Reading about the calculation of the Brauer group of rational numbers, the calculations of the group are extremely lengthy and technical. First of all, it will be very helpful to me if someone can explicitly give me a class which corresponds to a nontrivial member of the Brauer group of rational numbers. Secondly, I appreciate it if someone can sketch the calculation if we assume we know the Brauer group of every local field is $\mathbb{Q}/\mathbb{Z}$.
Brauer Group – Rational Numbers
ac.commutative-algebraalgebraic-number-theorybrauer-groupscentral-simple-algebrasnt.number-theory
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Okay, here's a few words about the relation between the $L$-series and the formal group. In general, if $F(X,Y)$ is the formal group law for $\hat G$, then there is an associated formal invariant differential $\omega(T)=P(T)dT$ given by $P(T)=F_X(0,T)^{-1}$. Formally integrating the power series $\omega(T)$ gives the formal logarithm $\ell(T)=\int_0^T\omega(T)$. The logarithm maps $\hat G$ to the additive formal group, so we can recover the formal group as $$ F(X,Y) = \ell^{-1}(\ell(X)+\ell(Y))$$. (See, e.g., Chapter IV of Arithmetic of Elliptic Curves for details.)
Now let $E$ be an elliptic curve and $\omega=dx/(2y+a_1x+a_3)$ be an invariant differential on $E$. If $E$ is modular, say corresponding to the cusp form $g(q)$, then we have (maybe up to a constant scaling factor) $\omega = g(q) dq/q = \sum_{n=1}^{\infty} a_nq^{n-1}$. Eichler-Shimura tell us that the coefficients of $g(q)$ are the coefficients of the $L$-series $L(s)=\sum_{n=1}^\infty a_n n^{-s}$. Integrating $\omega$ gives the elliptic logarithm, which is the function you denoted by $f$, i.e., $f(q)=\sum_{n=1}^\infty a_nq^n/n$, and then the formal group law on $E$ is $F(X,Y)=f^{-1}(f(X)+f(Y))$.
To me, the amazing thing here is that the Mellin transform of the invariant differential gives the $L$-series. Going from the invariant differential to the formal group law via the logarithm is quite natural.
In general, to compute the class group of a number field $K$ of degree $n$ and discriminant $\Delta$, we need to find some bound $N$ such that the class group of $K$ is generated by primes of norm at most $N$. Unconditionally, we have the Minkowski bound $$M_K = \sqrt{\lvert \Delta \rvert} \left( \frac{4}{\pi} \right)^{s} \frac{n!}{n^n},$$ where $s$ is the number of conjugate pairs of complex places of $K$. Since $M_K$ is proportional to the square root of the absolute discriminant, it is often infeasible to check all primes up to that bound. To my knowledge, no sizable asymptotic improvement on this bound has been proven unconditionally.
However, assuming GRH, we have a vastly stronger bound due to Bach (1990): under GRH, the class group is generated by primes of norm at most $$B_K = 12 \log^2 \lvert \Delta \rvert.$$ A similar bound, asymptotically weaker in theory but stronger in practice for many fields of interest, is due to Belabas, Diaz y Diaz, and Friedman (2008).
Let's take $\mathbb{Q}(\zeta_{29})$ as an example: the Minkowski bound is $14956520729 \approx 1.4 \times 10^{10}$, so we would need to check about 64 million primes to provably compute the class group—not totally infeasible with supercomputers but certainly a very computationally intensive task. In contrast, the Bach bound is $99190$, while the version implemented in Magma using results of Belabas, Diaz y Diaz, and Friedman gives a bound of $826$. With this latter bound, Magma can compute the class group in $1.8$ seconds on my laptop, assuming there are no elements of the class group whose representatives all have norm greater than $826$ (the existence of which would contradict GRH).
References:
- Bach, E. “Explicit bounds for primality testing and related problems”. Mathematics of Computation 55 (1990), no. 191, pp. 355–380. [MR1023756]
- Belabas, K., Diaz y Diaz, F., and Friedman, E. “Small generators of the ideal class group”. Mathematics of Computation 77 (2008), no. 262, pp. 1185–1197. [MR2373197]
Best Answer
What reference are you reading?
For a field $K$, every finite-dimensional central simple $K$-algebra $A$ is isomorphic to ${\rm M}_n(D)$ where $n$ is a positive integer and $D$ is a division ring with center $K$ where $\dim_K(D)$ is finite. The number $n$ is unique and $D$ is unique up to $K$-algeba isomorphism. The equivalence classes defining ${\rm Br}(K)$ just involve declaring ${\rm M}_n(D) \sim D$. That is, we identify two $A$'s if they have isomorphic $D$'s. Thus representatives for ${\rm Br}(K)$ are the finite-dimensional $K$-central division algebras, up to isomorphism.
