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Show that $langle,23 rangle =langle, 23 , alpha-10rangle^2 langle, 23, alpha-3 rangle$
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Let $alpha$ be a root of $x^3-x-1$ and $K=mathbbQ(alpha)$.
(i) Show that $langle,23 rangle =langle, 23 , alpha-10rangle^2 langle, 23, alpha-3 rangle$
(ii) Show that $langle,1 rangle =langle, 23, alpha-10, alpha-3 rangle $
I have shown $mathcalO_K=mathbbZ[alpha]$.
It's quite messy to simplify $langle, 23 , alpha-10rangle^2 langle, 23, alpha-3 rangle$ using $langle a,b rangle langle c, d rangle=langle ac, bd, ad, bc rangle$ unlike some quadratic extensions.
How do I show (i), (ii) using some theorems?
number-theory algebraic-number-theory
asked Jul 23 at 21:36
Andrew
31718
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up vote
4
down vote
favorite
Let $alpha$ be a root of $x^3-x-1$ and $K=mathbbQ(alpha)$.
(i) Show that $langle,23 rangle =langle, 23 , alpha-10rangle^2 langle, 23, alpha-3 rangle$
(ii) Show that $langle,1 rangle =langle, 23, alpha-10, alpha-3 rangle $
I have shown $mathcalO_K=mathbbZ[alpha]$.
It's quite messy to simplify $langle, 23 , alpha-10rangle^2 langle, 23, alpha-3 rangle$ using $langle a,b rangle langle c, d rangle=langle ac, bd, ad, bc rangle$ unlike some quadratic extensions.
How do I show (i), (ii) using some theorems?
number-theory algebraic-number-theory
asked Jul 23 at 21:36
Andrew
31718
I don't think (ii) is correct, because $(23,alpha-10,alpha-3)supset (23,-7)=(1)$.
– Takirion
Jul 23 at 22:04
True. It was indeed <1>. Thanks for catching the error.
– Andrew
Jul 23 at 22:15
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Let $alpha$ be a root of $x^3-x-1$ and $K=mathbbQ(alpha)$.
(i) Show that $langle,23 rangle =langle, 23 , alpha-10rangle^2 langle, 23, alpha-3 rangle$
(ii) Show that $langle,1 rangle =langle, 23, alpha-10, alpha-3 rangle $
I have shown $mathcalO_K=mathbbZ[alpha]$.
It's quite messy to simplify $langle, 23 , alpha-10rangle^2 langle, 23, alpha-3 rangle$ using $langle a,b rangle langle c, d rangle=langle ac, bd, ad, bc rangle$ unlike some quadratic extensions.
How do I show (i), (ii) using some theorems?
number-theory algebraic-number-theory
asked Jul 23 at 21:36
Andrew
31718
Let $alpha$ be a root of $x^3-x-1$ and $K=mathbbQ(alpha)$.
(i) Show that $langle,23 rangle =langle, 23 , alpha-10rangle^2 langle, 23, alpha-3 rangle$
(ii) Show that $langle,1 rangle =langle, 23, alpha-10, alpha-3 rangle $
I have shown $mathcalO_K=mathbbZ[alpha]$.
It's quite messy to simplify $langle, 23 , alpha-10rangle^2 langle, 23, alpha-3 rangle$ using $langle a,b rangle langle c, d rangle=langle ac, bd, ad, bc rangle$ unlike some quadratic extensions.
How do I show (i), (ii) using some theorems?
number-theory algebraic-number-theory
asked Jul 23 at 21:36
Andrew
31718
edited Jul 23 at 22:11
asked Jul 23 at 21:36
Andrew
31718
asked Jul 23 at 21:36
Andrew
31718
asked Jul 23 at 21:36
Andrew
31718
31718
I don't think (ii) is correct, because $(23,alpha-10,alpha-3)supset (23,-7)=(1)$.
– Takirion
Jul 23 at 22:04
True. It was indeed <1>. Thanks for catching the error.
– Andrew
Jul 23 at 22:15
add a comment |Â
I don't think (ii) is correct, because $(23,alpha-10,alpha-3)supset (23,-7)=(1)$.
– Takirion
Jul 23 at 22:04
True. It was indeed <1>. Thanks for catching the error.
– Andrew
Jul 23 at 22:15
I don't think (ii) is correct, because $(23,alpha-10,alpha-3)supset (23,-7)=(1)$.
– Takirion
Jul 23 at 22:04
I don't think (ii) is correct, because $(23,alpha-10,alpha-3)supset (23,-7)=(1)$.
