Let $\pi\ne1+i$ be a prime element of $\mathbb Z[i]$. I am interested in the ramification in the extension $\mathbb Q(i,\sqrt[4]\pi)/\mathbb Q(i)$, especially over $(1+i)$. I've tried for instance to use exact sequences of inertia groups, but the problem is I have just as little knowledge of ramification in the extension $\mathbb Q(i,\sqrt[4]\pi)/\mathbb Q$.
Does anyone have any tips on how to proceed to compute e.g. ramification indices? Thanks in advance!
On
The ramification at $1 + i$ is a little subtle. This is a local question, about the field $K := \mathbb Q_2(i)$.
The problem is to take a unit $u$ in the ring of integers $\mathcal O_K$, and understand the ramification in $K(u^{1/4})$.
The parallel question for $\mathbb Q_2$ is to understand ramification in $\mathbb Q_2(u^{1/2})$, when $u \in \mathbb Z_2^{\times}$. This depends on the congruence class of $u \bmod 4$. Understanding that case first should help you understand the case you are interested in.
PS. I began writing this several weeks ago, but just came back to it now. I see that Prof. Lubin has written a more extensive answer along the same lines.
On
continuing from @Lubin's answer
$G = \Bbb Q_2(i)^* / \Bbb Q_2(i)^{4*}$ is isomorphic to $(\Bbb Z/4\Bbb Z)^4$, with generators $(1+i), (i), (3+2i), (1+4i)$.
I picked $3+2i$ and $1+4i$ so that they are congruent to $1 \pmod {(1+i)^3}$. Furthermore, the norm of $3+2i$ is the farthest from $1$ possible (it is only $1 \pmod 4$), and the norm of $1+4i$ is $1 \pmod {16}$ (which is the best we can say because those elements are determined mod $(1+i)^7$).
The way $X^4-\pi$ factors over $\Bbb Q_2(i)$ tells us about the value of $g$ (the number of factors is the number of primes over $(1+i)$).
We have $g =4$ if and only $\pi$ is a $4th$ power in $\Bbb Q_2(i)$, that is if $\overline{\pi} = 1$, and $g \ge 2$ if and only if it is a square in $\Bbb Q_2(i)$, so we have a hierarchy $G_{g\ge 4} = \{1\} \subset G_{g\ge 2} = \{squares \} \subset G$ that tells us about $g$.
The extension is unramified ($e=1$) if and only if $\pi \equiv 1 \pmod {(1+i)^3}$ and $Norm(\pi) \equiv 1 \pmod {16}$, that is, if $\overline{\pi} \in G_{e \le 1} = \langle 1+4i \rangle$. Once again, there is a group hierarchy telling us the precise value of $e$, (maybe one can explicit the value of $e$ according to wether some values taken by the perfect pairing given by the $4$th power residue symbols are $1$ or not ? it should explain why they are groups, and might give us the indices in the hierarchy).
$G_{e\le 2}$ has to contain $G_{e \le 1}$ and $G_{g \ge 2}$ (because $efg=4$), and in fact it is generated by them. In the $1 \pmod {(1+i)^3}$ part, it contains the elements of norm $1 \pmod 8$, to which we add $(-1)$ and $2i$ from the unit and uniformizer parts.
Since the values for $e$ and $g$ determine the value of $f$, you end up with some nice triangular picture where those two hierarchies interact.
For example there are two inert cases ($g=e=1$), given by the classes $(1 \pm 4i)$.
Also, for the case of $\pi = 5$, it is not a square in $\Bbb Q_2(i)$, and $Norm(5)=25 \equiv 9 \pmod {16}$ so $g=1$ and $e=2$, hence $f=2$
On
Let $K = \Bbb Q_2(i)$. From local class field theory, there is a local Artin map $\Phi_K : K^* \to Gal(K^{ab}/K)$, such that
for any finite abelian extension $L$, $\Phi_K^{-1}(Gal(K^{ab}/L)) = Norm(L^*)$ (which induces an isomorphism $\Phi_{L/K}: K^*/Norm(L^*) \to Gal(L/K)$)
and $L/K$ is unramified if and only if the norm subgroup contains $\mathcal O^*$ (which induces an isomorphism $\pi^{\hat {\Bbb Z}} = \widehat{K^* / \mathcal O^*} \to Gal(K^{unr}/K) = Gal(\overline{\Bbb F_2} / \Bbb F_2) = Frob_2^{\hat {\Bbb Z}}$)
Then for a given extension with a given norm subgroup $H$, its largest unramified subextension corresponds to the smallest norm subgroup containing $H$ and $\mathcal O^*$, so you can read off the values of $e$ and $f$ as $e = [H\mathcal O^* : H]$ and $f = [K^* : H\mathcal O^*]$.
