|
6
|
|
edited Jun 1 2010 at 6:26 |
Yes, there are examples and Minkowski's proof for ${\mathbf Q}$ can be adapted to find a few of them. As with extensions Some examples of this kind among quadratic fields $F$, listed in increasing size of discriminant (in absolute value), are {\mathbf Q$Q}(\sqrt{-3}), we aim to show any proper extension \ \ {\mathbf Q}(i), \ \ {\mathbf Q}(\sqrt{5}), \ \ {\mathbf Q}(\sqrt{-7}), \ \ {\mathbf Q}(\sqrt{2}), \ \ {\mathbf Q}(\sqrt{-2}).A cubic and quartic field $F$ that will come out of the method I describe below are ${\mathbf Q}(\alpha)$ where $\alpha^3 - \alpha - 1 = 0$ and ${\mathbf Q}(\zeta_5)$. Now for the details. I suggest when reading this through for the first time that you keep a concrete example in mind, like $F = {\mathbf Q}(i)$. (That's what I did the first time I worked this out.) Over the rationals, Minkowski showed a number field has with degree larger than 1 must have a relative discriminant ideal that whose absolute value is not triviallarger than 1. If Over other number fields $F$ besides the rationals, the goal is a number field and to find sufficient conditions on $E$ is an F$ so that any finite extension , the $E/F$ with $[E:F] > 1$ has its discriminant ideal ${\mathfrak d}_{E/F}$ not equal to the unit ideal, and then a prime ideal factor will ramify in $E$. Rather than show ${\mathfrak d}_{E/F} \not= (1)$, we will look for a sufficient condition on $F$ which assures us that the norm of this ideal is related not 1. That means absolutely the same thing, but it's easier to work with ideal norms since they are positive integers rather than ideals, and moreover it lets us express the problem in terms of discriminants of number fields: the discriminants of $E$ and $F$ are related by $|d_E| $|d_E| = {\rm N}({\mathfrak d}_{E/F})|d_F|^{[E:F]}$. Therefore d}_{E/F})|d_F|^{[E:F]}.$$ So aiming to say show ${\mathfrak d}_{E/F} = \not= (1)$ is the same as saying avoiding $|d_E|^{1/[E:{\mathbf |d_E| = |d_F|^{[E:F]}$, which is the same as avoiding $$|d_E|^{1/[E:{\mathbf Q}]} = |d_F|^{1/[F:{\mathbf Q}]}$. Q}]}.$$ We want this notsufficient conditions on $F$ to happen if guarantee this equation for any proper finite extension $[E:F] > 1$E/F$ can't take place. Returning to the extension $E/F$, let $m = [F:{\mathbf Q}]$, so $[E:{\mathbf Q}] = [E:F][F:{\mathbf Q}] \geq 2m$ since $E$ is a larger field than $F$. Then Now if $E/F$ has had trivial discriminant ideal, we would have This hypothetical lower bound on the root discriminant of $F$ is larger than the proved Minkowski bound of $f(m)$, so any number field f(m)$. Suppose $F$ is a number field of degree $m$ whose root discriminant is less than $f(2m)$ f(2m)$. If $E/F$ is an example. Taking unramified at all primes in $m$th powers on F$ then $E$ and $F$ have equal root discriminants, so the inequality root discriminant of $E$ is less than $f(2m)$. However, we saw above that the root discriminant of $E$ is $\geq f(2m)$ when $[E:F] \geq 2$, so the only choice is $E = F$, i.e., no proper finite extension of $F$ can be unramified at all primes in $F$. Our goal now is to find examples of number fields $F$ with degree $m$ whose root discriminant is smaller than $f(2m)$: $|d_F|^{1/m} < f(2m)$shows we want to get. This is the same as Any $F$ which fits this condition will be an example. As a reality check, let $F$ be the rationals, so $m = 1$. We have $f(2) = \pi/2$ and $|d_{\mathbf Q}| = 1 < \pi/2$, so $\mathbf Q$ has no unramified extensions. We're on the right track.
