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Suppose $L/K$ is a Galois extension of number fields, with Galois group $G_{L/K}$. Write $\mathrm{Br}(L)^{G_{L/K}}$ for the subgroup of central simple algebras $A/L$ which are Galois-invariant; equivalently, these are the algebras such that for $v$ a place of $K$ and $w$ a place of $L$ with $w|v$, the local invariants $\mathrm{inv}_w(A)$ only depend on $v$. The Hochschild-Serre spectral sequence yields an exact sequence

$0\to \mathrm{Br}(L/K) \to \mathrm{Br}(K) \to \mathrm{Br}(L)^{G_{L/K}}\to H^3(G_{L/K},L^{\times}),$

where the third arrow is represented by the base change map $A\mapsto A\otimes_{K} L$, and the image of the fourth arrow is the "obstruction" preventing a Galois invariant algebra from arising via base change. Now, the $H^3$ appearing here is often zero; for example, it vanishes if all the Sylow subgroups of $G_{L/K}$ are cyclic. My questions:

  1. In the case $K=\mathbf{Q}$ (say), what is the simplest example of a Galois extension $L/\mathbf{Q}$ and a Galois-invariant central simple algebra over $L$ which is not a base change from any proper subfield? Does $G_{L/\mathbf{Q}}=A_4$ work?

  2. Is there a simple way to compute the obstruction map into $H^3$? It seems like the answer to this must naively be "no", since it cannot be as simple as "do such and such with the local invariants".

  3. How much of the above goes through without assuming $L/K$ is Galois?

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2 Answers 2

up vote 12 down vote accepted

[big edit]

(1) Let $L/K$ be as above. Take any non-identity element $\sigma \in G_{L/K}$, and let $F=L^{<\sigma>}$ be the fixed field of the cyclic subgroup generated by $\sigma$. By your comment above about the vanishing of $H^3$, every galois invariant CSA of $L$ is a base change of a CSA of $F$. Hence, there are no galois invariant CSA that are not a base change from any proper subfield.

But, if you fix $L$ and $K$, then there can be galois invariant CSA's of $L$ that are not a base change from $K$. Since the smallest non-cyclic group is $C_2\times C_2$, we would like to search for examples with such a galois group.

For $K=\mathbb{Q}$, probably, the simplest example is: $$L=\mathbb{Q}(\sqrt{-3},\sqrt{13}),\ (43)=P_1\cdot P_2\cdot P_3\cdot P_4,$$ $$u=\frac{1}{4}P_1+\frac{1}{4}P_2+\frac{1}{4}P_3+\frac{1}{4}P_4 \in \bigoplus_v Br(L_v)$$

I will prove below that this element of the Brauer group is not a base change, and that in fact a similar construction can be given for any extension.

(2) Near the end of Tate's "Global Class Field Theory" section in Cassels-Frohlich, there is a small paragraph devoted to $H^3(G_{L/K},L^\times)$:

"$H^3(G,L^\times)$ is cyclic of order $n/n_0$, the global degree divided by the lowest common multiple of local degrees, generated by $\delta u_{L/K}$ ($\delta:H^2(C_L)\rightarrow H^3(L^{\times})$), the "Teichmuller 3-class." ..."

Therefore, assume $n_0 < n$ for our galois extension $L/K$. Let $v_0$ be an unramified finite place of $K$ that splits completely (exists by Chebotarev's theorem). We can construct an element of the Brauer group similar to the one above: $$u := \sum_{w|v_0} \frac{1}{n} w$$

This is in fact an element of $Br(L)$ since the number of $w|v_0$ is $n$, so that $n\frac{1}{n} = 1 \in \mathbb{Z}$.

Proposition. For any prime $p$ that divides $\frac{n}{n_0}$, $\frac{n}{pn_0}\cdot u$ is not a base change.

From which we immediately get:

Corollary. The map $Br(L)^{G_{L/K}}\rightarrow H^3(L^\times)$ is onto.

Proof of proposition. Assume that $\frac{n}{pn_0}\cdot u$ is the base change of some $u'$, i.e. $$u' = \sum_v n_v v \mapsto \sum_v \sum_{w|v} [L_w : K_v] n_v w = \frac{n}{pn_0}\cdot u$$

Since for any $w|v_0$: $[L_w : K_{v_0}] = 1$, we must have $n_{v_0} = \frac{n}{pn_0}\cdot \frac{1}{n} = \frac{1}{pn_0}$. And since $\sum_v n_v \in \mathbb{Z}$, at least one other place $v_1$ has $$v_p(n_{v_1}) \le v_p(n_{v_0}) = v_p(\frac{1}{pn_0}) < 0$$

Where $v_p$ is the usual $p$-adic valuation. So, using that $n_0$ is the lcm of local degrees: $$v_p([L_{v_1}:K_{v_1}] n_{v_1}) \le v_p(n_0) + v_p(\frac{1}{pn_0}) = -1$$

Contradicting the zero coefficient of any $w|v_1$ in $u$.

A small computation shows that for the extension $\mathbb{Q}(\sqrt{-3},\sqrt{13})/\mathbb{Q}$, $n_0=2$, and that indeed this is the smallest (discriminant-wise) $C_2\times C_2$ such extension.

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Dror, thanks very much for these extremely helpful comments. A well-known gentleman emailed similar remarks to me, but I will be inclined to accept this answer (given a day or two). –  David Hansen Apr 10 '11 at 5:52
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I don't think that you have an obstruction on the level of Brauer groups. If you have a Galois-stable element over $L$, then you can choose some values of local invariants for its lift over $K$ such that their sum would be $0$ (in $\mathbb{Q}/\mathbb{Z}$). To this end you should lift your local invariants to elements of $\mathbb{Q}$ whose sum is $0$, and divide them by the degrees of the corresponding local extensions.

On the other hand, it seems that often the degree of any of the lifts (over $K$) is greater than the one of the original element (over $L$); hence on the level of algebras you can only lift a certain matrix algerbra over your original central divisible algebra.

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