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The Galois group acts on prime factors of an ideal of an integer ring as a permutation?
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Let $K/mathbb{Q}$ be a number field extension with ring of integers $mathcal{O}_K$. Let $Gal(K/mathbb{Q})$ be the Galois group of this extension, and for a rational prime $pinmathbb{Z}$ consider the ideal $p mathcal{O}_K$. It can be shown that $pmathcal{O}_K$ factorizes as a product of prime ideals in $mathcal{O}_K$, ie we have $$pmathcal{O}_K = prod_{i=1}^g mathfrak{p}_i^{e_{mathfrak{p}_i}}$$
This factorization is unique up to permutation of terms. I have a couple of questions regarding the action of $Gal(K/mathbb{Q})$ on the set of prime ideals $S= {mathfrak{p}_i: i =1,...,g}$:
- Why does the Galois group give a group action on this set?
- Assuming that we have a group action is it true that its elements act as permutations of S?
For (1), consider a prime $pinmathbb{Z}$ and let $mathfrak{p}$ be a prime over p in $mathcal{O}_K$. Take any $sigma in Gal(K/mathbb{Q})$, then $sigma(mathfrak{p})capmathbb{Z} subset mathbb{Z}$, and so $sigma(mathfrak{p}) cap mathbb{Z} = mathfrak{p} cap mathbb{Z}$ as $sigma$ fixes $mathbb{Q}$. Hence we see that $sigma(mathfrak{p})$ is a prime ideal in $mathcal{O}_K$ over $pinmathbb{Z}$. How to proceed from here?
For (2) it would suffice to show that any $sigma in Gal(K/mathbb{Q})$ is injective. So suppose that $mathfrak{p}_i neq mathfrak{p}_j$ are two prime ideals of $mathcal{O}_K$, such that $sigma(mathfrak{p}_i) = sigma(mathfrak{p}_j)$. Then as $sigma(mathfrak{p}_i)$ is an ideal in $mathcal{O}_K$, and so it is a fractional ideal, an inverse ideal $sigma(mathfrak{p}_i)^{-1}$ exists. Multiplying the above equation by this inverse gives $mathcal{O}_K = sigma(mathfrak{p}_i) sigma(mathfrak{p}_i)^{-1} = sigma(mathfrak{p}_j)sigma(mathfrak{p}_i)^{-1}$. Now as $sigma$ commutes with products, it also commutes with inverses, and so $sigma(mathfrak{p}_i)^{-1} = sigma(mathfrak{p}_i^{-1})$. Hence we find $mathcal{O}_K = sigma(mathfrak{p}_j mathfrak{p}_i^{-1})$. Now, I am a little confused if I can conclude from here that $mathfrak{p}_jmathfrak{p}_i^{-1} = mathcal{O}_K$?
Are my above two attempts correct? How can I finish them?
number-theory algebraic-number-theory
asked Jan 27 at 14:26
gengen
4722521
$endgroup$
add a comment |
$begingroup$
Let $K/mathbb{Q}$ be a number field extension with ring of integers $mathcal{O}_K$. Let $Gal(K/mathbb{Q})$ be the Galois group of this extension, and for a rational prime $pinmathbb{Z}$ consider the ideal $p mathcal{O}_K$. It can be shown that $pmathcal{O}_K$ factorizes as a product of prime ideals in $mathcal{O}_K$, ie we have $$pmathcal{O}_K = prod_{i=1}^g mathfrak{p}_i^{e_{mathfrak{p}_i}}$$
This factorization is unique up to permutation of terms. I have a couple of questions regarding the action of $Gal(K/mathbb{Q})$ on the set of prime ideals $S= {mathfrak{p}_i: i =1,...,g}$:
- Why does the Galois group give a group action on this set?
- Assuming that we have a group action is it true that its elements act as permutations of S?
For (1), consider a prime $pinmathbb{Z}$ and let $mathfrak{p}$ be a prime over p in $mathcal{O}_K$. Take any $sigma in Gal(K/mathbb{Q})$, then $sigma(mathfrak{p})capmathbb{Z} subset mathbb{Z}$, and so $sigma(mathfrak{p}) cap mathbb{Z} = mathfrak{p} cap mathbb{Z}$ as $sigma$ fixes $mathbb{Q}$. Hence we see that $sigma(mathfrak{p})$ is a prime ideal in $mathcal{O}_K$ over $pinmathbb{Z}$. How to proceed from here?
For (2) it would suffice to show that any $sigma in Gal(K/mathbb{Q})$ is injective. So suppose that $mathfrak{p}_i neq mathfrak{p}_j$ are two prime ideals of $mathcal{O}_K$, such that $sigma(mathfrak{p}_i) = sigma(mathfrak{p}_j)$. Then as $sigma(mathfrak{p}_i)$ is an ideal in $mathcal{O}_K$, and so it is a fractional ideal, an inverse ideal $sigma(mathfrak{p}_i)^{-1}$ exists. Multiplying the above equation by this inverse gives $mathcal{O}_K = sigma(mathfrak{p}_i) sigma(mathfrak{p}_i)^{-1} = sigma(mathfrak{p}_j)sigma(mathfrak{p}_i)^{-1}$. Now as $sigma$ commutes with products, it also commutes with inverses, and so $sigma(mathfrak{p}_i)^{-1} = sigma(mathfrak{p}_i^{-1})$. Hence we find $mathcal{O}_K = sigma(mathfrak{p}_j mathfrak{p}_i^{-1})$. Now, I am a little confused if I can conclude from here that $mathfrak{p}_jmathfrak{p}_i^{-1} = mathcal{O}_K$?
