Mixing
Factoring the ideal $(8)$ into a product of prime ideals in $mathbb{Q}(sqrt{-7})$
$begingroup$
I am trying to factor the ideal $(8)$ into a product of prime ideals in $mathbb{Q}(sqrt{-7})$.
I am not exactly sure how to go about doing this, and I feel I am missing some theory in the matter. What I have noted is that we have $$8=(1+sqrt{-7})(1-sqrt{-7})=2^3$$ and so I surmise the factorization will involve these numbers somehow. I also noted that $N(8)=64$, so the prime ideal factors will have norms that multiply out to $64.$ Unfortunately I cannot proceed further than this. Any hints or references are appreciated.
abstract-algebra algebraic-number-theory ideals factoring
edited Apr 8 '13 at 8:45
Martin Sleziak
44.8k10119272
asked Apr 8 '13 at 2:55
Coffee_TableCoffee_Table
1,98141630
$endgroup$
add a comment |
$begingroup$
I am trying to factor the ideal $(8)$ into a product of prime ideals in $mathbb{Q}(sqrt{-7})$.
I am not exactly sure how to go about doing this, and I feel I am missing some theory in the matter. What I have noted is that we have $$8=(1+sqrt{-7})(1-sqrt{-7})=2^3$$ and so I surmise the factorization will involve these numbers somehow. I also noted that $N(8)=64$, so the prime ideal factors will have norms that multiply out to $64.$ Unfortunately I cannot proceed further than this. Any hints or references are appreciated.
abstract-algebra algebraic-number-theory ideals factoring
edited Apr 8 '13 at 8:45
Martin Sleziak
44.8k10119272
asked Apr 8 '13 at 2:55
Coffee_TableCoffee_Table
1,98141630
$endgroup$
$begingroup$
I think that prime-ideals might be too specific tag, so I've changed it to the already existing tag ideals. (Feel free to edit the post again, if you have different opinion.)
$endgroup$
– Martin Sleziak
Apr 8 '13 at 8:47
$begingroup$
I'm agreed, no complaints from me.
$endgroup$
– Coffee_Table
Apr 8 '13 at 19:16
add a comment |
$begingroup$
I am trying to factor the ideal $(8)$ into a product of prime ideals in $mathbb{Q}(sqrt{-7})$.
I am not exactly sure how to go about doing this, and I feel I am missing some theory in the matter. What I have noted is that we have $$8=(1+sqrt{-7})(1-sqrt{-7})=2^3$$ and so I surmise the factorization will involve these numbers somehow. I also noted that $N(8)=64$, so the prime ideal factors will have norms that multiply out to $64.$ Unfortunately I cannot proceed further than this. Any hints or references are appreciated.
abstract-algebra algebraic-number-theory ideals factoring
edited Apr 8 '13 at 8:45
Martin Sleziak
44.8k10119272
asked Apr 8 '13 at 2:55
Coffee_TableCoffee_Table
1,98141630
$endgroup$
I am trying to factor the ideal $(8)$ into a product of prime ideals in $mathbb{Q}(sqrt{-7})$.
I am not exactly sure how to go about doing this, and I feel I am missing some theory in the matter. What I have noted is that we have $$8=(1+sqrt{-7})(1-sqrt{-7})=2^3$$ and so I surmise the factorization will involve these numbers somehow. I also noted that $N(8)=64$, so the prime ideal factors will have norms that multiply out to $64.$ Unfortunately I cannot proceed further than this. Any hints or references are appreciated.
