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How to count the number of elements in $(mathbbZ[i]/I^2014)otimes_mathbbZ[i](mathbbZ[i]/J^2014)$?
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Let $I,Junlhd mathbbZ[i]$ be the principal ideals generated by $7-i$ and $6i-7$, respectively. Find the number of elements in the $mathbbZ[i]$-module $A=(mathbbZ[i]/I^2014)otimes_mathbbZ[i](mathbbZ[i]/J^2014)$.
I know that $Acong mathbbZ[i]/(I^2014+J^2014)$, but I'm not sure where to go from here. I don't think there is anything special about the number 2014 here (the problem is from a past algebra exam question from that year), so I'm inclined to think that there is some trick I'm missing that could be applied for integers other than 2014 as well. My initial thought was to somehow use the fundamental theorem for modules over PID's in order to write $A$ as a direct sum, which would allow us to count the elements, but I can't quite get it to work. Any hints would be much appreciated!
commutative-algebra modules ideals tensor-products gaussian-integers
asked Aug 6 at 16:58
Arbutus
511515
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up vote
3
down vote
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Let $I,Junlhd mathbbZ[i]$ be the principal ideals generated by $7-i$ and $6i-7$, respectively. Find the number of elements in the $mathbbZ[i]$-module $A=(mathbbZ[i]/I^2014)otimes_mathbbZ[i](mathbbZ[i]/J^2014)$.
I know that $Acong mathbbZ[i]/(I^2014+J^2014)$, but I'm not sure where to go from here. I don't think there is anything special about the number 2014 here (the problem is from a past algebra exam question from that year), so I'm inclined to think that there is some trick I'm missing that could be applied for integers other than 2014 as well. My initial thought was to somehow use the fundamental theorem for modules over PID's in order to write $A$ as a direct sum, which would allow us to count the elements, but I can't quite get it to work. Any hints would be much appreciated!
commutative-algebra modules ideals tensor-products gaussian-integers
asked Aug 6 at 16:58
Arbutus
511515
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $I,Junlhd mathbbZ[i]$ be the principal ideals generated by $7-i$ and $6i-7$, respectively. Find the number of elements in the $mathbbZ[i]$-module $A=(mathbbZ[i]/I^2014)otimes_mathbbZ[i](mathbbZ[i]/J^2014)$.
I know that $Acong mathbbZ[i]/(I^2014+J^2014)$, but I'm not sure where to go from here. I don't think there is anything special about the number 2014 here (the problem is from a past algebra exam question from that year), so I'm inclined to think that there is some trick I'm missing that could be applied for integers other than 2014 as well. My initial thought was to somehow use the fundamental theorem for modules over PID's in order to write $A$ as a direct sum, which would allow us to count the elements, but I can't quite get it to work. Any hints would be much appreciated!
commutative-algebra modules ideals tensor-products gaussian-integers
asked Aug 6 at 16:58
Arbutus
511515
Let $I,Junlhd mathbbZ[i]$ be the principal ideals generated by $7-i$ and $6i-7$, respectively. Find the number of elements in the $mathbbZ[i]$-module $A=(mathbbZ[i]/I^2014)otimes_mathbbZ[i](mathbbZ[i]/J^2014)$.
I know that $Acong mathbbZ[i]/(I^2014+J^2014)$, but I'm not sure where to go from here. I don't think there is anything special about the number 2014 here (the problem is from a past algebra exam question from that year), so I'm inclined to think that there is some trick I'm missing that could be applied for integers other than 2014 as well. My initial thought was to somehow use the fundamental theorem for modules over PID's in order to write $A$ as a direct sum, which would allow us to count the elements, but I can't quite get it to work. Any hints would be much appreciated!
commutative-algebra modules ideals tensor-products gaussian-integers
asked Aug 6 at 16:58
Arbutus
511515
edited Aug 6 at 17:19
asked Aug 6 at 16:58
Arbutus
511515
asked Aug 6 at 16:58
Arbutus
511515
asked Aug 6 at 16:58
Arbutus
511515
511515
add a comment |Â
add a comment |Â
1 Answer
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Hint. for principal ideals $(a), (b)$ in a UFD, we have $(a)+(b) = (gcd(a,b))$.