Are my above two attempts correct? How can I finish them?
number-theory algebraic-number-theory
asked Jan 27 at 14:26
gengen
4722521
$endgroup$
$begingroup$
All of the elements of $operatorname{Gal}(K/Bbb Q)$ are injective by definition. You need to prove that the action of this group on the primes of $K$ is transitive.
$endgroup$
– ÍgjøgnumMeg
Jan 27 at 15:00
$begingroup$
@ÍgjøgnumMeg, why do I need to show transitivity?
$endgroup$
– gen
Jan 27 at 15:09
$begingroup$
Because a transitive action of a group $G$ on a set $X$ is one such that for every $x, y in X$ there exists $g in G$ such that $gx = y$. In this case you show that for every prime $mathfrak{p}_i$ there is a $sigma in operatorname{Gal}(K/Bbb Q)$ such that $sigma mathfrak{p}_i = mathfrak{p}_j$
$endgroup$
– ÍgjøgnumMeg
Jan 27 at 15:11
$begingroup$
Any time a group acts on a set, it permutes the elements of that set. So 2 is redundant.
$endgroup$
– RghtHndSd
Jan 27 at 15:56
$begingroup$
Thanks @RghtHndSd I see it now, this makes complete sense.
$endgroup$
– gen
Jan 27 at 16:01
add a comment |
$begingroup$
Let $K/mathbb{Q}$ be a number field extension with ring of integers $mathcal{O}_K$. Let $Gal(K/mathbb{Q})$ be the Galois group of this extension, and for a rational prime $pinmathbb{Z}$ consider the ideal $p mathcal{O}_K$. It can be shown that $pmathcal{O}_K$ factorizes as a product of prime ideals in $mathcal{O}_K$, ie we have $$pmathcal{O}_K = prod_{i=1}^g mathfrak{p}_i^{e_{mathfrak{p}_i}}$$
This factorization is unique up to permutation of terms. I have a couple of questions regarding the action of $Gal(K/mathbb{Q})$ on the set of prime ideals $S= {mathfrak{p}_i: i =1,...,g}$:
- Why does the Galois group give a group action on this set?
- Assuming that we have a group action is it true that its elements act as permutations of S?
For (1), consider a prime $pinmathbb{Z}$ and let $mathfrak{p}$ be a prime over p in $mathcal{O}_K$. Take any $sigma in Gal(K/mathbb{Q})$, then $sigma(mathfrak{p})capmathbb{Z} subset mathbb{Z}$, and so $sigma(mathfrak{p}) cap mathbb{Z} = mathfrak{p} cap mathbb{Z}$ as $sigma$ fixes $mathbb{Q}$. Hence we see that $sigma(mathfrak{p})$ is a prime ideal in $mathcal{O}_K$ over $pinmathbb{Z}$. How to proceed from here?
For (2) it would suffice to show that any $sigma in Gal(K/mathbb{Q})$ is injective. So suppose that $mathfrak{p}_i neq mathfrak{p}_j$ are two prime ideals of $mathcal{O}_K$, such that $sigma(mathfrak{p}_i) = sigma(mathfrak{p}_j)$. Then as $sigma(mathfrak{p}_i)$ is an ideal in $mathcal{O}_K$, and so it is a fractional ideal, an inverse ideal $sigma(mathfrak{p}_i)^{-1}$ exists. Multiplying the above equation by this inverse gives $mathcal{O}_K = sigma(mathfrak{p}_i) sigma(mathfrak{p}_i)^{-1} = sigma(mathfrak{p}_j)sigma(mathfrak{p}_i)^{-1}$. Now as $sigma$ commutes with products, it also commutes with inverses, and so $sigma(mathfrak{p}_i)^{-1} = sigma(mathfrak{p}_i^{-1})$. Hence we find $mathcal{O}_K = sigma(mathfrak{p}_j mathfrak{p}_i^{-1})$. Now, I am a little confused if I can conclude from here that $mathfrak{p}_jmathfrak{p}_i^{-1} = mathcal{O}_K$?
Are my above two attempts correct? How can I finish them?
number-theory algebraic-number-theory
asked Jan 27 at 14:26
gengen
4722521
$endgroup$
Let $K/mathbb{Q}$ be a number field extension with ring of integers $mathcal{O}_K$. Let $Gal(K/mathbb{Q})$ be the Galois group of this extension, and for a rational prime $pinmathbb{Z}$ consider the ideal $p mathcal{O}_K$. It can be shown that $pmathcal{O}_K$ factorizes as a product of prime ideals in $mathcal{O}_K$, ie we have $$pmathcal{O}_K = prod_{i=1}^g mathfrak{p}_i^{e_{mathfrak{p}_i}}$$
This factorization is unique up to permutation of terms. I have a couple of questions regarding the action of $Gal(K/mathbb{Q})$ on the set of prime ideals $S= {mathfrak{p}_i: i =1,...,g}$:
- Why does the Galois group give a group action on this set?