|
|
|
|
|
5
|
|
edited May 31 2010 at 3:11 |
This kind of argument using Minkowski's bound does work for a few cubic fields and quartic fields: for cubic fields it works as long as the discriminant of the field (in absolute value) is less than 31.39, and there are two such fields: ${\mathbf Q}(\alpha)$ and ${\mathbf Q}(\beta)$ where $\alpha^3 - \alpha - 1 = 0$ (discriminant -23) and $\beta^3 + \beta + 1 = 0$ (discriminant -31). The next smallest absolute value of a discriminant of a cubic field is 44, which is above the bound. For quartic fields we need the discriminant to be less than 158.32, and I know of three fields which work: ${\mathbf Q}(\gamma)$ where $\gamma^4 + 2\gamma^3 + 3\gamma + 1 = 0$ (discriminant 117), ${\mathbf Q}(\zeta_5)$ has discriminant 125, and ${\mathbf Q}(\zeta_{12})$ has discriminant 144.
|
|
|
|
|
4
|
|
edited May 31 2010 at 2:29 |
Yes, there are examples. As with extensions of $\mathbf Q$, we aim to show
any proper extension of the number field has a relative discriminant ideal
that is not trivial. If $F$ is a number field and $E$ is an extension,
the discriminant ideal ${\mathfrak d}_{E/F}$ is related to the discriminants of
$E$ and $F$ by $|d_E| = {\rm N}({\mathfrak d}_{E/F})|d_F|^{[E:F]}$.
Therefore to say ${\mathfrak d}_{E/F} = (1)$ is the same as saying $|d_E|^{1/[E:{\mathbf Q}]} = |d_F|^{1/[F:{\mathbf Q}]}$. We want this not to happen if $[E:F] > 1$.
For any number field $K$, the quantity $|d_K|^{1/[K:{\mathbf Q}]}$ is called the root discriminant of $K$. When $n = [K:{\mathbf Q}]$, Minkowski's lower bound on $|d_K|$ is
$$
|d_K| \geq \left(\frac{\pi}{4}\right)^n\frac{n^{2n}}{n!^2},
$$
so we get the root discriminant lower bound
$$
|d_K|^{1/n} \geq \left(\frac{\pi}{4}\right)\frac{n^{2}}{n!^{2/n}}.
$$
Call the right side $f(n)$, so the Minkowski bound says $|d_K|^{1/[K:{\mathbf Q}]} \geq f([K:{\mathbf Q}])$.
For $n = 1, 2, 3, 4$ the values of $f(n)$ are $.785, 1.570, 2.140, 2.565$, so we guess $f(n)$ is increasing and it's left as an exercise to prove that. (Hint: use the one-sided Stirling estimate $n! > (n/e)^n\sqrt{2\pi{n}}$.)
Returning to the extension $E/F$, let $m = [F:{\mathbf Q}]$, so $[E:{\mathbf Q}] = [E:F][F:{\mathbf Q}] \geq 2m$ since $E$ is a larger field than $F$. Then if
$E/F$ has trivial discriminant ideal,
$$
|d_F|^{1/[F:{\mathbf Q}]} = |d_E|^{1/[E:{\mathbf Q}]} \geq f([E:{\mathbf Q}]) \geq f(2m).
$$
This hypothetical lower bound on the root discriminant of $F$ is larger than the proved Minkowski bound of $f(m)$, so any number field $F$ of degree $m$
whose root discriminant is less than $f(2m)$ is an example.
Taking $m$th powers on the inequality $|d_F|^{1/m} < f(2m)$ shows we want to get
$$
|d_F| < f(2m)^m = \frac{\pi^mm^{2m}}{(2m)!}.
$$
Taking $m = 2$ we want $|d_F| < f(4)^2 = 6.57$ and the fields ${\mathbf Q}(i)$, ${\mathbf Q}(\sqrt{-3})$, and ${\mathbf Q}(\sqrt{5})$ all work.