Further hint: by calculating the norm of a Gaussian integer, one can guess what its prime factorization could be.
$$
Further hint: Depending on how much algebraic number theory you know, recall or prove using the smith normal form that the cardinality of $mathbb Z[i]/(a)$ equals the norm $N(a)$.
Solution.
$N(7-i)=50$ so its prime factors can only have norm $2$, $5$ or $25$. That is, they are $1+i$ and $5$, or $1+i$ and two prime divisors of $5$. It is not hard to see that $7-i neq pm i cdot(1+i)cdot5$, so it has to be of the form $pm i cdot (1+i) p_1^a_1p_2^a_2$, where $p_1,p_2=2pm i$ and $a_1+a_2=2$. By checking all possibilities, one finds $7-i = (1+i)(2-i)^2$. Similarly, $6i-7=i(2-i)(1+4i)$. Thus $I+J = (2-i)$, and the cardinality we're looking for is $N(2-i)^2014=5^2014$.
answered Aug 6 at 17:11
barto
13k32479
Hm, I think I understand most of your solution. Can you elaborate a bit more on how you found the factorizations of $7-i$ and $6i-7$? Also, shouldn't the final norm be 5^2014?
– Arbutus
Aug 10 at 22:39
1
I added a few more details. You are right about the final norm.
– barto
Aug 11 at 7:21
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Hint. for principal ideals $(a), (b)$ in a UFD, we have $(a)+(b) = (gcd(a,b))$.
Further hint: by calculating the norm of a Gaussian integer, one can guess what its prime factorization could be.
$$
Further hint: Depending on how much algebraic number theory you know, recall or prove using the smith normal form that the cardinality of $mathbb Z[i]/(a)$ equals the norm $N(a)$.
Solution.
$N(7-i)=50$ so its prime factors can only have norm $2$, $5$ or $25$. That is, they are $1+i$ and $5$, or $1+i$ and two prime divisors of $5$. It is not hard to see that $7-i neq pm i cdot(1+i)cdot5$, so it has to be of the form $pm i cdot (1+i) p_1^a_1p_2^a_2$, where $p_1,p_2=2pm i$ and $a_1+a_2=2$. By checking all possibilities, one finds $7-i = (1+i)(2-i)^2$. Similarly, $6i-7=i(2-i)(1+4i)$. Thus $I+J = (2-i)$, and the cardinality we're looking for is $N(2-i)^2014=5^2014$.
answered Aug 6 at 17:11
barto
13k32479
Hm, I think I understand most of your solution. Can you elaborate a bit more on how you found the factorizations of $7-i$ and $6i-7$? Also, shouldn't the final norm be 5^2014?
– Arbutus
Aug 10 at 22:39
1
I added a few more details. You are right about the final norm.
– barto
Aug 11 at 7:21
add a comment |Â
up vote
2
down vote
accepted
Hint. for principal ideals $(a), (b)$ in a UFD, we have $(a)+(b) = (gcd(a,b))$.
Further hint: by calculating the norm of a Gaussian integer, one can guess what its prime factorization could be.
$$
Further hint: Depending on how much algebraic number theory you know, recall or prove using the smith normal form that the cardinality of $mathbb Z[i]/(a)$ equals the norm $N(a)$.
Solution.
$N(7-i)=50$ so its prime factors can only have norm $2$, $5$ or $25$. That is, they are $1+i$ and $5$, or $1+i$ and two prime divisors of $5$. It is not hard to see that $7-i neq pm i cdot(1+i)cdot5$, so it has to be of the form $pm i cdot (1+i) p_1^a_1p_2^a_2$, where $p_1,p_2=2pm i$ and $a_1+a_2=2$. By checking all possibilities, one finds $7-i = (1+i)(2-i)^2$. Similarly, $6i-7=i(2-i)(1+4i)$. Thus $I+J = (2-i)$, and the cardinality we're looking for is $N(2-i)^2014=5^2014$.
answered Aug 6 at 17:11
barto
13k32479
Hm, I think I understand most of your solution. Can you elaborate a bit more on how you found the factorizations of $7-i$ and $6i-7$? Also, shouldn't the final norm be 5^2014?