- Assuming that we have a group action is it true that its elements act as permutations of S?
For (1), consider a prime $pinmathbb{Z}$ and let $mathfrak{p}$ be a prime over p in $mathcal{O}_K$. Take any $sigma in Gal(K/mathbb{Q})$, then $sigma(mathfrak{p})capmathbb{Z} subset mathbb{Z}$, and so $sigma(mathfrak{p}) cap mathbb{Z} = mathfrak{p} cap mathbb{Z}$ as $sigma$ fixes $mathbb{Q}$. Hence we see that $sigma(mathfrak{p})$ is a prime ideal in $mathcal{O}_K$ over $pinmathbb{Z}$. How to proceed from here?
For (2) it would suffice to show that any $sigma in Gal(K/mathbb{Q})$ is injective. So suppose that $mathfrak{p}_i neq mathfrak{p}_j$ are two prime ideals of $mathcal{O}_K$, such that $sigma(mathfrak{p}_i) = sigma(mathfrak{p}_j)$. Then as $sigma(mathfrak{p}_i)$ is an ideal in $mathcal{O}_K$, and so it is a fractional ideal, an inverse ideal $sigma(mathfrak{p}_i)^{-1}$ exists. Multiplying the above equation by this inverse gives $mathcal{O}_K = sigma(mathfrak{p}_i) sigma(mathfrak{p}_i)^{-1} = sigma(mathfrak{p}_j)sigma(mathfrak{p}_i)^{-1}$. Now as $sigma$ commutes with products, it also commutes with inverses, and so $sigma(mathfrak{p}_i)^{-1} = sigma(mathfrak{p}_i^{-1})$. Hence we find $mathcal{O}_K = sigma(mathfrak{p}_j mathfrak{p}_i^{-1})$. Now, I am a little confused if I can conclude from here that $mathfrak{p}_jmathfrak{p}_i^{-1} = mathcal{O}_K$?
Are my above two attempts correct? How can I finish them?
number-theory algebraic-number-theory
number-theory algebraic-number-theory
asked Jan 27 at 14:26
gengen
4722521
asked Jan 27 at 14:26
gengen
4722521
asked Jan 27 at 14:26
gengen
4722521
asked Jan 27 at 14:26
gengen
4722521
asked Jan 27 at 14:26
gengen
4722521
4722521
$begingroup$
All of the elements of $operatorname{Gal}(K/Bbb Q)$ are injective by definition. You need to prove that the action of this group on the primes of $K$ is transitive.
$endgroup$
– ÍgjøgnumMeg
Jan 27 at 15:00
$begingroup$
@ÍgjøgnumMeg, why do I need to show transitivity?
$endgroup$
– gen
Jan 27 at 15:09
$begingroup$
Because a transitive action of a group $G$ on a set $X$ is one such that for every $x, y in X$ there exists $g in G$ such that $gx = y$. In this case you show that for every prime $mathfrak{p}_i$ there is a $sigma in operatorname{Gal}(K/Bbb Q)$ such that $sigma mathfrak{p}_i = mathfrak{p}_j$
$endgroup$
– ÍgjøgnumMeg
Jan 27 at 15:11
$begingroup$
Any time a group acts on a set, it permutes the elements of that set. So 2 is redundant.
$endgroup$
– RghtHndSd
Jan 27 at 15:56
$begingroup$
Thanks @RghtHndSd I see it now, this makes complete sense.
$endgroup$
– gen
Jan 27 at 16:01
add a comment |
$begingroup$
All of the elements of $operatorname{Gal}(K/Bbb Q)$ are injective by definition. You need to prove that the action of this group on the primes of $K$ is transitive.
$endgroup$
– ÍgjøgnumMeg
Jan 27 at 15:00
$begingroup$
@ÍgjøgnumMeg, why do I need to show transitivity?
$endgroup$
– gen
Jan 27 at 15:09
$begingroup$
Because a transitive action of a group $G$ on a set $X$ is one such that for every $x, y in X$ there exists $g in G$ such that $gx = y$. In this case you show that for every prime $mathfrak{p}_i$ there is a $sigma in operatorname{Gal}(K/Bbb Q)$ such that $sigma mathfrak{p}_i = mathfrak{p}_j$
$endgroup$
– ÍgjøgnumMeg
Jan 27 at 15:11
$begingroup$
Any time a group acts on a set, it permutes the elements of that set. So 2 is redundant.
$endgroup$
– RghtHndSd
Jan 27 at 15:56
$begingroup$
Thanks @RghtHndSd I see it now, this makes complete sense.
$endgroup$
– gen
Jan 27 at 16:01
$begingroup$
All of the elements of $operatorname{Gal}(K/Bbb Q)$ are injective by definition. You need to prove that the action of this group on the primes of $K$ is transitive.
