Is there an elementary proof that there are infinitely many primes that are *not* completely split in an...
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I'm currently in the middle of teaching the adelic algebraic proofs of global class field theory. One of the intermediate lemmas that one shows is the following:
Lemma: if L/K is an abelian extension of number fields, then there are infinitely many primes of K that do not split competely in L.
Of course this is implied by Cebotarev's density theorem, but the adelic proof uses only algebra/topology and finiteness of class number/Dirichlet's units theorem.
There is a well-known elementary proof, (see eg this MO question) that there are infinitely many primes that are split in L/K. I was wondering whether there is also an elementary argument for infinitude of non-split primes in the extension? (As usual the notion of "elementary" is flexible, but I'm looking for something that uses a minimum of machinery.)
One possibility would be to distill the adelic proof into something algebraic, although this seems hard. Another option would be to look for ideals of O_K that are not norms from O_L: any such ideal must have a factor which does not split completely.
One of the answers to the MathOverflow question linked above does mention the paper Primes of degree one and algebraic cases of Čebotarev's theorem of Lenstra and Stevenhagen, which gives an elementary proof under the assumption that L contains a nontrivial ray class field of K. But it seems that one still needs to prove the first inequality in some form to use this.
nt.number-theory algebraic-number-theory class-field-theory number-fields elementary-proofs
asked 13 hours ago
Alison MillerAlison Miller
3,01812129
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add a comment |
$begingroup$
I'm currently in the middle of teaching the adelic algebraic proofs of global class field theory. One of the intermediate lemmas that one shows is the following:
Lemma: if L/K is an abelian extension of number fields, then there are infinitely many primes of K that do not split competely in L.
Of course this is implied by Cebotarev's density theorem, but the adelic proof uses only algebra/topology and finiteness of class number/Dirichlet's units theorem.
There is a well-known elementary proof, (see eg this MO question) that there are infinitely many primes that are split in L/K. I was wondering whether there is also an elementary argument for infinitude of non-split primes in the extension? (As usual the notion of "elementary" is flexible, but I'm looking for something that uses a minimum of machinery.)
One possibility would be to distill the adelic proof into something algebraic, although this seems hard. Another option would be to look for ideals of O_K that are not norms from O_L: any such ideal must have a factor which does not split completely.
One of the answers to the MathOverflow question linked above does mention the paper Primes of degree one and algebraic cases of Čebotarev's theorem of Lenstra and Stevenhagen, which gives an elementary proof under the assumption that L contains a nontrivial ray class field of K. But it seems that one still needs to prove the first inequality in some form to use this.
nt.number-theory algebraic-number-theory class-field-theory number-fields elementary-proofs
asked 13 hours ago
Alison MillerAlison Miller
3,01812129
$endgroup$
3
$begingroup$
If $K=mathbb{Q}$ this follows from Kronecker-Weber. Suppose $Lsubsetmathbb{Q}(zeta_n)$ corresponds to a proper subgroup $Hleq (mathbb{Z}/nmathbb{Z})^times$. A rational prime $pnmid n$ splits completely in $L$ iff $pmod n$ is in $H$. If $p_1,ldots,p_m$ are primes not dividing $n$ that do not split completely, we can pick $rin(mathbb{Z}/nmathbb{Z})^timesbackslash H$ and use the CRT to find a positive integer $Q$ congruent to $1$ mod $p_1cdots p_m$ and congruent to $r$ mod $n$. Then $Q$ must have a prime divisor that does not split completely, which is not $p_1,ldots,p_m$.
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– Julian Rosen
12 hours ago
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@JulianRosen Yes, and I believe that's the argument that Lenstra and Stevenhagen generalize to arbitrary ray class fields.
$endgroup$
– Alison Miller
12 hours ago
add a comment |
$begingroup$
I'm currently in the middle of teaching the adelic algebraic proofs of global class field theory. One of the intermediate lemmas that one shows is the following:
Lemma: if L/K is an abelian extension of number fields, then there are infinitely many primes of K that do not split competely in L.