If the root discriminant of $F$ is not below $f(2m)$, where $m = [F:{\mathbf Q}]$,
we can still squeeze out some information, namely an upper bound on the degree of an everywhere (= at finite places) unramified extension of $F$. For instance, ${\mathbf Q}(\sqrt{2})$ has discriminant 8, which is not below 6.57, so this argument doesn't show on its own that every proper extension of ${\mathbf Q}(\sqrt{2})$ is ramified at some prime in ${\mathbf Q}(\sqrt{2})$. However, we can bound the degree of such an extension. The root disc. of ${\mathbf Q}(\sqrt{2})$ is $\sqrt{8} \approx 2.828$, which lies between $f(4)$ and $f(5)$, so any proper unramified extension of ${\mathbf Q}(\sqrt{2})$ must be a quadratic extension of ${\mathbf Q}(\sqrt{2})$. Quadratic extensions are automatically abelian, so we would have an abelian extension of ${\mathbf Q}(\sqrt{2})$ unramified at no finite places, and there's no such thing by class field theory since ${\mathbf Q}(\sqrt{2})$ has narrow class number 1 (the class number not involving ramification at infinity). Therefore you can add ${\mathbf Q}(\sqrt{2})$ to the list of number fields with no proper extension unramified at all finite places. The same arguments apply to ${\mathbf Q}(\sqrt{-2})$ and ${\mathbf Q}(\sqrt{-7})$, whose root discriminants are also between $f(4)$ and $f(5)$.
Unfortunately, this method is really limited because although $f(n)$ is increasing,
it's actually bounded. By Stirling's formula, $f(n)$ has limit $\pi e^2/4 \approx 5.803$. So when a number field $F$ has root discriminant exceeding this value, this method won't give us any examples at all (you can't find an $m$ for which the root discriminant is below $f(2m)$). Presumably using [Edit: Using Odlyzko-type lower bounds on discriminants, one can produce some more examples. At least we got a fewexamples this simple way.as Torsten points out.]
|
|
|
|
3
|
|
edited May 30 2010 at 21:22 |
Yes, there are examples. As with extensions of $\mathbf Q$, we aim to show
any proper extension of the number field has a relative discriminant ideal
that is not trivial. If $F$ is a number field and $E$ is an extension,
the discriminant ideal ${\mathfrak d}_{E/F}$ is related to the discriminants of
$E$ and $F$ by $|d_E| = {\rm N}({\mathfrak d}_{E/F})|d_F|^{[E:F]}$.
Therefore to say ${\mathfrak d}_{E/F} = (1)$ is the same as saying $|d_E|^{1/[E:{\mathbf Q}]} = |d_F|^{1/[F:{\mathbf Q}]}$. We want this not to happen if $[E:F] > 1$.
For any number field $K$, the quantity $|d_K|^{1/[K:{\mathbf Q}]}$ is called the root discriminant of $K$. When $n = [K:{\mathbf Q}]$, Minkowski's lower bound on $|d_K|$ is
$$
|d_K| \geq \left(\frac{\pi}{4}\right)^n\frac{n^{2n}}{n!^2},
$$
so we get the root discriminant lower bound
$$
|d_K|^{1/n} \geq \left(\frac{\pi}{4}\right)\frac{n^{2}}{n!^{2/n}}.
$$
Call the right side $f(n)$, so the Minkowski bound says $|d_K|^{1/[K:{\mathbf Q}]} \geq f([K:{\mathbf Q}])$.
For $n = 1, 2, 3, 4$ the values of $f(n)$ are $.785, 1.570, 2.140, 2.565$, so we guess $f(n)$ is increasing and it's left as an exercise to prove that. (Hint: use the one-sided Stirling estimate $n! > (n/e)^n\sqrt{2\pi{n}}$.)
Returning to the extension $E/F$, let $m = [F:{\mathbf Q}]$, so $[E:{\mathbf Q}] = [E:F][F:{\mathbf Q}] \geq 2m$ since $E$ is a larger field than $F$. Then if
$E/F$ has trivial discriminant ideal,
$$
|d_F|^{1/[F:{\mathbf Q}]} = |d_E|^{1/[E:{\mathbf Q}]} \geq f([E:{\mathbf Q}]) \geq f(2m).
$$
This hypothetical lower bound on the root discriminant of $F$ is larger than the proved Minkowski bound of $f(m)$, so any number field $F$ of degree $m$
whose root discriminant is less than $f(2m)$ is an example.