$endgroup$
– ÍgjøgnumMeg
Jan 27 at 15:00
$begingroup$
All of the elements of $operatorname{Gal}(K/Bbb Q)$ are injective by definition. You need to prove that the action of this group on the primes of $K$ is transitive.
$endgroup$
– ÍgjøgnumMeg
Jan 27 at 15:00
$begingroup$
@ÍgjøgnumMeg, why do I need to show transitivity?
$endgroup$
– gen
Jan 27 at 15:09
$begingroup$
@ÍgjøgnumMeg, why do I need to show transitivity?
$endgroup$
– gen
Jan 27 at 15:09
$begingroup$
Because a transitive action of a group $G$ on a set $X$ is one such that for every $x, y in X$ there exists $g in G$ such that $gx = y$. In this case you show that for every prime $mathfrak{p}_i$ there is a $sigma in operatorname{Gal}(K/Bbb Q)$ such that $sigma mathfrak{p}_i = mathfrak{p}_j$
$endgroup$
– ÍgjøgnumMeg
Jan 27 at 15:11
$begingroup$
Because a transitive action of a group $G$ on a set $X$ is one such that for every $x, y in X$ there exists $g in G$ such that $gx = y$. In this case you show that for every prime $mathfrak{p}_i$ there is a $sigma in operatorname{Gal}(K/Bbb Q)$ such that $sigma mathfrak{p}_i = mathfrak{p}_j$
$endgroup$
– ÍgjøgnumMeg
Jan 27 at 15:11
$begingroup$
Any time a group acts on a set, it permutes the elements of that set. So 2 is redundant.
$endgroup$
– RghtHndSd
Jan 27 at 15:56
$begingroup$
Any time a group acts on a set, it permutes the elements of that set. So 2 is redundant.
$endgroup$
– RghtHndSd
Jan 27 at 15:56
$begingroup$
Thanks @RghtHndSd I see it now, this makes complete sense.
$endgroup$
– gen
Jan 27 at 16:01
$begingroup$
Thanks @RghtHndSd I see it now, this makes complete sense.
$endgroup$
– gen
Jan 27 at 16:01
add a comment |
1 Answer
1
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$begingroup$
For part $1$: you have a group action given by $cdot : operatorname{Gal}(K/Bbb Q) times S to S : sigma cdot mathfrak{p}_i = sigma(mathfrak{p}_i)$.
Clearly $operatorname{id}cdot mathfrak{p}_i = mathfrak{p}_i$ and given $sigma_1, sigma_2 in operatorname{Gal}(K/Bbb Q)$ you have $(sigma_1 circ sigma_2) cdot mathfrak{p}_i = sigma_1(sigma_2(mathfrak{p}_i))$. Note that $sigma_2(mathfrak{p}_i)in S$ because if $ab in sigma_2(mathfrak{p}_i)$ then $sigma_2^{-1}(ab) in mathfrak{p}_i$. Since $sigma_2$ is a homomorphism we have $sigma_2^{-1}(a)sigma_2^{-1}(b) in mathfrak{p}_i$, and since this is a prime ideal we have $sigma_2^{-1}(a) in mathfrak{p}_i$ or $sigma_2^{-1}(b) in mathfrak{p}_i$, and hence $a in sigma_2(mathfrak{p}_i)$ or $b in sigma_2(mathfrak{p}_i)$.
For part 2: we show that this action is transitive on the primes of $K$ lying over $p$ (i.e. that the action of $operatorname{Gal}(K/Bbb Q)$ just moves the primes around).
Suppose that $mathfrak{p}_i$ and $mathfrak{p}_j$ are not conjugate by an element of $operatorname{Gal}(K/Bbb Q)$ so that for all $sigma$, $sigma(mathfrak{p}_i) neq mathfrak{p}_j$. By the Chinese remainder theorem we can find a $beta in mathfrak{p}_i$ such that $beta notin mathfrak{p}_j$ for any $j neq i$. Let $b = Nbeta = prod_{sigma in operatorname{Gal}(K/Bbb Q)} sigma(beta)$. Then $b in Bbb Z cap mathfrak{p}_i = pmathcal{O}_K$. Now, for all $sigma in operatorname{Gal}(K/Bbb Q)$ we have $beta notin sigma^{-1}(mathfrak{p}_j)$ and so $sigma(beta) notin mathfrak{p}_j$. But we showed that $b in pmathcal{O}_K subset mathfrak{p}_j$ (since $mathfrak{p}_j mid p$) and since $b$ is a product of the $sigma(beta)$, this contradicts the primality of $mathfrak{p}_j$, so it must be that there is some $sigma in operatorname{Gal}(K/Bbb Q)$ such that $sigma(mathfrak{p}_i) = mathfrak{p}_j$.
answered Jan 27 at 15:39
ÍgjøgnumMegÍgjøgnumMeg
2,86011129
$endgroup$
$begingroup$
I might be wrong, but I'm not convinced that for part 2 you proved the right thing. Transitivity states something about the entire Galois group, whereas I only require that elements of it are permutations on S, ie. that's a condition on individual elems. Also, I'd find it helpful if you could spell out the 'standard' part of statement 1.