abstract-algebra algebraic-number-theory ideals factoring
abstract-algebra algebraic-number-theory ideals factoring
edited Apr 8 '13 at 8:45
Martin Sleziak
44.8k10119272
asked Apr 8 '13 at 2:55
Coffee_TableCoffee_Table
1,98141630
edited Apr 8 '13 at 8:45
Martin Sleziak
44.8k10119272
asked Apr 8 '13 at 2:55
Coffee_TableCoffee_Table
1,98141630
edited Apr 8 '13 at 8:45
Martin Sleziak
44.8k10119272
edited Apr 8 '13 at 8:45
Martin Sleziak
44.8k10119272
edited Apr 8 '13 at 8:45
Martin Sleziak
44.8k10119272
44.8k10119272
asked Apr 8 '13 at 2:55
Coffee_TableCoffee_Table
1,98141630
asked Apr 8 '13 at 2:55
Coffee_TableCoffee_Table
1,98141630
asked Apr 8 '13 at 2:55
Coffee_TableCoffee_Table
1,98141630
1,98141630
$begingroup$
I think that prime-ideals might be too specific tag, so I've changed it to the already existing tag ideals. (Feel free to edit the post again, if you have different opinion.)
$endgroup$
– Martin Sleziak
Apr 8 '13 at 8:47
$begingroup$
I'm agreed, no complaints from me.
$endgroup$
– Coffee_Table
Apr 8 '13 at 19:16
add a comment |
$begingroup$
I think that prime-ideals might be too specific tag, so I've changed it to the already existing tag ideals. (Feel free to edit the post again, if you have different opinion.)
$endgroup$
– Martin Sleziak
Apr 8 '13 at 8:47
$begingroup$
I'm agreed, no complaints from me.
$endgroup$
– Coffee_Table
Apr 8 '13 at 19:16
$begingroup$
I think that prime-ideals might be too specific tag, so I've changed it to the already existing tag ideals. (Feel free to edit the post again, if you have different opinion.)
$endgroup$
– Martin Sleziak
Apr 8 '13 at 8:47
$begingroup$
I think that prime-ideals might be too specific tag, so I've changed it to the already existing tag ideals. (Feel free to edit the post again, if you have different opinion.)
$endgroup$
– Martin Sleziak
Apr 8 '13 at 8:47
$begingroup$
I'm agreed, no complaints from me.
$endgroup$
– Coffee_Table
Apr 8 '13 at 19:16
$begingroup$
I'm agreed, no complaints from me.
$endgroup$
– Coffee_Table
Apr 8 '13 at 19:16
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $K=mathbb{Q}(sqrt{-7})$, so that $mathcal{O}_K=mathbb{Z}[frac{1+sqrt{-7}}{2}]$ because $-7equiv 1bmod 4$. Let $(8)$ denote the ideal generated by $8$ in $mathcal{O}_K$.
Because $(8)=(2)^3$, it will suffice to determine the factorization of $(2)$ in $mathcal{O}_K$, and then the factorization of $(8)$ will be the same with the exponents multiplied by $3$.
Note that $$(tfrac{1+sqrt{-7}}{2})(tfrac{1-sqrt{-7}}{2})=(2).$$
The norm of $frac{1+sqrt{-7}}{2}$ is $$Nmathopen{big(}tfrac{1+sqrt{-7}}{2}mathclose{big)}=mathopen{big(}tfrac{1}{2}mathclose{big)}^2+7mathopen{big(}tfrac{1}{2}mathclose{big)}^2=2,$$ and similarly with $frac{1-sqrt{-7}}{2}$, so that $mathcal{O}_K/mathopen{big(}frac{1+sqrt{-7}}{2}mathclose{big)}$ and $mathcal{O}_K/mathopen{big(}frac{1-sqrt{-7}}{2}mathclose{big)}$ have cardinality $2$, and they are therefore the field $mathbb{F}_2$.
Thus the ideals $mathopen{big(}frac{1+sqrt{-7}}{2}mathclose{big)}$ and $mathopen{big(}frac{1-sqrt{-7}}{2}mathclose{big)}$ are prime, so that
$$(8)=mathopen{big(}tfrac{1+sqrt{-7}}{2}mathclose{big)}^3mathopen{big(}tfrac{1-sqrt{-7}}{2}mathclose{big)}^3.$$
answered Apr 8 '13 at 3:05
Zev ChonolesZev Chonoles
110k16228429
$endgroup$
$begingroup$
To clarify: Would this argument be equally valid if one were to drop all mention of ideals and consider the problem in thr context of $mathcal{O}_K$ itself, with the element 8?