Taking $m$th powers on the inequality $|d_F|^{1/m} < f(2m)$ shows we want to get
$$
|d_F| < f(2m)^m = \frac{\pi^mm^{2m}}{(2m)!}.
$$
Taking $m = 2$ we want $|d_F| < f(4)^2 = 6.57$ and the fields ${\mathbf Q}(i)$, ${\mathbf Q}(\sqrt{-3})$, and ${\mathbf Q}(\sqrt{5})$ all work.
If the root discriminant of $F$ is not below $f(2m)$, where $m = [F:{\mathbf Q}]$,
we can still squeeze out some information, namely an upper bound on the degree of an everywhere (= at finite places) unramified extension of $F$. For instance, ${\mathbf Q}(\sqrt{2})$ has discriminant 8, which is not below 6.57, so this argument doesn't show on its own that every proper extension of ${\mathbf Q}(\sqrt{2})$ is ramified at some prime in ${\mathbf Q}(\sqrt{2})$. However, we can bound the degree of such an extension. The root disc. of ${\mathbf Q}(\sqrt{2})$ is $\sqrt{8} \approx 2.828$, which lies between $f(4)$ and $f(5)$, so any proper unramified extension of ${\mathbf Q}(\sqrt{2})$ must be a quadratic extension of ${\mathbf Q}(\sqrt{2})$. Quadratic extensions are automatically abelian, so we would have an abelian extension of ${\mathbf Q}(\sqrt{2})$ unramified at no finite places, and there's no such thing by class field theory since ${\mathbf Q}(\sqrt{2})$ has narrow class number 1 (the class number not involving ramification at infinity). Therefore you can add ${\mathbf Q}(\sqrt{2})$ to the list of number fields with no proper extension unramified at all finite places. The same arguments apply to ${\mathbf Q}(\sqrt{-2})$ and ${\mathbf Q}(\sqrt{-7})$, whose root discriminants are also between $f(4)$ and $f(5)$.
Unfortunately, this method is really limited because although $f(n)$ is increasing,
it's actually bounded. By Stirling's formula, $f(n)$ has limit $\pi e^2/4 \approx 5.803$. So when a number field $F$ has root discriminant exceeding this value, this method won't give us any examples at all (you can't find an $m$ for which the root discriminant is below $f(2m)$). There must be Presumably using Odlyzko-type lower bounds on discriminants one can produce more flexible methods that the argument used hereexamples. At least we got a few examples this simple way.
|
|
|
|
|
2
|
|
edited May 30 2010 at 21:01 |
Yes, there are examples. As with extensions of $\mathbf Q$, we aim to show
any proper extension of the number field has a relative discriminant ideal
that is not trivial. If $F$ is a number field and $E$ is an extension,
the discriminant ideal ${\mathfrak d}_{E/F}$ is related to the discriminants of
$E$ and $F$ by $|d_E| = {\rm N}({\mathfrak d}_{E/F})|d_F|^{[E:F]}$.
Therefore to say ${\mathfrak d}_{E/F} = (1)$ is the same as saying $|d_E|^{1/[E:{\mathbf Q}]} = |d_F|^{1/[F:{\mathbf Q}]}$. We want this not to happen if $[E:F] > 1$.
For any number field $K$, the quantity $|d_K|^{1/[K:{\mathbf Q}]}$ is called the root discriminant of $K$. When $n = [K:{\mathbf Q}]$, Minkowski's lower bound on $|d_K|$ is
$$
|d_K| \geq \left(\frac{\pi}{4}\right)^n\frac{n^{2n}}{n!^2},
$$
so we get the root discriminant lower bound
$$
|d_K|^{1/n} \geq \left(\frac{\pi}{4}\right)\frac{n^{2}}{n!^{2/n}}.
$$
Call the right side $f(n)$, so the Minkowski bound says $|d_K|^{1/[K:{\mathbf Q}]} \geq f([K:{\mathbf Q}])$.