$endgroup$
– gen
Jan 27 at 15:53
$begingroup$
An action is just a homomorphism $G to operatorname{Sym}(S)$ so it gives you a permutation anyway. I'm just showing you in this answer that the action of the group on the primes of $K$ "moves all the primes around".
$endgroup$
– ÍgjøgnumMeg
Jan 27 at 15:59
$begingroup$
Thanks for the clarification. I'm still a little confused why $sigma(mathfrak{p_i})in S$. You showed that it is an ideal, but that doesn't mean that $sigma(mathfrak{p_i}) = mathfrak{p}_j$ for some j.
$endgroup$
– gen
Jan 27 at 16:21
$begingroup$
That part just shows that $sigma(mathfrak{p}_i)$ is a prime ideal. That $sigma(mathfrak{p}_i) = mathfrak{p}_j$ is shown in part 2.
$endgroup$
– ÍgjøgnumMeg
Jan 27 at 16:25
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
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votes
$begingroup$
For part $1$: you have a group action given by $cdot : operatorname{Gal}(K/Bbb Q) times S to S : sigma cdot mathfrak{p}_i = sigma(mathfrak{p}_i)$.
Clearly $operatorname{id}cdot mathfrak{p}_i = mathfrak{p}_i$ and given $sigma_1, sigma_2 in operatorname{Gal}(K/Bbb Q)$ you have $(sigma_1 circ sigma_2) cdot mathfrak{p}_i = sigma_1(sigma_2(mathfrak{p}_i))$. Note that $sigma_2(mathfrak{p}_i)in S$ because if $ab in sigma_2(mathfrak{p}_i)$ then $sigma_2^{-1}(ab) in mathfrak{p}_i$. Since $sigma_2$ is a homomorphism we have $sigma_2^{-1}(a)sigma_2^{-1}(b) in mathfrak{p}_i$, and since this is a prime ideal we have $sigma_2^{-1}(a) in mathfrak{p}_i$ or $sigma_2^{-1}(b) in mathfrak{p}_i$, and hence $a in sigma_2(mathfrak{p}_i)$ or $b in sigma_2(mathfrak{p}_i)$.
For part 2: we show that this action is transitive on the primes of $K$ lying over $p$ (i.e. that the action of $operatorname{Gal}(K/Bbb Q)$ just moves the primes around).
Suppose that $mathfrak{p}_i$ and $mathfrak{p}_j$ are not conjugate by an element of $operatorname{Gal}(K/Bbb Q)$ so that for all $sigma$, $sigma(mathfrak{p}_i) neq mathfrak{p}_j$. By the Chinese remainder theorem we can find a $beta in mathfrak{p}_i$ such that $beta notin mathfrak{p}_j$ for any $j neq i$. Let $b = Nbeta = prod_{sigma in operatorname{Gal}(K/Bbb Q)} sigma(beta)$. Then $b in Bbb Z cap mathfrak{p}_i = pmathcal{O}_K$. Now, for all $sigma in operatorname{Gal}(K/Bbb Q)$ we have $beta notin sigma^{-1}(mathfrak{p}_j)$ and so $sigma(beta) notin mathfrak{p}_j$. But we showed that $b in pmathcal{O}_K subset mathfrak{p}_j$ (since $mathfrak{p}_j mid p$) and since $b$ is a product of the $sigma(beta)$, this contradicts the primality of $mathfrak{p}_j$, so it must be that there is some $sigma in operatorname{Gal}(K/Bbb Q)$ such that $sigma(mathfrak{p}_i) = mathfrak{p}_j$.
answered Jan 27 at 15:39
ÍgjøgnumMegÍgjøgnumMeg
2,86011129
$endgroup$
$begingroup$
I might be wrong, but I'm not convinced that for part 2 you proved the right thing. Transitivity states something about the entire Galois group, whereas I only require that elements of it are permutations on S, ie. that's a condition on individual elems. Also, I'd find it helpful if you could spell out the 'standard' part of statement 1.
$endgroup$
– gen
Jan 27 at 15:53
$begingroup$
An action is just a homomorphism $G to operatorname{Sym}(S)$ so it gives you a permutation anyway. I'm just showing you in this answer that the action of the group on the primes of $K$ "moves all the primes around".
$endgroup$
– ÍgjøgnumMeg
Jan 27 at 15:59
$begingroup$
Thanks for the clarification. I'm still a little confused why $sigma(mathfrak{p_i})in S$. You showed that it is an ideal, but that doesn't mean that $sigma(mathfrak{p_i}) = mathfrak{p}_j$ for some j.
$endgroup$
– gen
Jan 27 at 16:21
$begingroup$
That part just shows that $sigma(mathfrak{p}_i)$ is a prime ideal. That $sigma(mathfrak{p}_i) = mathfrak{p}_j$ is shown in part 2.
$endgroup$
– ÍgjøgnumMeg
Jan 27 at 16:25
add a comment |
$begingroup$
For part $1$: you have a group action given by $cdot : operatorname{Gal}(K/Bbb Q) times S to S : sigma cdot mathfrak{p}_i = sigma(mathfrak{p}_i)$.
