Mixing
Homomorphisms and Tensor Products
$begingroup$
Suppose $A$ and $B$ are finitely generated $mathbb{Z}$-modules. Then if $ operatorname{Hom}_mathbb{Z}(A,B) neq 0$ and $ operatorname{Hom}_mathbb{Z}(B,A)=0$ then $B otimes mathbb{Q} =0 $ and $Aotimes mathbb{Q} neq 0$. Can u guys help me out. I've tried using the fact that any element in the tensor product is of the form $sum botimes q$ and the properties of the module homomorphisms but I'm kinda stuck , should I use the fact they $A,B cong mathbb{Z}/(d_1)opluscdots oplus mathbb{Z}/(d_k)$ ?
abstract-algebra modules
edited Jan 27 at 22:41
Bernard
123k741117
asked Jan 27 at 22:27
Pedro SantosPedro Santos
1609
$endgroup$
add a comment |
$begingroup$
Suppose $A$ and $B$ are finitely generated $mathbb{Z}$-modules. Then if $ operatorname{Hom}_mathbb{Z}(A,B) neq 0$ and $ operatorname{Hom}_mathbb{Z}(B,A)=0$ then $B otimes mathbb{Q} =0 $ and $Aotimes mathbb{Q} neq 0$. Can u guys help me out. I've tried using the fact that any element in the tensor product is of the form $sum botimes q$ and the properties of the module homomorphisms but I'm kinda stuck , should I use the fact they $A,B cong mathbb{Z}/(d_1)opluscdots oplus mathbb{Z}/(d_k)$ ?
abstract-algebra modules
edited Jan 27 at 22:41
Bernard
123k741117
asked Jan 27 at 22:27
Pedro SantosPedro Santos
1609
$endgroup$
add a comment |
$begingroup$
Suppose $A$ and $B$ are finitely generated $mathbb{Z}$-modules. Then if $ operatorname{Hom}_mathbb{Z}(A,B) neq 0$ and $ operatorname{Hom}_mathbb{Z}(B,A)=0$ then $B otimes mathbb{Q} =0 $ and $Aotimes mathbb{Q} neq 0$. Can u guys help me out. I've tried using the fact that any element in the tensor product is of the form $sum botimes q$ and the properties of the module homomorphisms but I'm kinda stuck , should I use the fact they $A,B cong mathbb{Z}/(d_1)opluscdots oplus mathbb{Z}/(d_k)$ ?
abstract-algebra modules
edited Jan 27 at 22:41
Bernard
123k741117
asked Jan 27 at 22:27
Pedro SantosPedro Santos
1609
$endgroup$
Suppose $A$ and $B$ are finitely generated $mathbb{Z}$-modules. Then if $ operatorname{Hom}_mathbb{Z}(A,B) neq 0$ and $ operatorname{Hom}_mathbb{Z}(B,A)=0$ then $B otimes mathbb{Q} =0 $ and $Aotimes mathbb{Q} neq 0$. Can u guys help me out. I've tried using the fact that any element in the tensor product is of the form $sum botimes q$ and the properties of the module homomorphisms but I'm kinda stuck , should I use the fact they $A,B cong mathbb{Z}/(d_1)opluscdots oplus mathbb{Z}/(d_k)$ ?
abstract-algebra modules
abstract-algebra modules
edited Jan 27 at 22:41
Bernard
123k741117
asked Jan 27 at 22:27
Pedro SantosPedro Santos
1609
edited Jan 27 at 22:41
Bernard
123k741117
asked Jan 27 at 22:27
Pedro SantosPedro Santos
1609
edited Jan 27 at 22:41
Bernard
123k741117
edited Jan 27 at 22:41
Bernard
123k741117
edited Jan 27 at 22:41
Bernard
123k741117
123k741117
asked Jan 27 at 22:27
Pedro SantosPedro Santos
1609
asked Jan 27 at 22:27
Pedro SantosPedro Santos
1609
asked Jan 27 at 22:27
Pedro SantosPedro Santos
1609
1609
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Hint: You may write
$$ A cong bigoplus_{j=1}^m A_j, qquad B cong bigoplus_{k=1}^n B_k$$
where $A_j, B_j$ are either isomorphic to $mathbb{Z}^{d_j}$ or $mathbb{Z}/p_j^{r_j}mathbb{Z}$. Using
$$operatorname{Hom}_{mathbb{Z}} left(bigoplus_{j=1}^m A_j, bigoplus_{k=1}^n B_kright) cong bigoplus_{1leq j leq m, 1leq k leq n} operatorname{Hom}_{mathbb{Z}}(A_j, B_k)$$
it suffices to check when $operatorname{Hom}_{mathbb{Z}}(A_j, B_k) $ is trivial (or not).
