General form of $mathbb{Q}otimes M$ as $mathbb{Z}$-modules.
2
$begingroup$
Compute the following tensor products (as $mathbb{Z}$ modules):
$mathbb{Q}otimes mathbb{Z}/(n)$.
$mathbb{Q}otimesmathbb{Q}$.
$mathbb{Q}otimesmathbb{Q}/mathbb{Z}$.
$mathbb{Q}/mathbb{Z}otimes mathbb{Q}/mathbb{Z} $.
$mathbb{Q}otimes M$ where $M$ is any $mathbb{Z}$-module.
Here is what I have so far:
- It is clear that $mathbb{Q}otimes mathbb{Z}/(n)$ is not cyclic of order at most $n$ (but what is it?).
- The generators of this tensor product look like:
$$frac{p}{m}otimesfrac{q}{n}=pqleft(frac{1}{m}otimesfrac{1}{n}right) $$
where $m,n,p,qinmathbb{Z}$, $m,nneq 0$. Honestly, I can't recognize a usual group with the following property.
$mathbb{Q}/mathbb{Z}$ is a module where each element has finite order, and so we can expect that every element in the tensor product to have finite order.- My guess for this one is that $mathbb{Q}/mathbb{Z}otimes mathbb{Q}/mathbb{Z}cong mathbb{Q}/mathbb{Z}$ but only intuitively.
- Since $M$ is any module, what can we do?
I don't need a precise answer, just hints, except for the last part. What can we do when $M$ is any $mathbb{Z}$-module.
linear-algebra tensor-products
asked Dec 10 '18 at 18:49
UserAUserA
542216
$endgroup$
add a comment |
2
$begingroup$
Compute the following tensor products (as $mathbb{Z}$ modules):
$mathbb{Q}otimes mathbb{Z}/(n)$.
$mathbb{Q}otimesmathbb{Q}$.
$mathbb{Q}otimesmathbb{Q}/mathbb{Z}$.
$mathbb{Q}/mathbb{Z}otimes mathbb{Q}/mathbb{Z} $.
$mathbb{Q}otimes M$ where $M$ is any $mathbb{Z}$-module.
Here is what I have so far:
- It is clear that $mathbb{Q}otimes mathbb{Z}/(n)$ is not cyclic of order at most $n$ (but what is it?).
- The generators of this tensor product look like:
$$frac{p}{m}otimesfrac{q}{n}=pqleft(frac{1}{m}otimesfrac{1}{n}right) $$
where $m,n,p,qinmathbb{Z}$, $m,nneq 0$. Honestly, I can't recognize a usual group with the following property.
$mathbb{Q}/mathbb{Z}$ is a module where each element has finite order, and so we can expect that every element in the tensor product to have finite order.- My guess for this one is that $mathbb{Q}/mathbb{Z}otimes mathbb{Q}/mathbb{Z}cong mathbb{Q}/mathbb{Z}$ but only intuitively.
- Since $M$ is any module, what can we do?
I don't need a precise answer, just hints, except for the last part. What can we do when $M$ is any $mathbb{Z}$-module.
linear-algebra tensor-products
asked Dec 10 '18 at 18:49
UserAUserA
542216
$endgroup$
add a comment |
2
2
2
$begingroup$
Compute the following tensor products (as $mathbb{Z}$ modules):
$mathbb{Q}otimes mathbb{Z}/(n)$.
$mathbb{Q}otimesmathbb{Q}$.
$mathbb{Q}otimesmathbb{Q}/mathbb{Z}$.
$mathbb{Q}/mathbb{Z}otimes mathbb{Q}/mathbb{Z} $.
$mathbb{Q}otimes M$ where $M$ is any $mathbb{Z}$-module.
Here is what I have so far:
- It is clear that $mathbb{Q}otimes mathbb{Z}/(n)$ is not cyclic of order at most $n$ (but what is it?).
- The generators of this tensor product look like:
$$frac{p}{m}otimesfrac{q}{n}=pqleft(frac{1}{m}otimesfrac{1}{n}right) $$
where $m,n,p,qinmathbb{Z}$, $m,nneq 0$. Honestly, I can't recognize a usual group with the following property.
$mathbb{Q}/mathbb{Z}$ is a module where each element has finite order, and so we can expect that every element in the tensor product to have finite order.- My guess for this one is that $mathbb{Q}/mathbb{Z}otimes mathbb{Q}/mathbb{Z}cong mathbb{Q}/mathbb{Z}$ but only intuitively.
- Since $M$ is any module, what can we do?
I don't need a precise answer, just hints, except for the last part. What can we do when $M$ is any $mathbb{Z}$-module.
linear-algebra tensor-products
asked Dec 10 '18 at 18:49
UserAUserA
542216
$endgroup$
Compute the following tensor products (as $mathbb{Z}$ modules):
$mathbb{Q}otimes mathbb{Z}/(n)$.
$mathbb{Q}otimesmathbb{Q}$.
$mathbb{Q}otimesmathbb{Q}/mathbb{Z}$.
$mathbb{Q}/mathbb{Z}otimes mathbb{Q}/mathbb{Z} $.
$mathbb{Q}otimes M$ where $M$ is any $mathbb{Z}$-module.
Here is what I have so far:
- It is clear that $mathbb{Q}otimes mathbb{Z}/(n)$ is not cyclic of order at most $n$ (but what is it?).