Clearly $operatorname{id}cdot mathfrak{p}_i = mathfrak{p}_i$ and given $sigma_1, sigma_2 in operatorname{Gal}(K/Bbb Q)$ you have $(sigma_1 circ sigma_2) cdot mathfrak{p}_i = sigma_1(sigma_2(mathfrak{p}_i))$. Note that $sigma_2(mathfrak{p}_i)in S$ because if $ab in sigma_2(mathfrak{p}_i)$ then $sigma_2^{-1}(ab) in mathfrak{p}_i$. Since $sigma_2$ is a homomorphism we have $sigma_2^{-1}(a)sigma_2^{-1}(b) in mathfrak{p}_i$, and since this is a prime ideal we have $sigma_2^{-1}(a) in mathfrak{p}_i$ or $sigma_2^{-1}(b) in mathfrak{p}_i$, and hence $a in sigma_2(mathfrak{p}_i)$ or $b in sigma_2(mathfrak{p}_i)$.
For part 2: we show that this action is transitive on the primes of $K$ lying over $p$ (i.e. that the action of $operatorname{Gal}(K/Bbb Q)$ just moves the primes around).
Suppose that $mathfrak{p}_i$ and $mathfrak{p}_j$ are not conjugate by an element of $operatorname{Gal}(K/Bbb Q)$ so that for all $sigma$, $sigma(mathfrak{p}_i) neq mathfrak{p}_j$. By the Chinese remainder theorem we can find a $beta in mathfrak{p}_i$ such that $beta notin mathfrak{p}_j$ for any $j neq i$. Let $b = Nbeta = prod_{sigma in operatorname{Gal}(K/Bbb Q)} sigma(beta)$. Then $b in Bbb Z cap mathfrak{p}_i = pmathcal{O}_K$. Now, for all $sigma in operatorname{Gal}(K/Bbb Q)$ we have $beta notin sigma^{-1}(mathfrak{p}_j)$ and so $sigma(beta) notin mathfrak{p}_j$. But we showed that $b in pmathcal{O}_K subset mathfrak{p}_j$ (since $mathfrak{p}_j mid p$) and since $b$ is a product of the $sigma(beta)$, this contradicts the primality of $mathfrak{p}_j$, so it must be that there is some $sigma in operatorname{Gal}(K/Bbb Q)$ such that $sigma(mathfrak{p}_i) = mathfrak{p}_j$.
answered Jan 27 at 15:39
ÍgjøgnumMegÍgjøgnumMeg
2,86011129
$endgroup$
$begingroup$
I might be wrong, but I'm not convinced that for part 2 you proved the right thing. Transitivity states something about the entire Galois group, whereas I only require that elements of it are permutations on S, ie. that's a condition on individual elems. Also, I'd find it helpful if you could spell out the 'standard' part of statement 1.
$endgroup$
– gen
Jan 27 at 15:53
$begingroup$
An action is just a homomorphism $G to operatorname{Sym}(S)$ so it gives you a permutation anyway. I'm just showing you in this answer that the action of the group on the primes of $K$ "moves all the primes around".
$endgroup$
– ÍgjøgnumMeg
Jan 27 at 15:59
$begingroup$
Thanks for the clarification. I'm still a little confused why $sigma(mathfrak{p_i})in S$. You showed that it is an ideal, but that doesn't mean that $sigma(mathfrak{p_i}) = mathfrak{p}_j$ for some j.
$endgroup$
– gen
Jan 27 at 16:21
$begingroup$
That part just shows that $sigma(mathfrak{p}_i)$ is a prime ideal. That $sigma(mathfrak{p}_i) = mathfrak{p}_j$ is shown in part 2.
$endgroup$
– ÍgjøgnumMeg
Jan 27 at 16:25
add a comment |
$begingroup$
For part $1$: you have a group action given by $cdot : operatorname{Gal}(K/Bbb Q) times S to S : sigma cdot mathfrak{p}_i = sigma(mathfrak{p}_i)$.
Clearly $operatorname{id}cdot mathfrak{p}_i = mathfrak{p}_i$ and given $sigma_1, sigma_2 in operatorname{Gal}(K/Bbb Q)$ you have $(sigma_1 circ sigma_2) cdot mathfrak{p}_i = sigma_1(sigma_2(mathfrak{p}_i))$. Note that $sigma_2(mathfrak{p}_i)in S$ because if $ab in sigma_2(mathfrak{p}_i)$ then $sigma_2^{-1}(ab) in mathfrak{p}_i$. Since $sigma_2$ is a homomorphism we have $sigma_2^{-1}(a)sigma_2^{-1}(b) in mathfrak{p}_i$, and since this is a prime ideal we have $sigma_2^{-1}(a) in mathfrak{p}_i$ or $sigma_2^{-1}(b) in mathfrak{p}_i$, and hence $a in sigma_2(mathfrak{p}_i)$ or $b in sigma_2(mathfrak{p}_i)$.
For part 2: we show that this action is transitive on the primes of $K$ lying over $p$ (i.e. that the action of $operatorname{Gal}(K/Bbb Q)$ just moves the primes around).