After that we can use
$$ A otimes mathbb{Q} cong bigoplus_{j=1}^m (A_j otimes mathbb{Q}), qquad B otimes mathbb{Q} cong bigoplus_{k=1}^n (B_k otimes mathbb{Q}) $$
Furthermore, we have for $dneq 0$ (just use the fact that $[a]otimes c= [da]otimes (c/d)=0$)
$$ (mathbb{Z}/d mathbb{Z}) otimes mathbb{Q} cong 0$$
and (using $z otimes q = 1otimes (zg)$)
$$ mathbb{Z} otimes mathbb{Q} cong mathbb{Q} qquad text{and thus} qquad mathbb{Z}^d otimes mathbb{Q} cong mathbb{Q}^d.$$
edited Jan 29 at 13:49
Fabio Lucchini
9,51111426
answered Jan 27 at 23:04
Severin SchravenSeverin Schraven
6,5751935
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
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votes
$begingroup$
Hint: You may write
$$ A cong bigoplus_{j=1}^m A_j, qquad B cong bigoplus_{k=1}^n B_k$$
where $A_j, B_j$ are either isomorphic to $mathbb{Z}^{d_j}$ or $mathbb{Z}/p_j^{r_j}mathbb{Z}$. Using
$$operatorname{Hom}_{mathbb{Z}} left(bigoplus_{j=1}^m A_j, bigoplus_{k=1}^n B_kright) cong bigoplus_{1leq j leq m, 1leq k leq n} operatorname{Hom}_{mathbb{Z}}(A_j, B_k)$$
it suffices to check when $operatorname{Hom}_{mathbb{Z}}(A_j, B_k) $ is trivial (or not).
After that we can use
$$ A otimes mathbb{Q} cong bigoplus_{j=1}^m (A_j otimes mathbb{Q}), qquad B otimes mathbb{Q} cong bigoplus_{k=1}^n (B_k otimes mathbb{Q}) $$
Furthermore, we have for $dneq 0$ (just use the fact that $[a]otimes c= [da]otimes (c/d)=0$)
$$ (mathbb{Z}/d mathbb{Z}) otimes mathbb{Q} cong 0$$
and (using $z otimes q = 1otimes (zg)$)
$$ mathbb{Z} otimes mathbb{Q} cong mathbb{Q} qquad text{and thus} qquad mathbb{Z}^d otimes mathbb{Q} cong mathbb{Q}^d.$$
edited Jan 29 at 13:49
Fabio Lucchini
9,51111426
answered Jan 27 at 23:04
Severin SchravenSeverin Schraven
6,5751935
$endgroup$
add a comment |
$begingroup$
Hint: You may write
$$ A cong bigoplus_{j=1}^m A_j, qquad B cong bigoplus_{k=1}^n B_k$$
where $A_j, B_j$ are either isomorphic to $mathbb{Z}^{d_j}$ or $mathbb{Z}/p_j^{r_j}mathbb{Z}$. Using
$$operatorname{Hom}_{mathbb{Z}} left(bigoplus_{j=1}^m A_j, bigoplus_{k=1}^n B_kright) cong bigoplus_{1leq j leq m, 1leq k leq n} operatorname{Hom}_{mathbb{Z}}(A_j, B_k)$$
it suffices to check when $operatorname{Hom}_{mathbb{Z}}(A_j, B_k) $ is trivial (or not).