$endgroup$
– Coffee_Table
Apr 8 '13 at 3:14
$begingroup$
@Coffee_Table Might I ask what you are referring to? Maybe you talk about idèles and adèles? Thanks in advance.
$endgroup$
– awllower
Apr 9 '13 at 0:30
$begingroup$
@awllower : I am referring to neither, just plain old ideals. Assuming you are referring to my comment right above (and not my original question), I mean to say that we could find the prime factorization of the number 8 in the ring $mathcal{O}_K$ in an almost identical manner as the one described in the answer above, rather than finding the prime factorization of the ideal $(8)$.
$endgroup$
– Coffee_Table
Apr 9 '13 at 1:22
$begingroup$
@Coffee_Table I see.Thanks for the clarification then.
$endgroup$
– awllower
Apr 9 '13 at 3:52
add a comment |
$begingroup$
As you noted, $N(8) = 64$. Also note that $N(1 pm sqrt{-7}) = 8$ and $N(2) = 4$. Since the ring of algebraic integers of $textbf Q(sqrt{-7})$ is said to be a unique factorization domain and, as you noticed, $(1 - sqrt{-7})(1 + sqrt{-7}) = 2^3 = 8$, this must mean that the two apparently distinct factorizations are in fact incomplete factorizations, just as $4^3$ would be an incomplete factorization of $64$ in $textbf Z$.
As it turns out, $$frac{1 + sqrt{-7}}{2}$$ is an algebraic integer and it belongs in this domain, since its minimal polynomial is $x^2 - x + 2$. Therefore, as numbers, we have $$left( frac{1}{2} - frac{sqrt{-7}}{2} right)^3 left( frac{1}{2} + frac{sqrt{-7}}{2} right)^3 = 8.$$
However, for ideals, we need to verify that $$leftlangle frac{1}{2} - frac{sqrt{-7}}{2} rightrangle not subseteq leftlangle frac{1}{2} + frac{sqrt{-7}}{2} rightrangle$$ nor vice-versa. A couple of divisions will quickly confirm that $langle 2 rangle$ is a splitting, not ramifying ideal. Therefore, $$langle 8 rangle = leftlangle frac{1}{2} - frac{sqrt{-7}}{2} rightrangle^3 leftlangle frac{1}{2} + frac{sqrt{-7}}{2} rightrangle^3.$$
answered Jan 18 at 22:32
David R.David R.
3961728
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $K=mathbb{Q}(sqrt{-7})$, so that $mathcal{O}_K=mathbb{Z}[frac{1+sqrt{-7}}{2}]$ because $-7equiv 1bmod 4$. Let $(8)$ denote the ideal generated by $8$ in $mathcal{O}_K$.
Because $(8)=(2)^3$, it will suffice to determine the factorization of $(2)$ in $mathcal{O}_K$, and then the factorization of $(8)$ will be the same with the exponents multiplied by $3$.
Note that $$(tfrac{1+sqrt{-7}}{2})(tfrac{1-sqrt{-7}}{2})=(2).$$
The norm of $frac{1+sqrt{-7}}{2}$ is $$Nmathopen{big(}tfrac{1+sqrt{-7}}{2}mathclose{big)}=mathopen{big(}tfrac{1}{2}mathclose{big)}^2+7mathopen{big(}tfrac{1}{2}mathclose{big)}^2=2,$$ and similarly with $frac{1-sqrt{-7}}{2}$, so that $mathcal{O}_K/mathopen{big(}frac{1+sqrt{-7}}{2}mathclose{big)}$ and $mathcal{O}_K/mathopen{big(}frac{1-sqrt{-7}}{2}mathclose{big)}$ have cardinality $2$, and they are therefore the field $mathbb{F}_2$.