Suppose that $mathfrak{p}_i$ and $mathfrak{p}_j$ are not conjugate by an element of $operatorname{Gal}(K/Bbb Q)$ so that for all $sigma$, $sigma(mathfrak{p}_i) neq mathfrak{p}_j$. By the Chinese remainder theorem we can find a $beta in mathfrak{p}_i$ such that $beta notin mathfrak{p}_j$ for any $j neq i$. Let $b = Nbeta = prod_{sigma in operatorname{Gal}(K/Bbb Q)} sigma(beta)$. Then $b in Bbb Z cap mathfrak{p}_i = pmathcal{O}_K$. Now, for all $sigma in operatorname{Gal}(K/Bbb Q)$ we have $beta notin sigma^{-1}(mathfrak{p}_j)$ and so $sigma(beta) notin mathfrak{p}_j$. But we showed that $b in pmathcal{O}_K subset mathfrak{p}_j$ (since $mathfrak{p}_j mid p$) and since $b$ is a product of the $sigma(beta)$, this contradicts the primality of $mathfrak{p}_j$, so it must be that there is some $sigma in operatorname{Gal}(K/Bbb Q)$ such that $sigma(mathfrak{p}_i) = mathfrak{p}_j$.
answered Jan 27 at 15:39
ÍgjøgnumMegÍgjøgnumMeg
2,86011129
$endgroup$
For part $1$: you have a group action given by $cdot : operatorname{Gal}(K/Bbb Q) times S to S : sigma cdot mathfrak{p}_i = sigma(mathfrak{p}_i)$.
Clearly $operatorname{id}cdot mathfrak{p}_i = mathfrak{p}_i$ and given $sigma_1, sigma_2 in operatorname{Gal}(K/Bbb Q)$ you have $(sigma_1 circ sigma_2) cdot mathfrak{p}_i = sigma_1(sigma_2(mathfrak{p}_i))$. Note that $sigma_2(mathfrak{p}_i)in S$ because if $ab in sigma_2(mathfrak{p}_i)$ then $sigma_2^{-1}(ab) in mathfrak{p}_i$. Since $sigma_2$ is a homomorphism we have $sigma_2^{-1}(a)sigma_2^{-1}(b) in mathfrak{p}_i$, and since this is a prime ideal we have $sigma_2^{-1}(a) in mathfrak{p}_i$ or $sigma_2^{-1}(b) in mathfrak{p}_i$, and hence $a in sigma_2(mathfrak{p}_i)$ or $b in sigma_2(mathfrak{p}_i)$.
For part 2: we show that this action is transitive on the primes of $K$ lying over $p$ (i.e. that the action of $operatorname{Gal}(K/Bbb Q)$ just moves the primes around).
Suppose that $mathfrak{p}_i$ and $mathfrak{p}_j$ are not conjugate by an element of $operatorname{Gal}(K/Bbb Q)$ so that for all $sigma$, $sigma(mathfrak{p}_i) neq mathfrak{p}_j$. By the Chinese remainder theorem we can find a $beta in mathfrak{p}_i$ such that $beta notin mathfrak{p}_j$ for any $j neq i$. Let $b = Nbeta = prod_{sigma in operatorname{Gal}(K/Bbb Q)} sigma(beta)$. Then $b in Bbb Z cap mathfrak{p}_i = pmathcal{O}_K$. Now, for all $sigma in operatorname{Gal}(K/Bbb Q)$ we have $beta notin sigma^{-1}(mathfrak{p}_j)$ and so $sigma(beta) notin mathfrak{p}_j$. But we showed that $b in pmathcal{O}_K subset mathfrak{p}_j$ (since $mathfrak{p}_j mid p$) and since $b$ is a product of the $sigma(beta)$, this contradicts the primality of $mathfrak{p}_j$, so it must be that there is some $sigma in operatorname{Gal}(K/Bbb Q)$ such that $sigma(mathfrak{p}_i) = mathfrak{p}_j$.
answered Jan 27 at 15:39
ÍgjøgnumMegÍgjøgnumMeg
2,86011129
edited Jan 27 at 16:01
answered Jan 27 at 15:39
ÍgjøgnumMegÍgjøgnumMeg
2,86011129
answered Jan 27 at 15:39
ÍgjøgnumMegÍgjøgnumMeg
2,86011129
answered Jan 27 at 15:39
ÍgjøgnumMegÍgjøgnumMeg
2,86011129
2,86011129
$begingroup$
I might be wrong, but I'm not convinced that for part 2 you proved the right thing. Transitivity states something about the entire Galois group, whereas I only require that elements of it are permutations on S, ie. that's a condition on individual elems. Also, I'd find it helpful if you could spell out the 'standard' part of statement 1.
$endgroup$
– gen
Jan 27 at 15:53
$begingroup$
An action is just a homomorphism $G to operatorname{Sym}(S)$ so it gives you a permutation anyway. I'm just showing you in this answer that the action of the group on the primes of $K$ "moves all the primes around".
$endgroup$
– ÍgjøgnumMeg
Jan 27 at 15:59
$begingroup$
Thanks for the clarification. I'm still a little confused why $sigma(mathfrak{p_i})in S$. You showed that it is an ideal, but that doesn't mean that $sigma(mathfrak{p_i}) = mathfrak{p}_j$ for some j.