Of course this is implied by Cebotarev's density theorem, but the adelic proof uses only algebra/topology and finiteness of class number/Dirichlet's units theorem.
There is a well-known elementary proof, (see eg this MO question) that there are infinitely many primes that are split in L/K. I was wondering whether there is also an elementary argument for infinitude of non-split primes in the extension? (As usual the notion of "elementary" is flexible, but I'm looking for something that uses a minimum of machinery.)
One possibility would be to distill the adelic proof into something algebraic, although this seems hard. Another option would be to look for ideals of O_K that are not norms from O_L: any such ideal must have a factor which does not split completely.
One of the answers to the MathOverflow question linked above does mention the paper Primes of degree one and algebraic cases of Čebotarev's theorem of Lenstra and Stevenhagen, which gives an elementary proof under the assumption that L contains a nontrivial ray class field of K. But it seems that one still needs to prove the first inequality in some form to use this.
nt.number-theory algebraic-number-theory class-field-theory number-fields elementary-proofs
asked 13 hours ago
Alison MillerAlison Miller
3,01812129
$endgroup$
I'm currently in the middle of teaching the adelic algebraic proofs of global class field theory. One of the intermediate lemmas that one shows is the following:
Lemma: if L/K is an abelian extension of number fields, then there are infinitely many primes of K that do not split competely in L.
Of course this is implied by Cebotarev's density theorem, but the adelic proof uses only algebra/topology and finiteness of class number/Dirichlet's units theorem.
There is a well-known elementary proof, (see eg this MO question) that there are infinitely many primes that are split in L/K. I was wondering whether there is also an elementary argument for infinitude of non-split primes in the extension? (As usual the notion of "elementary" is flexible, but I'm looking for something that uses a minimum of machinery.)
One possibility would be to distill the adelic proof into something algebraic, although this seems hard. Another option would be to look for ideals of O_K that are not norms from O_L: any such ideal must have a factor which does not split completely.
One of the answers to the MathOverflow question linked above does mention the paper Primes of degree one and algebraic cases of Čebotarev's theorem of Lenstra and Stevenhagen, which gives an elementary proof under the assumption that L contains a nontrivial ray class field of K. But it seems that one still needs to prove the first inequality in some form to use this.
nt.number-theory algebraic-number-theory class-field-theory number-fields elementary-proofs
nt.number-theory algebraic-number-theory class-field-theory number-fields elementary-proofs
asked 13 hours ago
Alison MillerAlison Miller
3,01812129
asked 13 hours ago
Alison MillerAlison Miller
3,01812129
edited 10 hours ago
Alison Miller
asked 13 hours ago
Alison MillerAlison Miller
3,01812129
asked 13 hours ago
Alison MillerAlison Miller
3,01812129
asked 13 hours ago
Alison MillerAlison Miller
3,01812129
3,01812129
3
$begingroup$
If $K=mathbb{Q}$ this follows from Kronecker-Weber. Suppose $Lsubsetmathbb{Q}(zeta_n)$ corresponds to a proper subgroup $Hleq (mathbb{Z}/nmathbb{Z})^times$. A rational prime $pnmid n$ splits completely in $L$ iff $pmod n$ is in $H$. If $p_1,ldots,p_m$ are primes not dividing $n$ that do not split completely, we can pick $rin(mathbb{Z}/nmathbb{Z})^timesbackslash H$ and use the CRT to find a positive integer $Q$ congruent to $1$ mod $p_1cdots p_m$ and congruent to $r$ mod $n$. Then $Q$ must have a prime divisor that does not split completely, which is not $p_1,ldots,p_m$.
$endgroup$
– Julian Rosen
12 hours ago
$begingroup$
@JulianRosen Yes, and I believe that's the argument that Lenstra and Stevenhagen generalize to arbitrary ray class fields.