- The generators of this tensor product look like:
$$frac{p}{m}otimesfrac{q}{n}=pqleft(frac{1}{m}otimesfrac{1}{n}right) $$
where $m,n,p,qinmathbb{Z}$, $m,nneq 0$. Honestly, I can't recognize a usual group with the following property.
$mathbb{Q}/mathbb{Z}$ is a module where each element has finite order, and so we can expect that every element in the tensor product to have finite order.- My guess for this one is that $mathbb{Q}/mathbb{Z}otimes mathbb{Q}/mathbb{Z}cong mathbb{Q}/mathbb{Z}$ but only intuitively.
- Since $M$ is any module, what can we do?
I don't need a precise answer, just hints, except for the last part. What can we do when $M$ is any $mathbb{Z}$-module.
linear-algebra tensor-products
linear-algebra tensor-products
asked Dec 10 '18 at 18:49
UserAUserA
542216
asked Dec 10 '18 at 18:49
UserAUserA
542216
asked Dec 10 '18 at 18:49
UserAUserA
542216
asked Dec 10 '18 at 18:49
UserAUserA
542216
asked Dec 10 '18 at 18:49
UserAUserA
542216
542216
add a comment |
add a comment |
1 Answer
1
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1
$begingroup$
Some hints to get you started:
- For $Bbb Q otimes Bbb Z/(n)$, note that
$$
q otimes k = nfrac qn otimes k = frac qn otimes nk = frac qn otimes 0
$$ - For $p,q in Bbb Q$, we have $p otimes q = (pq) otimes 1$
- For $p in Bbb Q$ and $[q] in Bbb Q/Bbb Z$, note that $$
p otimes [frac ab] = (b frac pb) otimes [frac ab] = frac pb
otimes [a] = 0$$
- Does the same trick work for 4?
- Look for the common pattern in parts 1,2,3.
answered Dec 10 '18 at 19:36
OmnomnomnomOmnomnomnom
128k791184
$endgroup$
add a comment |
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1 Answer
1
active
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votes
1 Answer
1
active
oldest
votes
active
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1
$begingroup$
Some hints to get you started:
- For $Bbb Q otimes Bbb Z/(n)$, note that
$$
q otimes k = nfrac qn otimes k = frac qn otimes nk = frac qn otimes 0
$$ - For $p,q in Bbb Q$, we have $p otimes q = (pq) otimes 1$
- For $p in Bbb Q$ and $[q] in Bbb Q/Bbb Z$, note that $$
p otimes [frac ab] = (b frac pb) otimes [frac ab] = frac pb
otimes [a] = 0$$
- Does the same trick work for 4?
- Look for the common pattern in parts 1,2,3.
answered Dec 10 '18 at 19:36
OmnomnomnomOmnomnomnom
128k791184
$endgroup$
add a comment |
1
$begingroup$
Some hints to get you started:
- For $Bbb Q otimes Bbb Z/(n)$, note that
$$
q otimes k = nfrac qn otimes k = frac qn otimes nk = frac qn otimes 0
$$ - For $p,q in Bbb Q$, we have $p otimes q = (pq) otimes 1$
- For $p in Bbb Q$ and $[q] in Bbb Q/Bbb Z$, note that $$
p otimes [frac ab] = (b frac pb) otimes [frac ab] = frac pb
otimes [a] = 0$$
- Does the same trick work for 4?
- Look for the common pattern in parts 1,2,3.
answered Dec 10 '18 at 19:36
OmnomnomnomOmnomnomnom
128k791184
$endgroup$
add a comment |
1
1
1
$begingroup$
Some hints to get you started:
- For $Bbb Q otimes Bbb Z/(n)$, note that
$$
q otimes k = nfrac qn otimes k = frac qn otimes nk = frac qn otimes 0
$$ - For $p,q in Bbb Q$, we have $p otimes q = (pq) otimes 1$
- For $p in Bbb Q$ and $[q] in Bbb Q/Bbb Z$, note that $$
p otimes [frac ab] = (b frac pb) otimes [frac ab] = frac pb
otimes [a] = 0$$
- Does the same trick work for 4?
- Look for the common pattern in parts 1,2,3.
answered Dec 10 '18 at 19:36
OmnomnomnomOmnomnomnom
128k791184
$endgroup$
Some hints to get you started:
- For $Bbb Q otimes Bbb Z/(n)$, note that
$$
q otimes k = nfrac qn otimes k = frac qn otimes nk = frac qn otimes 0
$$ - For $p,q in Bbb Q$, we have $p otimes q = (pq) otimes 1$
- For $p in Bbb Q$ and $[q] in Bbb Q/Bbb Z$, note that $$
p otimes [frac ab] = (b frac pb) otimes [frac ab] = frac pb
otimes [a] = 0$$
- Does the same trick work for 4?
- Look for the common pattern in parts 1,2,3.
answered Dec 10 '18 at 19:36
OmnomnomnomOmnomnomnom
128k791184
edited Dec 10 '18 at 19:44
answered Dec 10 '18 at 19:36
OmnomnomnomOmnomnomnom
128k791184
answered Dec 10 '18 at 19:36
OmnomnomnomOmnomnomnom
128k791184
answered Dec 10 '18 at 19:36
OmnomnomnomOmnomnomnom
128k791184
128k791184
add a comment |
add a comment |
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