Thus the ideals $mathopen{big(}frac{1+sqrt{-7}}{2}mathclose{big)}$ and $mathopen{big(}frac{1-sqrt{-7}}{2}mathclose{big)}$ are prime, so that
$$(8)=mathopen{big(}tfrac{1+sqrt{-7}}{2}mathclose{big)}^3mathopen{big(}tfrac{1-sqrt{-7}}{2}mathclose{big)}^3.$$
answered Apr 8 '13 at 3:05
Zev ChonolesZev Chonoles
110k16228429
$endgroup$
$begingroup$
To clarify: Would this argument be equally valid if one were to drop all mention of ideals and consider the problem in thr context of $mathcal{O}_K$ itself, with the element 8?
$endgroup$
– Coffee_Table
Apr 8 '13 at 3:14
$begingroup$
@Coffee_Table Might I ask what you are referring to? Maybe you talk about idèles and adèles? Thanks in advance.
$endgroup$
– awllower
Apr 9 '13 at 0:30
$begingroup$
@awllower : I am referring to neither, just plain old ideals. Assuming you are referring to my comment right above (and not my original question), I mean to say that we could find the prime factorization of the number 8 in the ring $mathcal{O}_K$ in an almost identical manner as the one described in the answer above, rather than finding the prime factorization of the ideal $(8)$.
$endgroup$
– Coffee_Table
Apr 9 '13 at 1:22
$begingroup$
@Coffee_Table I see.Thanks for the clarification then.
$endgroup$
– awllower
Apr 9 '13 at 3:52
add a comment |
$begingroup$
Let $K=mathbb{Q}(sqrt{-7})$, so that $mathcal{O}_K=mathbb{Z}[frac{1+sqrt{-7}}{2}]$ because $-7equiv 1bmod 4$. Let $(8)$ denote the ideal generated by $8$ in $mathcal{O}_K$.
Because $(8)=(2)^3$, it will suffice to determine the factorization of $(2)$ in $mathcal{O}_K$, and then the factorization of $(8)$ will be the same with the exponents multiplied by $3$.
Note that $$(tfrac{1+sqrt{-7}}{2})(tfrac{1-sqrt{-7}}{2})=(2).$$
The norm of $frac{1+sqrt{-7}}{2}$ is $$Nmathopen{big(}tfrac{1+sqrt{-7}}{2}mathclose{big)}=mathopen{big(}tfrac{1}{2}mathclose{big)}^2+7mathopen{big(}tfrac{1}{2}mathclose{big)}^2=2,$$ and similarly with $frac{1-sqrt{-7}}{2}$, so that $mathcal{O}_K/mathopen{big(}frac{1+sqrt{-7}}{2}mathclose{big)}$ and $mathcal{O}_K/mathopen{big(}frac{1-sqrt{-7}}{2}mathclose{big)}$ have cardinality $2$, and they are therefore the field $mathbb{F}_2$.
Thus the ideals $mathopen{big(}frac{1+sqrt{-7}}{2}mathclose{big)}$ and $mathopen{big(}frac{1-sqrt{-7}}{2}mathclose{big)}$ are prime, so that
$$(8)=mathopen{big(}tfrac{1+sqrt{-7}}{2}mathclose{big)}^3mathopen{big(}tfrac{1-sqrt{-7}}{2}mathclose{big)}^3.$$
answered Apr 8 '13 at 3:05
Zev ChonolesZev Chonoles
110k16228429
$endgroup$
$begingroup$
To clarify: Would this argument be equally valid if one were to drop all mention of ideals and consider the problem in thr context of $mathcal{O}_K$ itself, with the element 8?
$endgroup$
– Coffee_Table
Apr 8 '13 at 3:14
$begingroup$
@Coffee_Table Might I ask what you are referring to? Maybe you talk about idèles and adèles? Thanks in advance.