The group law on ${\rm Br}(K)$ uses the tensor product of representatives: $[A][A'] = [A \otimes_K A']$ (isomorphic central simple $K$-algebras have isomorphic tensor products over $K$). Since we can represent each element of ${\rm Br}(K)$ by a unique finite-dimensional $K$-central division algebra up to isomorphism, in terms of $K$-central division algebras you could say the group law in ${\rm Br}(K)$ is $[D][D'] = [D'']$ where $D \otimes_K D' \cong {\rm M}_n(D'')$ for some $n$ and $D''$. (Note: a tensor product of two $K$-central division rings need not be a division ring, e.g., ${\mathbf H} \otimes_{\mathbf R} {\mathbf H} \cong {\rm M}_{4}(\mathbf R)$, so in ${\rm Br}(\mathbf R) = \{[\mathbf R],[\mathbf H]\} =\{1,[\mathbf H]\}$ we have $[\mathbf H]^2 = 1$.)
To ask for a nontrivial element of ${\rm Br}(K)$ is to ask for a finite-dimensional $K$-central division algebra other than $K$ itself. The simplest examples besides $K$ are the quaternion algebras over $K$, which are defined to be the $4$-dimensional central simple $K$-algebras. Every quaternion algebra over $K$ that's not isomorphic to ${\rm M}_2(K)$ will be a division ring. (For some $K$ there are no quaternion algebras over $K$ except for ${\rm M}_2(K)$, such as when $K$ is algebraically closed or finite, in fact in these cases ${\rm Br}(K)$ is trivial: all finite-dimensional $K$-central simple algebras are just the rings ${\rm M}_n(K)$ up to isomorphism.) A concrete account of quaternion algebras, which avoids working in characteristic $2$ (even though the general theory can handle this case if you develop it in a more abstract way) is here. In ${\rm Br}(K)$, the quaternion algebras (including the trivial or "split" case ${\rm M}_2(K)$) are representatives of the $2$-torsion subgroup: every element of ${\rm Br}(K)$ with order dividing $2$ is represented by a unique quaternion algebra over $K$, so the tensor product operation is a group law on quaternion algebras over $K$.
It's very helpful to be able to understand quaternion algebras well before trying to deal with the full Brauer group of a field, just as quadratic fields are a good testing ground to understand theorems about general number fields.
The construction called cyclic algebras gives you an $n^2$-dimensional $K$-central simple algebra from a cyclic Galois extension $L/K$ of degree $n$. (The Hamilton quaternions come in this way from the quadratic extension $\mathbf C/\mathbf R$: $\mathbf H = \mathbf C + \mathbf Cj$ where $j^2 = -1$ and $jz = \overline{z}j$ for all $z \in \mathbf C$.) Number fields and local fields have lots of cyclic Galois extensions (think of the unramified extensions of $\mathbf Q_p$ or suitable subfields of cyclotomic extensions of $\mathbf Q$), so you can get lots of central simple algebras of number fields and local fields this way. An account of cyclic algebras is in Section 14 of Chapter 5 of Lam's A First Course in Noncommutative Rings, and in that section there is an example of a $9$-dimensional division ring over $\mathbf Q$ built from a cyclic cubic extension of $\mathbf Q$ (the cubic subfield of the $7$th cyclotomic field). Also see Definition 7.10 and Theorem 7.13 here. There is a more general way of building $n^2$-dimensional central simple algebras from degree-$n$ Galois extensions (not necessarily having a cyclic Galois group) called a crossed product algebra. It is not always clear when these constructions are nontrivial, meaning they are not isomorphic to ${\rm M}_n(K)$
The general concepts I have mentioned here can be applied to an arbitrary base field $K$: $\mathbf Q$, $\mathbf R$, $\mathbf Q_p$, finite fields, and so on. The fact that ${\rm Br}(\mathbf Q_p) \cong \mathbf Q/\mathbf Z$ amounts to showing every $n^2$-dimensional division ring over $\mathbf Q_p$ can be built as a cyclic algebra using the degree-$n$ unramified extension of $\mathbf Q_p$ (which is a cyclic extension). See Theorem 7.14 here. Figuring out when two such constructions can be isomorphic, in terms of the parameters that show up in the construction, turns out to be the same as asking what the elements of order $n$ look like in $\mathbf Q/\mathbf Z$. A detailed proof of the calculation of ${\rm Br}(K)$ when $K$ is a local field is in Jacobson's Basic Algebra II: see the sections on division rings and the Brauer group near the end of Chapter $9$.