For $n = 1, 2, 3, 4$ the values of $f(n)$ are $.785, 1.570, 2.140, 2.565$, so we guess $f(n)$ is increasing and it's left as an exercise to prove that. (Hint: use the one-sided Stirling estimate $n! > (n/e)^n\sqrt{2\pi{n}}$.)
Returning to the extension $E/F$, let $m = [F:{\mathbf Q}]$, so $[E:{\mathbf Q}] = [E:F][F:{\mathbf Q}] \geq 2m$ since $E$ is a larger field than $F$. Then if
$E/F$ has trivial discriminant ideal,
$$
|d_F|^{1/[F:{\mathbf Q}]} = |d_E|^{1/[E:{\mathbf Q}]} \geq f([E:{\mathbf Q}]) \geq f(2m).
$$
This hypothetical lower bound on the root discriminant of $F$ is larger than the proved Minkowski bound of $f(m)$, so any number field $F$ of degree $m$
whose root discriminant is less than $f(2m)$ is an example.
Taking $m$th powers on the inequality $|d_F|^{1/m} < f(2m)$ shows we want to get
$$
|d_F| < f(2m)^m = \frac{\pi^mm^{2m}}{(2m)!}.
$$
Taking $m = 2$ we want $|d_F| < f(4)^2 = 6.57$ and the fields ${\mathbf Q}(i)$, ${\mathbf Q}(\sqrt{-3})$, and ${\mathbf Q}(\sqrt{5})$ all work.
If the root discriminant of $F$ is not below $f(2m)$, where $m = [F:{\mathbf Q}]$,
we can still squeeze out some information, namely an upper bound on the degree of an everywhere (= at finite places) unramified extension of $F$. For instance, ${\mathbf Q}(\sqrt{2})$ has discriminant 8, which is not below 6.57, so this argument doesn't show on its own that every proper extension of ${\mathbf Q}(\sqrt{2})$ is ramified at some prime in ${\mathbf Q}(\sqrt{2})$. However, we can bound the degree of such an extension. The root disc. of ${\mathbf Q}(\sqrt{2})$ is $\sqrt{8} \approx 2.828$, which lies between $f(4)$ and $f(5)$, so any proper unramified extension of ${\mathbf Q}(\sqrt{2})$ must be a quadratic extension of ${\mathbf Q}(\sqrt{2})$. Quadratic extensions are automatically abelian, so we would have an abelian extension of ${\mathbf Q}(\sqrt{2})$ unramified at no finite places, and there's no such thing by class field theory since ${\mathbf Q}(\sqrt{2})$ has narrow class number 1 (the class number not involving ramification at infinity). Therefore you can add ${\mathbf Q}(\sqrt{2})$ to the list of number fields with no proper extension unramified at all finite places. The same arguments apply to ${\mathbf Q}(\sqrt{-2})$ and ${\mathbf Q}(\sqrt{-7})$, whose root discriminants are also between $f(4)$ and $f(5)$.
Unfortunately, this method is really limited because although $f(n)$ is increasing,
it's actually bounded. By Stirling's formula, $f(n)$ has limit $\pi e^2/4 \approx 5.803$. So when a number field $F$ has root discriminant exceeding this value, this method won't give us any examples at all (you can't find an $m$ for which the root discriminant is below $f(2m)$). I don't know how to generate examples by any other methodThere must be more flexible methods that the argument used here. At least we got a few examples this way.
|
|
|
|
|
1
|
|
answered May 30 2010 at 20:45 |
Yes, there are examples. As with extensions of $\mathbf Q$, we aim to show
any proper extension of the number field has a relative discriminant ideal
that is not trivial. If $F$ is a number field and $E$ is an extension,
the discriminant ideal ${\mathfrak d}_{E/F}$ is related to the discriminants of
$E$ and $F$ by $|d_E| = {\rm N}({\mathfrak d}_{E/F})|d_F|^{[E:F]}$.