$endgroup$
– awllower
Apr 9 '13 at 0:30
$begingroup$
@awllower : I am referring to neither, just plain old ideals. Assuming you are referring to my comment right above (and not my original question), I mean to say that we could find the prime factorization of the number 8 in the ring $mathcal{O}_K$ in an almost identical manner as the one described in the answer above, rather than finding the prime factorization of the ideal $(8)$.
$endgroup$
– Coffee_Table
Apr 9 '13 at 1:22
$begingroup$
@Coffee_Table I see.Thanks for the clarification then.
$endgroup$
– awllower
Apr 9 '13 at 3:52
add a comment |
$begingroup$
Let $K=mathbb{Q}(sqrt{-7})$, so that $mathcal{O}_K=mathbb{Z}[frac{1+sqrt{-7}}{2}]$ because $-7equiv 1bmod 4$. Let $(8)$ denote the ideal generated by $8$ in $mathcal{O}_K$.
Because $(8)=(2)^3$, it will suffice to determine the factorization of $(2)$ in $mathcal{O}_K$, and then the factorization of $(8)$ will be the same with the exponents multiplied by $3$.
Note that $$(tfrac{1+sqrt{-7}}{2})(tfrac{1-sqrt{-7}}{2})=(2).$$
The norm of $frac{1+sqrt{-7}}{2}$ is $$Nmathopen{big(}tfrac{1+sqrt{-7}}{2}mathclose{big)}=mathopen{big(}tfrac{1}{2}mathclose{big)}^2+7mathopen{big(}tfrac{1}{2}mathclose{big)}^2=2,$$ and similarly with $frac{1-sqrt{-7}}{2}$, so that $mathcal{O}_K/mathopen{big(}frac{1+sqrt{-7}}{2}mathclose{big)}$ and $mathcal{O}_K/mathopen{big(}frac{1-sqrt{-7}}{2}mathclose{big)}$ have cardinality $2$, and they are therefore the field $mathbb{F}_2$.
Thus the ideals $mathopen{big(}frac{1+sqrt{-7}}{2}mathclose{big)}$ and $mathopen{big(}frac{1-sqrt{-7}}{2}mathclose{big)}$ are prime, so that
$$(8)=mathopen{big(}tfrac{1+sqrt{-7}}{2}mathclose{big)}^3mathopen{big(}tfrac{1-sqrt{-7}}{2}mathclose{big)}^3.$$
answered Apr 8 '13 at 3:05
Zev ChonolesZev Chonoles
110k16228429
$endgroup$
Let $K=mathbb{Q}(sqrt{-7})$, so that $mathcal{O}_K=mathbb{Z}[frac{1+sqrt{-7}}{2}]$ because $-7equiv 1bmod 4$. Let $(8)$ denote the ideal generated by $8$ in $mathcal{O}_K$.
Because $(8)=(2)^3$, it will suffice to determine the factorization of $(2)$ in $mathcal{O}_K$, and then the factorization of $(8)$ will be the same with the exponents multiplied by $3$.
Note that $$(tfrac{1+sqrt{-7}}{2})(tfrac{1-sqrt{-7}}{2})=(2).$$
The norm of $frac{1+sqrt{-7}}{2}$ is $$Nmathopen{big(}tfrac{1+sqrt{-7}}{2}mathclose{big)}=mathopen{big(}tfrac{1}{2}mathclose{big)}^2+7mathopen{big(}tfrac{1}{2}mathclose{big)}^2=2,$$ and similarly with $frac{1-sqrt{-7}}{2}$, so that $mathcal{O}_K/mathopen{big(}frac{1+sqrt{-7}}{2}mathclose{big)}$ and $mathcal{O}_K/mathopen{big(}frac{1-sqrt{-7}}{2}mathclose{big)}$ have cardinality $2$, and they are therefore the field $mathbb{F}_2$.