Now consider $L = K(\{\sqrt[4] \pi \mid \pi \in K^*\})$. Here, $Gal(L/K)$ is naturally isomorphic to $(K^*/K^{*4})^\vee = \hom(K^*, U_4 = \{1,i,-1,-i\}) $ and since its norm subgroup has to contain $K^{*4}$, it has to be that exactly to make the size match.
Then you obtain a perfect pairing $K^*/K^{*4} \to K/K^{*4} \to U_4$ defined as $\langle x, y \rangle = \Phi_{L/K}(x)(\sqrt[4]y)/(\sqrt[4]y) \in \{1,i,-1,-i\}$.
From the properties of $\Phi$ it follows that
$\langle x,y \rangle = 1$ iff $\Phi_{L/K}(x)$ fixes $K(\sqrt[4]y)$ iff $x \in Norm (K(\sqrt[4]y)^*)$, in particular, since $\sqrt[4]y$ has norm $-y$, $\langle -x,x \rangle = 1$. Then also
$\langle x,y \rangle \langle x,z \rangle = \langle x,yz \rangle$ and $\langle x,z \rangle \langle y,z \rangle = \langle xy,z \rangle$, which is enough to deduce the reciprocity law $\langle x,y \rangle \langle y,x \rangle = 1$.
Those properties are enough to quickly determine the pairing up to complex conjugation, and it is enough to compute orthogonals for what follows.
If $y \in K^*/K^{*4}$, then its orthogonal $y^\bot = \{x \mid \langle x,y \rangle = 1\}$ is the norm subgroup of the extension $K(\sqrt[4] y)$, and the parameters you're looking for are $e = [y^\bot \mathcal O^* : y^\bot], f = [K^* : y^\bot \mathcal O^*], g = 4/ef = 4/[K^* : y^\bot]$.
It turns out that ${\mathcal O^*}^\bot$ is the subgroup generated by $(1+4i)$ (which means that the maximal unramified subextension of $L$ is obtained by taking a fourth root of $1+4i$ ; $\mathcal O^{* \bot}$ is also the cyclic $4$-group closest to $1$ for the $(1+i)$-adic metric), so dualizing everything, we finally get :
$f(y) = |\langle y \rangle \cap \langle 1+4i \rangle |$
$e(y) = [\langle y \rangle : \langle y \rangle \cap \langle 1+4i \rangle] = |\langle y \rangle| / f$
$g(y) = 4/|\langle y \rangle|$
Notice that for $d=1,2,4$,
$d$ divides $g(y)$ iff $ord(y)$ divides $(4/d)$ iff $y^{4/d} = 1$,
and $e(y)$ divides $d$ iff $y^d \in \langle 1+4i \rangle $ iff $y=zt$ with $z \in \langle 1+4i \rangle$ and $t^d=1$, so we have the nice following subgroups :
$G_{1 \mid g} = K^*/K^{*4}$
$G_{2 \mid g} = K^{*2}/K^{*4}$
$G_{4 \mid g} = \{1\}$
$G_{e \mid 1} = \mathcal O^{*\bot} = \langle 1+4i \rangle$
$G_{e \mid 2} = G_{e \mid 1} G_{2 \mid g}$
$G_{e \mid 4} = G_{e \mid 1} G_{1 \mid g}$
If you want to know "how much" the extensions ramify (when they do), or equivalently, what is the power of $(1+i)$ in the relative discriminant of the extensions, or where do you have to look to get the integral closure of $\Bbb Z_2[i,\sqrt[4] \pi]$, it should be determined by how much $\langle y \rangle$ intersects the various subgroups $(1 + (1+i)^k\mathcal O_K)^\bot$ for $k=2 \ldots 6$ (so far we have only looked at the case where $k=1$)
I find this question somewhat frustrating, ’cause I haven’t been able to answer it to a degree of completeness that satisfies me, but at least I can help. I don’t see any way of doing this in a few lines.
First, since you’re interested only in the ramification and splitting of the prime $\mathfrak m=(1+i)$ in your extension $K=k(\pi^{1/4})$, where $k=\Bbb Q(i)$, the question is purely local, and we can feel free to localize and complete and thus work over $k'=\Bbb Q_2(i)$.
