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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 $f(5)$.

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 wide 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 $f(5)$.

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 $[E:F] > 1$.

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 $$ |d_F| < f(2m)^m = \frac{\pi^mm^{2m}}{(2m)!}. $$

Yes, there are examples and Minkowski's proof for ${\mathbf Q}$ can be adapted to find a few of them. Some examples of this kind among quadratic fields $F$, listed in increasing size of discriminant (in absolute value), are $$ {\mathbf Q}(\sqrt{-3}), \ \ {\mathbf Q}(i), \ \ {\mathbf Q}(\sqrt{5}), \ \ {\mathbf Q}(\sqrt{-7}), \ \ {\mathbf Q}(\sqrt{2}), \ \ {\mathbf Q}(\sqrt{-2}). $$ 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 with degree larger than 1 must have a discriminant whose absolute value is larger than 1. Over other number fields $F$ besides the rationals, the goal is to find sufficient conditions on $F$ so that any finite extension $E/F$ 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 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 $$|d_E| = {\rm N}({\mathfrak d}_{E/F})|d_F|^{[E:F]}.$$ So aiming to show ${\mathfrak d}_{E/F} \not= (1)$ is the same as avoiding $|d_E| = |d_F|^{[E:F]}$, which is the same as avoiding $$|d_E|^{1/[E:{\mathbf Q}]} = |d_F|^{1/[F:{\mathbf Q}]}.$$ We want sufficient conditions on $F$ to guarantee this equation for any proper finite extension $E/F$ 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 $F$. Now if $E/F$ had trivial discriminant ideal, we would have $$ |d_F|^{1/[F:{\mathbf Q}]} = |d_E|^{1/[E:{\mathbf Q}]} \geq f([E:{\mathbf Q}]) \geq f(2m). $$ This hypothetical lower bound on the root discriminant of $F$ is larger than the proved Minkowski bound of $f(m)$. Suppose $F$ is a number field of degree $m$ whose root discriminant is less than $f(2m)$. If $E/F$ is unramified at all primes in $F$ then $E$ and $F$ have equal root discriminants, so the 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 $f(2m)$: $|d_F|^{1/m} < f(2m)$. This is the same as $$ |d_F| < f(2m)^m = \frac{\pi^mm^{2m}}{(2m)!}. $$ 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.

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.