Thus the ideals $mathopen{big(}frac{1+sqrt{-7}}{2}mathclose{big)}$ and $mathopen{big(}frac{1-sqrt{-7}}{2}mathclose{big)}$ are prime, so that
$$(8)=mathopen{big(}tfrac{1+sqrt{-7}}{2}mathclose{big)}^3mathopen{big(}tfrac{1-sqrt{-7}}{2}mathclose{big)}^3.$$
answered Apr 8 '13 at 3:05
Zev ChonolesZev Chonoles
110k16228429
answered Apr 8 '13 at 3:05
Zev ChonolesZev Chonoles
110k16228429
answered Apr 8 '13 at 3:05
Zev ChonolesZev Chonoles
110k16228429
answered Apr 8 '13 at 3:05
Zev ChonolesZev Chonoles
110k16228429
110k16228429
$begingroup$
To clarify: Would this argument be equally valid if one were to drop all mention of ideals and consider the problem in thr context of $mathcal{O}_K$ itself, with the element 8?
$endgroup$
– Coffee_Table
Apr 8 '13 at 3:14
$begingroup$
@Coffee_Table Might I ask what you are referring to? Maybe you talk about idèles and adèles? Thanks in advance.
$endgroup$
– awllower
Apr 9 '13 at 0:30
$begingroup$
@awllower : I am referring to neither, just plain old ideals. Assuming you are referring to my comment right above (and not my original question), I mean to say that we could find the prime factorization of the number 8 in the ring $mathcal{O}_K$ in an almost identical manner as the one described in the answer above, rather than finding the prime factorization of the ideal $(8)$.
$endgroup$
– Coffee_Table
Apr 9 '13 at 1:22
$begingroup$
@Coffee_Table I see.Thanks for the clarification then.
$endgroup$
– awllower
Apr 9 '13 at 3:52
add a comment |
$begingroup$
To clarify: Would this argument be equally valid if one were to drop all mention of ideals and consider the problem in thr context of $mathcal{O}_K$ itself, with the element 8?
$endgroup$
– Coffee_Table
Apr 8 '13 at 3:14
$begingroup$
@Coffee_Table Might I ask what you are referring to? Maybe you talk about idèles and adèles? Thanks in advance.
$endgroup$
– awllower
Apr 9 '13 at 0:30
$begingroup$
@awllower : I am referring to neither, just plain old ideals. Assuming you are referring to my comment right above (and not my original question), I mean to say that we could find the prime factorization of the number 8 in the ring $mathcal{O}_K$ in an almost identical manner as the one described in the answer above, rather than finding the prime factorization of the ideal $(8)$.
$endgroup$
– Coffee_Table
Apr 9 '13 at 1:22
$begingroup$
@Coffee_Table I see.Thanks for the clarification then.
$endgroup$
– awllower
Apr 9 '13 at 3:52
$begingroup$
To clarify: Would this argument be equally valid if one were to drop all mention of ideals and consider the problem in thr context of $mathcal{O}_K$ itself, with the element 8?
$endgroup$
– Coffee_Table
Apr 8 '13 at 3:14
$begingroup$
To clarify: Would this argument be equally valid if one were to drop all mention of ideals and consider the problem in thr context of $mathcal{O}_K$ itself, with the element 8?
$endgroup$
– Coffee_Table
Apr 8 '13 at 3:14
$begingroup$
@Coffee_Table Might I ask what you are referring to? Maybe you talk about idèles and adèles? Thanks in advance.
$endgroup$
– awllower
Apr 9 '13 at 0:30
$begingroup$
@Coffee_Table Might I ask what you are referring to? Maybe you talk about idèles and adèles? Thanks in advance.
$endgroup$
– awllower
Apr 9 '13 at 0:30
$begingroup$
@awllower : I am referring to neither, just plain old ideals. Assuming you are referring to my comment right above (and not my original question), I mean to say that we could find the prime factorization of the number 8 in the ring $mathcal{O}_K$ in an almost identical manner as the one described in the answer above, rather than finding the prime factorization of the ideal $(8)$.
