphism $$Gto G/t(G)to (G/t(G))otimes_{mathbb Z}mathbb Qtomathbb Q.$$



Remark. If $H$ is a torsion-free abelian group, its finitely generated subgroups are free, so flat. Since $H$ is the colimit of its finitely generated subgroups, it is itself flat, and tensoring the exact sequence $0tomathbb Ztomathbb Q$ with $H$ gives an exact sequence $0to Hotimes_{mathbb Z}mathbb Z=Hto Hotimes_{mathbb Z}mathbb Q$. Doing this for $H=G/t(G)$ shows $G/t(G)$ embeds in $(G/t(G))otimes_{mathbb Z}mathbb Q$, as claimed above.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    ♦ Is there another solution for the question which is not containing tensor?
    $endgroup$
    – karparvar
    Dec 18 '16 at 18:14











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









4












$begingroup$

Let $G=prod_{ngeq2}mathbb Z_n$ and let $t(G)$ be the torsion subgroup, which is properly contained in $G$ (the element $(1,1,1,dots)$ is not in $t(G)$, for example) Then $G/t(G)$ is a torsion-free abelian group, which therefore embeds into its localization $(G/t(G))otimes_{mathbb Z}mathbb Q$, which is a non-zero rational vector space, and in fact generates it as a vector space. There is a non-zero $mathbb Q$-linear map $(G/t(G))otimes_{mathbb Z}mathbb Qtomathbb Q$ (here the Choice Police will observe that we are using the axiom of choice, of course...). Composing, we get a non-zero morphism $$Gto G/t(G)to (G/t(G))otimes_{mathbb Z}mathbb Qtomathbb Q.$$



Remark. If $H$ is a torsion-free abelian group, its finitely generated subgroups are free, so flat. Since $H$ is the colimit of its finitely generated subgroups, it is itself flat, and tensoring the exact sequence $0tomathbb Ztomathbb Q$ with $H$ gives an exact sequence $0to Hotimes_{mathbb Z}mathbb Z=Hto Hotimes_{mathbb Z}mathbb Q$. Doing this for $H=G/t(G)$ shows $G/t(G)$ embeds in $(G/t(G))otimes_{mathbb Z}mathbb Q$, as claimed above.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    ♦ Is there another solution for the question which is not containing tensor?
    $endgroup$
    – karparvar
    Dec 18 '16 at 18:14
















4












$begingroup$

Let $G=prod_{ngeq2}mathbb Z_n$ and let $t(G)$ be the torsion subgroup, which is properly contained in $G$ (the element $(1,1,1,dots)$ is not in $t(G)$, for example) Then $G/t(G)$ is a torsion-free abelian group, which therefore embeds into its localization $(G/t(G))otimes_{mathbb Z}mathbb Q$, which is a non-zero rational vector space, and in fact generates it as a vector space. There is a non-zero $mathbb Q$-linear map $(G/t(G))otimes_{mathbb Z}mathbb Qtomathbb Q$ (here the Choice Police will observe that we are using the axiom of choice, of course...). Composing, we get a non-zero morphism $$Gto G/t(G)to (G/t(G))otimes_{mathbb Z}mathbb Qtomathbb Q.$$



Remark. If $H$ is a torsion-free abelian group, its finitely generated subgroups are free, so flat. Since $H$ is the colimit of its finitely generated subgroups, it is itself flat, and tensoring the exact sequence $0tomathbb Ztomathbb Q$ with $H$ gives an exact sequence $0to Hotimes_{mathbb Z}mathbb Z=Hto Hotimes_{mathbb Z}mathbb Q$. Doing this for $H=G/t(G)$ shows $G/t(G)$ embeds in $(G/t(G))otimes_{mathbb Z}mathbb Q$, as claimed above.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    ♦ Is there another solution for the question which is not containing tensor?
    $endgroup$
    – karparvar
    Dec 18 '16 at 18:14














4












4








4





$begingroup$

Let $G=prod_{ngeq2}mathbb Z_n$ and let $t(G)$ be the torsion subgroup, which is properly contained in $G$ (the element $(1,1,1,dots)$ is not in $t(G)$, for example) Then $G/t(G)$ is a torsion-free abelian group, which therefore embeds into its localization $(G/t(G))otimes_{mathbb Z}mathbb Q$, which is a non-zero rational vector space, and in fact generates it as a vector space. There is a non-zero $mathbb Q$-linear map $(G/t(G))otimes_{mathbb Z}mathbb Qtomathbb Q$ (here the Choice Police will observe that we are using the axiom of choice, of course...). Composing, we get a non-zero morphism $$Gto G/t(G)to (G/t(G))otimes_{mathbb Z}mathbb Qtomathbb Q.$$



Remark. If $H$ is a torsion-free abelian group, its finitely generated subgroups are free, so flat. Since $H$ is the colimit of its finitely generated subgroups, it is itself flat, and tensoring the exact sequence $0tomathbb Ztomathbb Q$ with $H$ gives an exact sequence $0to Hotimes_{mathbb Z}mathbb Z=Hto Hotimes_{mathbb Z}mathbb Q$. Doing this for $H=G/t(G)$ shows $G/t(G)$ embeds in $(G/t(G))otimes_{mathbb Z}mathbb Q$, as claimed above.






share|cite|improve this answer











$endgroup$



Let $G=prod_{ngeq2}mathbb Z_n$ and let $t(G)$ be the torsion subgroup, which is properly contained in $G$ (the element $(1,1,1,dots)$ is not in $t(G)$, for example) Then $G/t(G)$ is a torsion-free abelian group, which therefore embeds into its localization $(G/t(G))otimes_{mathbb Z}mathbb Q$, which is a non-zero rational vector space, and in fact generates it as a vector space. There is a non-zero $mathbb Q$-linear map $(G/t(G))otimes_{mathbb Z}mathbb Qtomathbb Q$ (here the Choice Police will observe that we are using the axiom of choice, of course...). Composing, we get a non-zero morphism $$Gto G/t(G)to (G/t(G))otimes_{mathbb Z}mathbb Qtomathbb Q.$$



Remark. If $H$ is a torsion-free abelian group, its finitely generated subgroups are free, so flat. Since $H$ is the colimit of its finitely generated subgroups, it is itself flat, and tensoring the exact sequence $0tomathbb Ztomathbb Q$ with $H$ gives an exact sequence $0to Hotimes_{mathbb Z}mathbb Z=Hto Hotimes_{mathbb Z}mathbb Q$. Doing this for $H=G/t(G)$ shows $G/t(G)$ embeds in $(G/t(G))otimes_{mathbb Z}mathbb Q$, as claimed above.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Mar 19 '12 at 5:12

























answered Mar 19 '12 at 4:53









Mariano Suárez-ÁlvarezMariano Suárez-Álvarez

111k7157288




111k7157288












  • $begingroup$
    ♦ Is there another solution for the question which is not containing tensor?
    $endgroup$
    – karparvar
    Dec 18 '16 at 18:14


















  • $begingroup$
    ♦ Is there another solution for the question which is not containing tensor?
    $endgroup$
    – karparvar
    Dec 18 '16 at 18:14
















$begingroup$
♦ Is there another solution for the question which is not containing tensor?
$endgroup$
– karparvar
Dec 18 '16 at 18:14




$begingroup$
♦ Is there another solution for the question which is not containing tensor?
$endgroup$
– karparvar
Dec 18 '16 at 18:14


















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