After that we can use
$$ A otimes mathbb{Q} cong bigoplus_{j=1}^m (A_j otimes mathbb{Q}), qquad B otimes mathbb{Q} cong bigoplus_{k=1}^n (B_k otimes mathbb{Q}) $$
Furthermore, we have for $dneq 0$ (just use the fact that $[a]otimes c= [da]otimes (c/d)=0$)
$$ (mathbb{Z}/d mathbb{Z}) otimes mathbb{Q} cong 0$$
and (using $z otimes q = 1otimes (zg)$)
$$ mathbb{Z} otimes mathbb{Q} cong mathbb{Q} qquad text{and thus} qquad mathbb{Z}^d otimes mathbb{Q} cong mathbb{Q}^d.$$
edited Jan 29 at 13:49
Fabio Lucchini
9,51111426
answered Jan 27 at 23:04
Severin SchravenSeverin Schraven
6,5751935
$endgroup$
add a comment |
$begingroup$
Hint: You may write
$$ A cong bigoplus_{j=1}^m A_j, qquad B cong bigoplus_{k=1}^n B_k$$
where $A_j, B_j$ are either isomorphic to $mathbb{Z}^{d_j}$ or $mathbb{Z}/p_j^{r_j}mathbb{Z}$. Using
$$operatorname{Hom}_{mathbb{Z}} left(bigoplus_{j=1}^m A_j, bigoplus_{k=1}^n B_kright) cong bigoplus_{1leq j leq m, 1leq k leq n} operatorname{Hom}_{mathbb{Z}}(A_j, B_k)$$
it suffices to check when $operatorname{Hom}_{mathbb{Z}}(A_j, B_k) $ is trivial (or not).
After that we can use
$$ A otimes mathbb{Q} cong bigoplus_{j=1}^m (A_j otimes mathbb{Q}), qquad B otimes mathbb{Q} cong bigoplus_{k=1}^n (B_k otimes mathbb{Q}) $$
Furthermore, we have for $dneq 0$ (just use the fact that $[a]otimes c= [da]otimes (c/d)=0$)
$$ (mathbb{Z}/d mathbb{Z}) otimes mathbb{Q} cong 0$$
and (using $z otimes q = 1otimes (zg)$)
$$ mathbb{Z} otimes mathbb{Q} cong mathbb{Q} qquad text{and thus} qquad mathbb{Z}^d otimes mathbb{Q} cong mathbb{Q}^d.$$
edited Jan 29 at 13:49
Fabio Lucchini
9,51111426
answered Jan 27 at 23:04
Severin SchravenSeverin Schraven
6,5751935
$endgroup$
Hint: You may write
$$ A cong bigoplus_{j=1}^m A_j, qquad B cong bigoplus_{k=1}^n B_k$$
where $A_j, B_j$ are either isomorphic to $mathbb{Z}^{d_j}$ or $mathbb{Z}/p_j^{r_j}mathbb{Z}$. Using
$$operatorname{Hom}_{mathbb{Z}} left(bigoplus_{j=1}^m A_j, bigoplus_{k=1}^n B_kright) cong bigoplus_{1leq j leq m, 1leq k leq n} operatorname{Hom}_{mathbb{Z}}(A_j, B_k)$$
it suffices to check when $operatorname{Hom}_{mathbb{Z}}(A_j, B_k) $ is trivial (or not).
After that we can use
$$ A otimes mathbb{Q} cong bigoplus_{j=1}^m (A_j otimes mathbb{Q}), qquad B otimes mathbb{Q} cong bigoplus_{k=1}^n (B_k otimes mathbb{Q}) $$
Furthermore, we have for $dneq 0$ (just use the fact that $[a]otimes c= [da]otimes (c/d)=0$)
$$ (mathbb{Z}/d mathbb{Z}) otimes mathbb{Q} cong 0$$
and (using $z otimes q = 1otimes (zg)$)
$$ mathbb{Z} otimes mathbb{Q} cong mathbb{Q} qquad text{and thus} qquad mathbb{Z}^d otimes mathbb{Q} cong mathbb{Q}^d.$$
edited Jan 29 at 13:49
Fabio Lucchini
9,51111426
answered Jan 27 at 23:04
Severin SchravenSeverin Schraven
6,5751935
edited Jan 29 at 13:49
Fabio Lucchini
9,51111426
edited Jan 29 at 13:49
Fabio Lucchini
9,51111426
edited Jan 29 at 13:49
Fabio Lucchini
9,51111426
9,51111426
answered Jan 27 at 23:04
Severin SchravenSeverin Schraven
6,5751935
answered Jan 27 at 23:04
Severin SchravenSeverin Schraven
6,5751935
answered Jan 27 at 23:04
Severin SchravenSeverin Schraven
6,5751935
6,5751935
add a comment |
add a comment |
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