$endgroup$
– Alison Miller
12 hours ago
add a comment |
3
$begingroup$
If $K=mathbb{Q}$ this follows from Kronecker-Weber. Suppose $Lsubsetmathbb{Q}(zeta_n)$ corresponds to a proper subgroup $Hleq (mathbb{Z}/nmathbb{Z})^times$. A rational prime $pnmid n$ splits completely in $L$ iff $pmod n$ is in $H$. If $p_1,ldots,p_m$ are primes not dividing $n$ that do not split completely, we can pick $rin(mathbb{Z}/nmathbb{Z})^timesbackslash H$ and use the CRT to find a positive integer $Q$ congruent to $1$ mod $p_1cdots p_m$ and congruent to $r$ mod $n$. Then $Q$ must have a prime divisor that does not split completely, which is not $p_1,ldots,p_m$.
$endgroup$
– Julian Rosen
12 hours ago
$begingroup$
@JulianRosen Yes, and I believe that's the argument that Lenstra and Stevenhagen generalize to arbitrary ray class fields.
$endgroup$
– Alison Miller
12 hours ago
3
3
$begingroup$
If $K=mathbb{Q}$ this follows from Kronecker-Weber. Suppose $Lsubsetmathbb{Q}(zeta_n)$ corresponds to a proper subgroup $Hleq (mathbb{Z}/nmathbb{Z})^times$. A rational prime $pnmid n$ splits completely in $L$ iff $pmod n$ is in $H$. If $p_1,ldots,p_m$ are primes not dividing $n$ that do not split completely, we can pick $rin(mathbb{Z}/nmathbb{Z})^timesbackslash H$ and use the CRT to find a positive integer $Q$ congruent to $1$ mod $p_1cdots p_m$ and congruent to $r$ mod $n$. Then $Q$ must have a prime divisor that does not split completely, which is not $p_1,ldots,p_m$.
$endgroup$
– Julian Rosen
12 hours ago
$begingroup$
If $K=mathbb{Q}$ this follows from Kronecker-Weber. Suppose $Lsubsetmathbb{Q}(zeta_n)$ corresponds to a proper subgroup $Hleq (mathbb{Z}/nmathbb{Z})^times$. A rational prime $pnmid n$ splits completely in $L$ iff $pmod n$ is in $H$. If $p_1,ldots,p_m$ are primes not dividing $n$ that do not split completely, we can pick $rin(mathbb{Z}/nmathbb{Z})^timesbackslash H$ and use the CRT to find a positive integer $Q$ congruent to $1$ mod $p_1cdots p_m$ and congruent to $r$ mod $n$. Then $Q$ must have a prime divisor that does not split completely, which is not $p_1,ldots,p_m$.
$endgroup$
– Julian Rosen
12 hours ago
$begingroup$
@JulianRosen Yes, and I believe that's the argument that Lenstra and Stevenhagen generalize to arbitrary ray class fields.
$endgroup$
– Alison Miller
12 hours ago
$begingroup$
@JulianRosen Yes, and I believe that's the argument that Lenstra and Stevenhagen generalize to arbitrary ray class fields.
$endgroup$
– Alison Miller
12 hours ago
add a comment |
1 Answer
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I mention this as an answer since it is too long for comments. I do not know what the adelic proof assumes. Suppose that all but finitely many primes of $K$ split completely in $L$. Suppose $d$ is the degree of $L$ over $K$. Then the zeta function of $L$ is the $d$-th power of the zeta function of $K$, up to finitely many factors. But the Zeta functions of $L$ and $K$ have only a simple pole at $s=1$ (implied by finiteness of class number ...). Hence $d=1$ and $L=K$.
answered 13 hours ago
VenkataramanaVenkataramana
9,09412951
$endgroup$
$begingroup$
Good point! That's certainly simpler than the adelic proof.
$endgroup$
– Alison Miller
12 hours ago
add a comment |
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I mention this as an answer since it is too long for comments. I do not know what the adelic proof assumes. Suppose that all but finitely many primes of $K$ split completely in $L$. Suppose $d$ is the degree of $L$ over $K$. Then the zeta function of $L$ is the $d$-th power of the zeta function of $K$, up to finitely many factors. But the Zeta functions of $L$ and $K$ have only a simple pole at $s=1$ (implied by finiteness of class number ...). Hence $d=1$ and $L=K$.
answered 13 hours ago
VenkataramanaVenkataramana
9,09412951
$endgroup$
$begingroup$
Good point! That's certainly simpler than the adelic proof.