Therefore to say ${\mathfrak d}_{E/F} = (1)$ is the same as saying $|d_E|^{1/[E:{\mathbf Q}]} = |d_F|^{1/[F:{\mathbf Q}]}$. We want this not to happen if $[E:F] > 1$.
For any number field $K$, the quantity $|d_K|^{1/[K:{\mathbf Q}]}$ is called the root discriminant of $K$. When $n = [K:{\mathbf Q}]$, Minkowski's lower bound on $|d_K|$ is
$$
|d_K| \geq \left(\frac{\pi}{4}\right)^n\frac{n^{2n}}{n!^2},
$$
so we get the root discriminant lower bound
$$
|d_K|^{1/n} \geq \left(\frac{\pi}{4}\right)\frac{n^{2}}{n!^{2/n}}.
$$
Call the right side $f(n)$, so the Minkowski bound says $|d_K|^{1/[K:{\mathbf Q}]} \geq f([K:{\mathbf Q}])$.
For $n = 1, 2, 3, 4$ the values of $f(n)$ are $.785, 1.570, 2.140, 2.565$, so we guess $f(n)$ is increasing and it's left as an exercise to prove that. (Hint: use the one-sided Stirling estimate $n! > (n/e)^n\sqrt{2\pi{n}}$.)
Returning to the extension $E/F$, let $m = [F:{\mathbf Q}]$, so $[E:{\mathbf Q}] = [E:F][F:{\mathbf Q}] \geq 2m$ since $E$ is a larger field than $F$. Then if
$E/F$ has trivial discriminant ideal,
$$
|d_F|^{1/[F:{\mathbf Q}]} = |d_E|^{1/[E:{\mathbf Q}]} \geq f([E:{\mathbf Q}]) \geq f(2m).
$$
This hypothetical lower bound on the root discriminant of $F$ is larger than the proved Minkowski bound of $f(m)$, so any number field $F$ of degree $m$
whose root discriminant is less than $f(2m)$ is an example.
Taking $m$th powers on the inequality $|d_F|^{1/m} < f(2m)$ shows we want to get
$$
|d_F| < f(2m)^m = \frac{\pi^mm^{2m}}{(2m)!}.
$$
Taking $m = 2$ we want $|d_F| < f(4)^2 = 6.57$ and the fields ${\mathbf Q}(i)$, ${\mathbf Q}(\sqrt{-3})$, and ${\mathbf Q}(\sqrt{5})$ all work.
If the root discriminant of $F$ is not below $f(2m)$, where $m = [F:{\mathbf Q}]$,
we can still squeeze out some information, namely an upper bound on the degree of an everywhere (= at finite places) unramified extension of $F$. For instance, ${\mathbf Q}(\sqrt{2})$ has discriminant 8, which is not below 6.57, so this argument doesn't show on its own that every proper extension of ${\mathbf Q}(\sqrt{2})$ is ramified at some prime in ${\mathbf Q}(\sqrt{2})$. However, we can bound the degree of such an extension. The root disc. of ${\mathbf Q}(\sqrt{2})$ is $\sqrt{8} \approx 2.828$, which lies between $f(4)$ and $f(5)$, so any proper unramified extension of ${\mathbf Q}(\sqrt{2})$ must be a quadratic extension of ${\mathbf Q}(\sqrt{2})$. Quadratic extensions are automatically abelian, so we would have an abelian extension of ${\mathbf Q}(\sqrt{2})$ unramified at no finite places, and there's no such thing by class field theory since ${\mathbf Q}(\sqrt{2})$ has narrow class number 1 (the class number not involving ramification at infinity). Therefore you can add ${\mathbf Q}(\sqrt{2})$ to the list of number fields with no proper extension unramified at all finite places.
Unfortunately, this method is really limited because although $f(n)$ is increasing,
it's actually bounded. By Stirling's formula, $f(n)$ has limit $\pi e^2/4 \approx 5.803$. So when a number field $F$ has root discriminant exceeding this value, this method won't give us any examples at all (you can't find an $m$ for which the root discriminant is below $f(2m)$). I don't know how to generate examples by any other method. At least we got a few examples this way.
|
|
|