$endgroup$
– gen
Jan 27 at 16:21
$begingroup$
That part just shows that $sigma(mathfrak{p}_i)$ is a prime ideal. That $sigma(mathfrak{p}_i) = mathfrak{p}_j$ is shown in part 2.
$endgroup$
– ÍgjøgnumMeg
Jan 27 at 16:25
add a comment |
$begingroup$
I might be wrong, but I'm not convinced that for part 2 you proved the right thing. Transitivity states something about the entire Galois group, whereas I only require that elements of it are permutations on S, ie. that's a condition on individual elems. Also, I'd find it helpful if you could spell out the 'standard' part of statement 1.
$endgroup$
– gen
Jan 27 at 15:53
$begingroup$
An action is just a homomorphism $G to operatorname{Sym}(S)$ so it gives you a permutation anyway. I'm just showing you in this answer that the action of the group on the primes of $K$ "moves all the primes around".
$endgroup$
– ÍgjøgnumMeg
Jan 27 at 15:59
$begingroup$
Thanks for the clarification. I'm still a little confused why $sigma(mathfrak{p_i})in S$. You showed that it is an ideal, but that doesn't mean that $sigma(mathfrak{p_i}) = mathfrak{p}_j$ for some j.
$endgroup$
– gen
Jan 27 at 16:21
$begingroup$
That part just shows that $sigma(mathfrak{p}_i)$ is a prime ideal. That $sigma(mathfrak{p}_i) = mathfrak{p}_j$ is shown in part 2.
$endgroup$
– ÍgjøgnumMeg
Jan 27 at 16:25
$begingroup$
I might be wrong, but I'm not convinced that for part 2 you proved the right thing. Transitivity states something about the entire Galois group, whereas I only require that elements of it are permutations on S, ie. that's a condition on individual elems. Also, I'd find it helpful if you could spell out the 'standard' part of statement 1.
$endgroup$
– gen
Jan 27 at 15:53
$begingroup$
I might be wrong, but I'm not convinced that for part 2 you proved the right thing. Transitivity states something about the entire Galois group, whereas I only require that elements of it are permutations on S, ie. that's a condition on individual elems. Also, I'd find it helpful if you could spell out the 'standard' part of statement 1.
$endgroup$
– gen
Jan 27 at 15:53
$begingroup$
An action is just a homomorphism $G to operatorname{Sym}(S)$ so it gives you a permutation anyway. I'm just showing you in this answer that the action of the group on the primes of $K$ "moves all the primes around".
$endgroup$
– ÍgjøgnumMeg
Jan 27 at 15:59
$begingroup$
An action is just a homomorphism $G to operatorname{Sym}(S)$ so it gives you a permutation anyway. I'm just showing you in this answer that the action of the group on the primes of $K$ "moves all the primes around".
$endgroup$
– ÍgjøgnumMeg
Jan 27 at 15:59
$begingroup$
Thanks for the clarification. I'm still a little confused why $sigma(mathfrak{p_i})in S$. You showed that it is an ideal, but that doesn't mean that $sigma(mathfrak{p_i}) = mathfrak{p}_j$ for some j.
$endgroup$
– gen
Jan 27 at 16:21
$begingroup$
Thanks for the clarification. I'm still a little confused why $sigma(mathfrak{p_i})in S$. You showed that it is an ideal, but that doesn't mean that $sigma(mathfrak{p_i}) = mathfrak{p}_j$ for some j.
$endgroup$
– gen
Jan 27 at 16:21
$begingroup$
That part just shows that $sigma(mathfrak{p}_i)$ is a prime ideal. That $sigma(mathfrak{p}_i) = mathfrak{p}_j$ is shown in part 2.
$endgroup$
– ÍgjøgnumMeg
Jan 27 at 16:25
$begingroup$
That part just shows that $sigma(mathfrak{p}_i)$ is a prime ideal. That $sigma(mathfrak{p}_i) = mathfrak{p}_j$ is shown in part 2.
$endgroup$
– ÍgjøgnumMeg
Jan 27 at 16:25
add a comment |
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All of the elements of $operatorname{Gal}(K/Bbb Q)$ are injective by definition. You need to prove that the action of this group on the primes of $K$ is transitive.
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– ÍgjøgnumMeg
Jan 27 at 15:00
$begingroup$
@ÍgjøgnumMeg, why do I need to show transitivity?
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– gen
Jan 27 at 15:09
$begingroup$
Because a transitive action of a group $G$ on a set $X$ is one such that for every $x, y in X$ there exists $g in G$ such that $gx = y$. In this case you show that for every prime $mathfrak{p}_i$ there is a $sigma in operatorname{Gal}(K/Bbb Q)$ such that $sigma mathfrak{p}_i = mathfrak{p}_j$
$endgroup$
– ÍgjøgnumMeg
Jan 27 at 15:11
$begingroup$
Any time a group acts on a set, it permutes the elements of that set. So 2 is redundant.
$endgroup$
– RghtHndSd
Jan 27 at 15:56
$begingroup$
Thanks @RghtHndSd I see it now, this makes complete sense.
$endgroup$
– gen
Jan 27 at 16:01