$endgroup$
– Coffee_Table
Apr 9 '13 at 1:22
$begingroup$
@awllower : I am referring to neither, just plain old ideals. Assuming you are referring to my comment right above (and not my original question), I mean to say that we could find the prime factorization of the number 8 in the ring $mathcal{O}_K$ in an almost identical manner as the one described in the answer above, rather than finding the prime factorization of the ideal $(8)$.
$endgroup$
– Coffee_Table
Apr 9 '13 at 1:22
$begingroup$
@Coffee_Table I see.Thanks for the clarification then.
$endgroup$
– awllower
Apr 9 '13 at 3:52
$begingroup$
@Coffee_Table I see.Thanks for the clarification then.
$endgroup$
– awllower
Apr 9 '13 at 3:52
add a comment |
$begingroup$
As you noted, $N(8) = 64$. Also note that $N(1 pm sqrt{-7}) = 8$ and $N(2) = 4$. Since the ring of algebraic integers of $textbf Q(sqrt{-7})$ is said to be a unique factorization domain and, as you noticed, $(1 - sqrt{-7})(1 + sqrt{-7}) = 2^3 = 8$, this must mean that the two apparently distinct factorizations are in fact incomplete factorizations, just as $4^3$ would be an incomplete factorization of $64$ in $textbf Z$.
As it turns out, $$frac{1 + sqrt{-7}}{2}$$ is an algebraic integer and it belongs in this domain, since its minimal polynomial is $x^2 - x + 2$. Therefore, as numbers, we have $$left( frac{1}{2} - frac{sqrt{-7}}{2} right)^3 left( frac{1}{2} + frac{sqrt{-7}}{2} right)^3 = 8.$$
However, for ideals, we need to verify that $$leftlangle frac{1}{2} - frac{sqrt{-7}}{2} rightrangle not subseteq leftlangle frac{1}{2} + frac{sqrt{-7}}{2} rightrangle$$ nor vice-versa. A couple of divisions will quickly confirm that $langle 2 rangle$ is a splitting, not ramifying ideal. Therefore, $$langle 8 rangle = leftlangle frac{1}{2} - frac{sqrt{-7}}{2} rightrangle^3 leftlangle frac{1}{2} + frac{sqrt{-7}}{2} rightrangle^3.$$
answered Jan 18 at 22:32
David R.David R.
3961728
$endgroup$
add a comment |
$begingroup$
As you noted, $N(8) = 64$. Also note that $N(1 pm sqrt{-7}) = 8$ and $N(2) = 4$. Since the ring of algebraic integers of $textbf Q(sqrt{-7})$ is said to be a unique factorization domain and, as you noticed, $(1 - sqrt{-7})(1 + sqrt{-7}) = 2^3 = 8$, this must mean that the two apparently distinct factorizations are in fact incomplete factorizations, just as $4^3$ would be an incomplete factorization of $64$ in $textbf Z$.
As it turns out, $$frac{1 + sqrt{-7}}{2}$$ is an algebraic integer and it belongs in this domain, since its minimal polynomial is $x^2 - x + 2$. Therefore, as numbers, we have $$left( frac{1}{2} - frac{sqrt{-7}}{2} right)^3 left( frac{1}{2} + frac{sqrt{-7}}{2} right)^3 = 8.$$
However, for ideals, we need to verify that $$leftlangle frac{1}{2} - frac{sqrt{-7}}{2} rightrangle not subseteq leftlangle frac{1}{2} + frac{sqrt{-7}}{2} rightrangle$$ nor vice-versa. A couple of divisions will quickly confirm that $langle 2 rangle$ is a splitting, not ramifying ideal. Therefore, $$langle 8 rangle = leftlangle frac{1}{2} - frac{sqrt{-7}}{2} rightrangle^3 leftlangle frac{1}{2} + frac{sqrt{-7}}{2} rightrangle^3.$$
answered Jan 18 at 22:32
David R.David R.
