Let $p >0$ be an odd prime and let $\mathbb{K} = \mathbb{Q}(\zeta) \subseteq \mathbb{C}$ with $\zeta$ a primitive $p$th root of unity. There is a unique subfield $\mathbb{Q} \subseteq \mathbb{F} \subseteq \mathbb{K}$ satisfying $[\mathbb{F}:\mathbb{Q}]=2$. Specifically $\mathbb{F} = \mathbb{Q}(\alpha)$ where $\alpha^2 = (-1)^{\frac{p-1}{2}}p$.


If $p \equiv -1 \pmod{4}$ then $[\mathbb{K}:\mathbb{F}] = (p-1)/2$ is odd so $\mathrm{N}\_{\mathbb{K/F}}(-1) = (-1)^{[\mathbb{K}:\mathbb{F}]} = -1$, where $\mathrm{N}\_{\mathbb{K/F}} : \mathbb{K} \to \mathbb{F}$ is the field norm.



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> When $p \equiv 1 \pmod{4}$ why is $-1$ not in the image of $\mathrm{N}\_{\mathbb{K/F}}$?
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This should be very elementary and for reasons coming from representation theory of finite groups I know this statement is true. However I'd like a straightforward number theory argument for this.


My basic idea was the following. We have $[\mathbb{K}:\mathbb{Q}] = p-1$ and by assumption $4 \mid p-1$ so there exists a unique subfield $\mathbb{Q} \subseteq\mathbb{E} \subseteq \mathbb{K}$ with $[\mathbb{E}:\mathbb{Q}] = 4$. By transitivity of norms, if $-1$ is in the image of $\mathrm{N}\_{\mathbb{K/F}}$ then it's also in the image of $\mathrm{N}\_{\mathbb{E/F}}$. Hence, it suffices to show it's not in the image of $\mathrm{N}\_{\mathbb{E/F}}$. Here is where I got a bit stuck as I wasn't sure what the field $\mathbb{E}$ is exactly. One can get a basis by taking Galois sums of $\zeta$. There might also be a way to use the discriminent.


I have the same question in the local case. So assume $\mathbb{K} = \mathbb{Q}\_{\ell}(\zeta)$ with $\ell > 0$ a prime and $\zeta \in \overline{\mathbb{Q}}\_{\ell}$ a primitive $p$th root of unity. Looking in Serre's Local Fields the Galois group $\mathrm{Gal}(\mathbb{K}/\mathbb{Q}\_{\ell})$ should still be cyclic. So assume $2$ divides $[\mathbb{K}:\mathbb{Q}\_{\ell}]$ then there is a unique subfield $\mathbb{Q}\_{\ell}\subseteq\mathbb{F} \subseteq \mathbb{K}$ with $[\mathbb{F}:\mathbb{Q}\_{\ell}]=2$. The following should be true:



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> If $\ell \neq p$ then $-1$ is in the image of the norm map $\mathrm{N}\_{\mathbb{K/F}}$. If $\ell = p$ then $-1$ is in the image of the norm map $\mathrm{N}\_{\mathbb{K/F}}$ if and only if $p \equiv -1 \pmod{4}$.
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By a block theory argument from finite groups I know the statement when $\ell \neq p$ is true. Hasse's Norm Theorem would then imply that the $\ell = p$ case agrees with the global case. However this all feels far too overblown. There should be an elementary number theory argument for all of this.


Apologies in advance if this is all too elementary. I am sure these answers and arguments are well known.



When is $-1$ in the image of a field norm?
*Global question*: If $p = 1 \bmod 4$, then the element $-1 \in F^\times$ is not a norm from $K^\times$, because $F^\times$ is totally real and $K^\times$ is totally complex; thus $-1$ is not a local norm at either of the infinite places of $F$, and hence cannot be a global norm either.


*Local question, $\ell \ne p$*: If $\ell \ne p$ then the fields $K$ and $F$ you define are both unramified extensions of $\mathbf{Q}\_\ell$. Hence $N\_{K/F}(O\_K^\times) = O\_F^\times$, and in particular contains $-1$. (This surjectivity result for the norm map on units follows from the analogous result for the residue fields, which is easy to prove by hand, see slide 6 of [these lecture notes by Garrett](http://www-users.math.umn.edu/%7Egarrett/m/number_theory/overheads/noth-04-09-2012.pdf).)


*Local question, $\ell = p$*: As pointed out by KConrad in the comments, if you take $\omega$ a $(p-1)$-st root of unity in $\mathbf{Q}\_p^\times \subseteq F^\times$, then the composite $F^\times \hookrightarrow K^\times \xrightarrow{N\_{K/F}} F^\times$ is raising to the $(p-1)/2$-th power; so its image contains $\omega^{(p-1)/2}$ for every $(p-1)$-st root of unity, i.e. it always contains $-1$.


(You were correct in deducing from Hasse's theorem that if $-1$ is not a norm globally then it had to fail to be a norm locally at some place. However, the bad place is not $p$ but $\infty$.)


