Hom of the direct product of $mathbb{Z}_{n}$ to the rationals is nonzero.
$begingroup$
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
asked Mar 19 '12 at 1:58
user6495user6495
1,7341518
$endgroup$
add a comment |
$begingroup$
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
asked Mar 19 '12 at 1:58
user6495user6495
1,7341518
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
asked Mar 19 '12 at 1:58
user6495user6495
1,7341518
$endgroup$
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
asked Mar 19 '12 at 1:58
user6495user6495
1,7341518
asked Mar 19 '12 at 1:58
user6495user6495
1,7341518
edited Dec 18 '18 at 15:40
user593746
asked Mar 19 '12 at 1:58
user6495user6495
1,7341518
asked Mar 19 '12 at 1:58
user6495user6495
1,7341518
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Mar 19 '12 at 4:53
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
111k7157288
$endgroup$
$begingroup$
♦ Is there another solution for the question which is not containing tensor?
$endgroup$
– karparvar
Dec 18 '16 at 18:14
add a comment |
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1 Answer
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1 Answer
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$begingroup$
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.
answered Mar 19 '12 at 4:53
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
111k7157288
$endgroup$
$begingroup$
♦ Is there another solution for the question which is not containing tensor?
$endgroup$
– karparvar
Dec 18 '16 at 18:14
add a comment |
$begingroup$
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.
answered Mar 19 '12 at 4:53
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
111k7157288
$endgroup$
$begingroup$
♦ Is there another solution for the question which is not containing tensor?
$endgroup$
– karparvar
Dec 18 '16 at 18:14
add a comment |
$begingroup$
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.
answered Mar 19 '12 at 4:53
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
111k7157288
$endgroup$
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.
answered Mar 19 '12 at 4:53
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
111k7157288
edited Mar 19 '12 at 5:12
answered Mar 19 '12 at 4:53
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
111k7157288
answered Mar 19 '12 at 4:53
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
111k7157288
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
add a comment |
$begingroup$
♦ 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
add a comment |
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$begingroup$
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