$endgroup$
– Alison Miller
12 hours ago
add a comment |
$begingroup$
I mention this as an answer since it is too long for comments. I do not know what the adelic proof assumes. Suppose that all but finitely many primes of $K$ split completely in $L$. Suppose $d$ is the degree of $L$ over $K$. Then the zeta function of $L$ is the $d$-th power of the zeta function of $K$, up to finitely many factors. But the Zeta functions of $L$ and $K$ have only a simple pole at $s=1$ (implied by finiteness of class number ...). Hence $d=1$ and $L=K$.
answered 13 hours ago
VenkataramanaVenkataramana
9,09412951
$endgroup$
$begingroup$
Good point! That's certainly simpler than the adelic proof.
$endgroup$
– Alison Miller
12 hours ago
add a comment |
$begingroup$
I mention this as an answer since it is too long for comments. I do not know what the adelic proof assumes. Suppose that all but finitely many primes of $K$ split completely in $L$. Suppose $d$ is the degree of $L$ over $K$. Then the zeta function of $L$ is the $d$-th power of the zeta function of $K$, up to finitely many factors. But the Zeta functions of $L$ and $K$ have only a simple pole at $s=1$ (implied by finiteness of class number ...). Hence $d=1$ and $L=K$.
answered 13 hours ago
VenkataramanaVenkataramana
9,09412951
$endgroup$
I mention this as an answer since it is too long for comments. I do not know what the adelic proof assumes. Suppose that all but finitely many primes of $K$ split completely in $L$. Suppose $d$ is the degree of $L$ over $K$. Then the zeta function of $L$ is the $d$-th power of the zeta function of $K$, up to finitely many factors. But the Zeta functions of $L$ and $K$ have only a simple pole at $s=1$ (implied by finiteness of class number ...). Hence $d=1$ and $L=K$.
answered 13 hours ago
VenkataramanaVenkataramana
9,09412951
answered 13 hours ago
VenkataramanaVenkataramana
9,09412951
answered 13 hours ago
VenkataramanaVenkataramana
9,09412951
answered 13 hours ago
VenkataramanaVenkataramana
9,09412951
9,09412951
$begingroup$
Good point! That's certainly simpler than the adelic proof.
$endgroup$
– Alison Miller
12 hours ago
add a comment |
$begingroup$
Good point! That's certainly simpler than the adelic proof.
$endgroup$
– Alison Miller
12 hours ago
$begingroup$
Good point! That's certainly simpler than the adelic proof.
$endgroup$
– Alison Miller
12 hours ago
$begingroup$
Good point! That's certainly simpler than the adelic proof.
$endgroup$
– Alison Miller
12 hours ago
add a comment |
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$begingroup$
If $K=mathbb{Q}$ this follows from Kronecker-Weber. Suppose $Lsubsetmathbb{Q}(zeta_n)$ corresponds to a proper subgroup $Hleq (mathbb{Z}/nmathbb{Z})^times$. A rational prime $pnmid n$ splits completely in $L$ iff $pmod n$ is in $H$. If $p_1,ldots,p_m$ are primes not dividing $n$ that do not split completely, we can pick $rin(mathbb{Z}/nmathbb{Z})^timesbackslash H$ and use the CRT to find a positive integer $Q$ congruent to $1$ mod $p_1cdots p_m$ and congruent to $r$ mod $n$. Then $Q$ must have a prime divisor that does not split completely, which is not $p_1,ldots,p_m$.
$endgroup$
– Julian Rosen
12 hours ago
$begingroup$
@JulianRosen Yes, and I believe that's the argument that Lenstra and Stevenhagen generalize to arbitrary ray class fields.
$endgroup$
– Alison Miller
12 hours ago