Second, I’m sure you know that since the extension $K\supset k$ is Galois, all the primes above $\mathfrak m$ have the same ramification index and residue field degree. So we have $efg=4$, where as usual, $e$ is the ramification index, $f$ is the residue field degree, and $g$ is the number of primes of $K$ above $\mathfrak m$. So there are apparently six cases: $e=4$ (totally ramified), $f=4$ (inert), $g=4$ (totally split), and the three mixed cases $e=f=2$, $e=g=2$, and $f=g=2$. Since the extension is cyclic, all six occur.
Third, we’re working with a Kummer extension, and we can use the facts and techniques of Kummer theory.
We could get to work right now on your problem, but I think it’s useful to look at two simpler analogous situations. If your base is $\Bbb Q$ instead, and ask about the splitting and ramification above $2$ of the extension $\Bbb Q(\sqrt n)$, where $n$ is odd (in all this, primality of your $\pi$ is not significant), then the answer depends entirely on the congruence of $n$ modulo $8$: if $n\equiv1$, then $2$ splits; if $n\equiv3$ or $7$, then $2$ ramifies; and if $n\equiv5$, the prime $2$ remains prime, i.e. is inert. You can see this very easily by passing to the complete situation, and ask about what happens when you adjoin $\sqrt m$ to $\Bbb Z_2$. Kummer extension, the quadratic extensions are told to you by $\Bbb Q_2^*/{\Bbb Q_2^*}^2$. Since you’re interested only in the square roots of odd numbers, you're reduced to looking at $\Bbb Z_2^*/{\Bbb Z_2^*}^2$. As you probably know, the multiplicative structure of $\Bbb Z_2^*$ is isomorphic to the additive group $C_2\oplus\Bbb Z_2$, where $C_m$ is the cyclic group of order $m$. The two generators are $-1$ and $5$. The squares here are therefore of index $4$, and you easily see that the subgroup $1+8\Bbb Z_2$ is contained in the squares but also has index $4$, so that ${\Bbb Z_2^*}^2=1+8\Bbb Z_2$ The four representatives of the quotient are $\{\pm1,\pm3\}$, corresponding to the four cases I mentioned above.
Let’s do the same for quadratic extensions of $k'=\Bbb Q_2(i)$ generated by the square roots of “odd” elements, that is numbers $\pi$ not divisible by $1+i$. I’ll call the ring of integers here $\mathfrak o$, equal to $\Bbb Z_2[\varpi]$, where $\varpi=1+i$ is a prime element in the local ring $\mathfrak o$; I’ll call the ideal that $\varpi$ generates $\mathfrak m$ again. Now, since the residue field is just $\Bbb F_2$, the units of $\mathfrak o$ are again $1+\mathfrak m$, which now has a multiplicative structure isomorphic to the additive group $C_4\oplus\Bbb Z_2\oplus\Bbb Z_2$. The squares therefore are of index $8$, and again $1+4\mathfrak m$ is contained in the squares, but has index $16$; the nonsquare not in $1+4\mathfrak m$ is $-1$, so that the squares are $\langle-1,1+4\mathfrak m\rangle$. So the splitting and ramification of $\mathfrak m$ in $K=k(\sqrt\pi)$ depends entirely on the congruence of $\pi$ modulo the multiplicative subgroup $\langle-1,1+4\mathfrak m\rangle$ of $1+\mathfrak m=\mathfrak o^*$. For instance, if $\pi\equiv1$, i.e. if $\pi\in{\mathfrak o^*}^2$, then $(1+i)$ splits; if $\pi\equiv5$, $(1+i)$ remains inert in the extension; and in all other cases, $(1+i)$ ramifies.
Finally, the question that you did ask. Now, because we’re working with a Kummer extension of degree $4$, you need to worry about ${k^*}^4$ as a subgroup of $k^*$, but because of your restriction to the fourth roots of odd Gaussian numbers, you need to examine $\mathfrak o^*=1+\mathfrak m$ and the subroup of its fourth powers. Now the subgroup $1+8\mathfrak m\subset1+\mathfrak m$ happens to be equal to ${\mathfrak o^*}^4$, the index being $64$. Again, if $\pi$ lands on the identity element of the quotient group, i.e. if $\pi\equiv1\pmod{1+8\mathfrak m}$, then $(1+i)$ splits completely in the extension. You can try any particular one of the other $63$ classes to see what the behavior is, depending on how the polynomial $X^4-\pi$ splits over $k'$. Because there are so many mixed cases, though, I don’t know what sensible general statements one may make. And I feel bad that I haven’t identified the (unique) class in the quotient group that gives you the inert case, except to check that it’s not $\sqrt[4]5\,$: that one gives a mixed case, $e=1$, $f=g=2$, as I recall.