– Arbutus
Aug 10 at 22:39
1
I added a few more details. You are right about the final norm.
– barto
Aug 11 at 7:21
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Hint. for principal ideals $(a), (b)$ in a UFD, we have $(a)+(b) = (gcd(a,b))$.
Further hint: by calculating the norm of a Gaussian integer, one can guess what its prime factorization could be.
$$
Further hint: Depending on how much algebraic number theory you know, recall or prove using the smith normal form that the cardinality of $mathbb Z[i]/(a)$ equals the norm $N(a)$.
Solution.
$N(7-i)=50$ so its prime factors can only have norm $2$, $5$ or $25$. That is, they are $1+i$ and $5$, or $1+i$ and two prime divisors of $5$. It is not hard to see that $7-i neq pm i cdot(1+i)cdot5$, so it has to be of the form $pm i cdot (1+i) p_1^a_1p_2^a_2$, where $p_1,p_2=2pm i$ and $a_1+a_2=2$. By checking all possibilities, one finds $7-i = (1+i)(2-i)^2$. Similarly, $6i-7=i(2-i)(1+4i)$. Thus $I+J = (2-i)$, and the cardinality we're looking for is $N(2-i)^2014=5^2014$.
answered Aug 6 at 17:11
barto
13k32479
Hint. for principal ideals $(a), (b)$ in a UFD, we have $(a)+(b) = (gcd(a,b))$.
Further hint: by calculating the norm of a Gaussian integer, one can guess what its prime factorization could be.
$$
Further hint: Depending on how much algebraic number theory you know, recall or prove using the smith normal form that the cardinality of $mathbb Z[i]/(a)$ equals the norm $N(a)$.
Solution.
$N(7-i)=50$ so its prime factors can only have norm $2$, $5$ or $25$. That is, they are $1+i$ and $5$, or $1+i$ and two prime divisors of $5$. It is not hard to see that $7-i neq pm i cdot(1+i)cdot5$, so it has to be of the form $pm i cdot (1+i) p_1^a_1p_2^a_2$, where $p_1,p_2=2pm i$ and $a_1+a_2=2$. By checking all possibilities, one finds $7-i = (1+i)(2-i)^2$. Similarly, $6i-7=i(2-i)(1+4i)$. Thus $I+J = (2-i)$, and the cardinality we're looking for is $N(2-i)^2014=5^2014$.
answered Aug 6 at 17:11
barto
13k32479
edited Aug 11 at 7:21
answered Aug 6 at 17:11
barto
13k32479
answered Aug 6 at 17:11
barto
13k32479
answered Aug 6 at 17:11
barto
13k32479
13k32479
Hm, I think I understand most of your solution. Can you elaborate a bit more on how you found the factorizations of $7-i$ and $6i-7$? Also, shouldn't the final norm be 5^2014?
– Arbutus
Aug 10 at 22:39
1
I added a few more details. You are right about the final norm.
– barto
Aug 11 at 7:21
add a comment |Â
Hm, I think I understand most of your solution. Can you elaborate a bit more on how you found the factorizations of $7-i$ and $6i-7$? Also, shouldn't the final norm be 5^2014?
– Arbutus
Aug 10 at 22:39
1
I added a few more details. You are right about the final norm.
– barto
Aug 11 at 7:21
Hm, I think I understand most of your solution. Can you elaborate a bit more on how you found the factorizations of $7-i$ and $6i-7$? Also, shouldn't the final norm be 5^2014?
– Arbutus
Aug 10 at 22:39
Hm, I think I understand most of your solution. Can you elaborate a bit more on how you found the factorizations of $7-i$ and $6i-7$? Also, shouldn't the final norm be 5^2014?
– Arbutus
Aug 10 at 22:39
1
1
I added a few more details. You are right about the final norm.
– barto
Aug 11 at 7:21
I added a few more details. You are right about the final norm.
– barto
Aug 11 at 7:21
add a comment |Â
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