– Takirion
Jul 23 at 22:04
True. It was indeed <1>. Thanks for catching the error.
– Andrew
Jul 23 at 22:15
True. It was indeed <1>. Thanks for catching the error.
– Andrew
Jul 23 at 22:15
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
4
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accepted
To find the ideal prime decomposition of $23mathcal O_K$ all you need is to factorize $x^3-x-1$ in $mathbbF_23$. We have the following:
$$x^3-x-1 = (x-10)^2(x-3) text in mathbbF_23$$
Hence we get $23mathcal O_K = (23,alpha-10)^2(23,alpha-3)$
This result is known as the Dedekind's Theorem and you can find more about it here.
For the second part, as noted in the comments we have that:
$$-7 = (alpha - 10) - (alpha - 3) in (23, alpha-10, alpha-3)$$
$$1 = 4cdot23 + 13 cdot (-7) in (23, alpha-10, alpha-3)$$
Hence we get that $(1) subseteq (23,alpha-10)^2(23,alpha-3)$. The other inclusion is trivial.
answered Jul 24 at 1:07
Stefan4024
28k53074
You are right, actually I could find the prime ideal factorization using the theorem. I'm wondering how you factorized $x^3-x-1$ in $mathbbF_23$. Did you do it by inspection or did you use Berlekamp’s Algorithm?
– Andrew
Jul 24 at 1:40
@Andrew in fact I used a math engine to find th factorization. Nevertheless I feel inspection should be the way to go here. You can easily find that 3 is a root and then just apply the quadratic formula upon division by $(x-3)$
– Stefan4024
Jul 24 at 10:20
I got it. Thanks a lot!
– Andrew
Jul 26 at 2:19
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
To find the ideal prime decomposition of $23mathcal O_K$ all you need is to factorize $x^3-x-1$ in $mathbbF_23$. We have the following:
$$x^3-x-1 = (x-10)^2(x-3) text in mathbbF_23$$
Hence we get $23mathcal O_K = (23,alpha-10)^2(23,alpha-3)$
This result is known as the Dedekind's Theorem and you can find more about it here.
For the second part, as noted in the comments we have that:
$$-7 = (alpha - 10) - (alpha - 3) in (23, alpha-10, alpha-3)$$
$$1 = 4cdot23 + 13 cdot (-7) in (23, alpha-10, alpha-3)$$
Hence we get that $(1) subseteq (23,alpha-10)^2(23,alpha-3)$. The other inclusion is trivial.
answered Jul 24 at 1:07
Stefan4024
28k53074
You are right, actually I could find the prime ideal factorization using the theorem. I'm wondering how you factorized $x^3-x-1$ in $mathbbF_23$. Did you do it by inspection or did you use Berlekamp’s Algorithm?
– Andrew
Jul 24 at 1:40
@Andrew in fact I used a math engine to find th factorization. Nevertheless I feel inspection should be the way to go here. You can easily find that 3 is a root and then just apply the quadratic formula upon division by $(x-3)$
– Stefan4024
Jul 24 at 10:20
I got it. Thanks a lot!
– Andrew
Jul 26 at 2:19
add a comment |Â
up vote
4
down vote
accepted
To find the ideal prime decomposition of $23mathcal O_K$ all you need is to factorize $x^3-x-1$ in $mathbbF_23$. We have the following:
$$x^3-x-1 = (x-10)^2(x-3) text in mathbbF_23$$
Hence we get $23mathcal O_K = (23,alpha-10)^2(23,alpha-3)$
This result is known as the Dedekind's Theorem and you can find more about it here.
For the second part, as noted in the comments we have that:
$$-7 = (alpha - 10) - (alpha - 3) in (23, alpha-10, alpha-3)$$
$$1 = 4cdot23 + 13 cdot (-7) in (23, alpha-10, alpha-3)$$
Hence we get that $(1) subseteq (23,alpha-10)^2(23,alpha-3)$. The other inclusion is trivial.
answered Jul 24 at 1:07
Stefan4024
28k53074
You are right, actually I could find the prime ideal factorization using the theorem. I'm wondering how you factorized $x^3-x-1$ in $mathbbF_23$. Did you do it by inspection or did you use Berlekamp’s Algorithm?
– Andrew
Jul 24 at 1:40
@Andrew in fact I used a math engine to find th factorization. Nevertheless I feel inspection should be the way to go here. You can easily find that 3 is a root and then just apply the quadratic formula upon division by $(x-3)$
– Stefan4024
Jul 24 at 10:20
I got it. Thanks a lot!