3961728
$endgroup$
add a comment |
$begingroup$
As you noted, $N(8) = 64$. Also note that $N(1 pm sqrt{-7}) = 8$ and $N(2) = 4$. Since the ring of algebraic integers of $textbf Q(sqrt{-7})$ is said to be a unique factorization domain and, as you noticed, $(1 - sqrt{-7})(1 + sqrt{-7}) = 2^3 = 8$, this must mean that the two apparently distinct factorizations are in fact incomplete factorizations, just as $4^3$ would be an incomplete factorization of $64$ in $textbf Z$.
As it turns out, $$frac{1 + sqrt{-7}}{2}$$ is an algebraic integer and it belongs in this domain, since its minimal polynomial is $x^2 - x + 2$. Therefore, as numbers, we have $$left( frac{1}{2} - frac{sqrt{-7}}{2} right)^3 left( frac{1}{2} + frac{sqrt{-7}}{2} right)^3 = 8.$$
However, for ideals, we need to verify that $$leftlangle frac{1}{2} - frac{sqrt{-7}}{2} rightrangle not subseteq leftlangle frac{1}{2} + frac{sqrt{-7}}{2} rightrangle$$ nor vice-versa. A couple of divisions will quickly confirm that $langle 2 rangle$ is a splitting, not ramifying ideal. Therefore, $$langle 8 rangle = leftlangle frac{1}{2} - frac{sqrt{-7}}{2} rightrangle^3 leftlangle frac{1}{2} + frac{sqrt{-7}}{2} rightrangle^3.$$
answered Jan 18 at 22:32
David R.David R.
3961728
$endgroup$
As you noted, $N(8) = 64$. Also note that $N(1 pm sqrt{-7}) = 8$ and $N(2) = 4$. Since the ring of algebraic integers of $textbf Q(sqrt{-7})$ is said to be a unique factorization domain and, as you noticed, $(1 - sqrt{-7})(1 + sqrt{-7}) = 2^3 = 8$, this must mean that the two apparently distinct factorizations are in fact incomplete factorizations, just as $4^3$ would be an incomplete factorization of $64$ in $textbf Z$.
As it turns out, $$frac{1 + sqrt{-7}}{2}$$ is an algebraic integer and it belongs in this domain, since its minimal polynomial is $x^2 - x + 2$. Therefore, as numbers, we have $$left( frac{1}{2} - frac{sqrt{-7}}{2} right)^3 left( frac{1}{2} + frac{sqrt{-7}}{2} right)^3 = 8.$$
However, for ideals, we need to verify that $$leftlangle frac{1}{2} - frac{sqrt{-7}}{2} rightrangle not subseteq leftlangle frac{1}{2} + frac{sqrt{-7}}{2} rightrangle$$ nor vice-versa. A couple of divisions will quickly confirm that $langle 2 rangle$ is a splitting, not ramifying ideal. Therefore, $$langle 8 rangle = leftlangle frac{1}{2} - frac{sqrt{-7}}{2} rightrangle^3 leftlangle frac{1}{2} + frac{sqrt{-7}}{2} rightrangle^3.$$
answered Jan 18 at 22:32
David R.David R.
3961728
answered Jan 18 at 22:32
David R.David R.
3961728
answered Jan 18 at 22:32
David R.David R.
3961728
answered Jan 18 at 22:32
David R.David R.
3961728
3961728
add a comment |
add a comment |
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I think that prime-ideals might be too specific tag, so I've changed it to the already existing tag ideals. (Feel free to edit the post again, if you have different opinion.)
$endgroup$
– Martin Sleziak
Apr 8 '13 at 8:47
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I'm agreed, no complaints from me.
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– Coffee_Table
Apr 8 '13 at 19:16