– Andrew
Jul 26 at 2:19
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
To find the ideal prime decomposition of $23mathcal O_K$ all you need is to factorize $x^3-x-1$ in $mathbbF_23$. We have the following:
$$x^3-x-1 = (x-10)^2(x-3) text in mathbbF_23$$
Hence we get $23mathcal O_K = (23,alpha-10)^2(23,alpha-3)$
This result is known as the Dedekind's Theorem and you can find more about it here.
For the second part, as noted in the comments we have that:
$$-7 = (alpha - 10) - (alpha - 3) in (23, alpha-10, alpha-3)$$
$$1 = 4cdot23 + 13 cdot (-7) in (23, alpha-10, alpha-3)$$
Hence we get that $(1) subseteq (23,alpha-10)^2(23,alpha-3)$. The other inclusion is trivial.
answered Jul 24 at 1:07
Stefan4024
28k53074
To find the ideal prime decomposition of $23mathcal O_K$ all you need is to factorize $x^3-x-1$ in $mathbbF_23$. We have the following:
$$x^3-x-1 = (x-10)^2(x-3) text in mathbbF_23$$
Hence we get $23mathcal O_K = (23,alpha-10)^2(23,alpha-3)$
This result is known as the Dedekind's Theorem and you can find more about it here.
For the second part, as noted in the comments we have that:
$$-7 = (alpha - 10) - (alpha - 3) in (23, alpha-10, alpha-3)$$
$$1 = 4cdot23 + 13 cdot (-7) in (23, alpha-10, alpha-3)$$
Hence we get that $(1) subseteq (23,alpha-10)^2(23,alpha-3)$. The other inclusion is trivial.
answered Jul 24 at 1:07
Stefan4024
28k53074
answered Jul 24 at 1:07
Stefan4024
28k53074
answered Jul 24 at 1:07
Stefan4024
28k53074
answered Jul 24 at 1:07
Stefan4024
28k53074
28k53074
You are right, actually I could find the prime ideal factorization using the theorem. I'm wondering how you factorized $x^3-x-1$ in $mathbbF_23$. Did you do it by inspection or did you use Berlekamp’s Algorithm?
– Andrew
Jul 24 at 1:40
@Andrew in fact I used a math engine to find th factorization. Nevertheless I feel inspection should be the way to go here. You can easily find that 3 is a root and then just apply the quadratic formula upon division by $(x-3)$
– Stefan4024
Jul 24 at 10:20
I got it. Thanks a lot!
– Andrew
Jul 26 at 2:19
add a comment |Â
You are right, actually I could find the prime ideal factorization using the theorem. I'm wondering how you factorized $x^3-x-1$ in $mathbbF_23$. Did you do it by inspection or did you use Berlekamp’s Algorithm?
– Andrew
Jul 24 at 1:40
@Andrew in fact I used a math engine to find th factorization. Nevertheless I feel inspection should be the way to go here. You can easily find that 3 is a root and then just apply the quadratic formula upon division by $(x-3)$
– Stefan4024
Jul 24 at 10:20
I got it. Thanks a lot!
– Andrew
Jul 26 at 2:19
You are right, actually I could find the prime ideal factorization using the theorem. I'm wondering how you factorized $x^3-x-1$ in $mathbbF_23$. Did you do it by inspection or did you use Berlekamp’s Algorithm?
– Andrew
Jul 24 at 1:40
You are right, actually I could find the prime ideal factorization using the theorem. I'm wondering how you factorized $x^3-x-1$ in $mathbbF_23$. Did you do it by inspection or did you use Berlekamp’s Algorithm?
– Andrew
Jul 24 at 1:40
@Andrew in fact I used a math engine to find th factorization. Nevertheless I feel inspection should be the way to go here. You can easily find that 3 is a root and then just apply the quadratic formula upon division by $(x-3)$
– Stefan4024
Jul 24 at 10:20
@Andrew in fact I used a math engine to find th factorization. Nevertheless I feel inspection should be the way to go here. You can easily find that 3 is a root and then just apply the quadratic formula upon division by $(x-3)$
– Stefan4024
Jul 24 at 10:20
I got it. Thanks a lot!
– Andrew
Jul 26 at 2:19
I got it. Thanks a lot!
– Andrew
Jul 26 at 2:19
add a comment |Â
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I don't think (ii) is correct, because $(23,alpha-10,alpha-3)supset (23,-7)=(1)$.
– Takirion
Jul 23 at 22:04
True. It was indeed <1>. Thanks for catching the error.
– Andrew
Jul 23 at 22:15