= Solution
We first prove the simple Artinian case of the <Artin–Wedderburn theorem>. This part does not need finite dimension over $k$. Choose a minimal nonzero <left ideal> $L$, using the descending chain condition. Simplicity gives $LA=A$. Also $L^2\ne0$: otherwise $LAL=(LA)L=L$ would be zero, since each $lal'$ lies in $L^2$. Choose $x\in L$ for which right multiplication $r_x:L\to L$ is nonzero. By simplicity of $L$, it is an automorphism. Choose $e\in L$ with $ex=x$. Then $(e^2-e)x=0$, and injectivity of $r_x$ gives $e^2=e$. Thus $L=Ae$.
Right multiplication by $e$ splits the left regular <module> as $A=Ae\oplus A(1-e)$. Repeat within the remaining <left ideal>: if it contains a minimal <left ideal> $Af$ generated by an <idempotent>, the map $v\mapsto vf$ splits off $Af$. Each nonzero remaining complement is a strictly smaller <left ideal>, so the Artinian condition terminates the process. Therefore the regular <module> is a finite <direct sum> of minimal <left ideals>.
All these simple summands are isomorphic. For <primitive idempotents> $e,f$, simplicity gives $AeA=A$, so $fAe\ne0$; otherwise $fAeA=fA$ would be zero. A nonzero element of $fAe$ defines a nonzero homomorphism $Af\to Ae$, which is an isomorphism between <simple modules>. Thus $A\cong L^{\oplus n}$ as a left <module>. Put $\Delta=\operatorname{End}_A(L)$, a <division algebra> by the <Schur lemma>. Right multiplication gives
$$
A^{\mathrm{op}}\cong\operatorname{End}_A(A)
\cong M_n(\Delta).
$$
Taking opposites, with <matrix> transposition reversing products, proves
$$
\boxed{A\cong M_n(D),\qquad D=\Delta^{\mathrm{op}}\text{ a division algebra}.}
$$
The construction is $k$-linear and embeds $k$ into the center of $D$.
A <central simple algebra> is a finite-dimensional simple $k$-algebra whose center is exactly $k$. The <Brauer group> consists of such algebras modulo <matrix> stabilization: $A,B$ represent the same class if $M_r(A)\cong M_s(B)$ for some positive integers $r,s$. Its operation and inverse are
$$
\boxed{[A][B]=[A\otimes_kB],\qquad 1=[k],\qquad [A]^{-1}=[A^{\mathrm{op}}].}
$$
The <tensor product> is central simple, and $A\otimes A^{\mathrm{op}}\cong\operatorname{End}_k(A)$ by $(a\otimes b)(x)=axb$, explaining the inverse. Both facts can be checked after a <splitting field of a central simple algebra> extension, where they are matrix-algebra identities, and descended by faithful scalar extension. Every class has a unique central division-algebra representative, by the regular-module description above.
To prove infinitude over $\mathbb Q$ explicitly, put $E=\mathbb Q(i)$ and, for $a\ne0$, let
$$
A_a=(-1,a)_{\mathbb Q}=E\oplus Ej,\qquad j^2=a,\quad jz=\overline z\,j.
$$
This is a <quaternion algebra>. It is central simple: the center is $\mathbb Q$, and after adjoining $i$ its generators have the faithful <matrix> model
$$
i\longmapsto\operatorname{diag}(i,-i),\qquad
j\longmapsto\begin{pmatrix}0&a\\1&0\end{pmatrix}.
$$
These generate $M_2(E)$; consequently a nonzero proper <two-sided ideal> over $\mathbb Q$ is impossible. The quaternion conjugation gives
$$
N(z+w j)=N_{E/\mathbb Q}(z)-aN_{E/\mathbb Q}(w).
$$
A nonzero element has an inverse exactly when this norm is nonzero. Thus $A_a$ is a <division algebra> when $a$ is not an $E/\mathbb Q$ norm; if $a$ is a norm it has a nonzero zero divisor and is $M_2(\mathbb Q)$ by the classification just proved.
For a prime $p\equiv3\pmod4$, every nonzero rational sum $x^2+y^2$ has even $p$-adic valuation. Divide out the minimum valuation of $x,y$; cancellation modulo $p$ is impossible because $-1$ is not a square modulo $p$. Therefore $p$ is not an $E/\mathbb Q$ norm, and $A_p$ is a nontrivial division-algebra Brauer class. This is the <quaternion norm obstruction at primes congruent to three modulo four>.
We also prove the product rule needed to distinguish classes. In $A_a\otimes A_b$, write the generators as $i_1,j_1,i_2,j_2$. The commuting subalgebras generated respectively by
$$
i_1,\ j_1j_2\qquad\text{and}\qquad i_1i_2,\ j_2
$$
are $A_{ab}$ and the split <quaternion algebra> $(1,b)_{\mathbb Q}\cong M_2(\mathbb Q)$. The latter has the <matrix> model $\operatorname{diag}(1,-1)$ and $\left(\begin{smallmatrix}0&b\\1&0\end{smallmatrix}\right)$. Together the two subalgebras generate the whole <tensor product>, since they recover $i_2$ and $j_1$. Dimensions are sixteen on both sides, giving
$$
\boxed{A_a\otimes A_b\cong A_{ab}\otimes M_2(\mathbb Q),\qquad [A_a][A_b]=[A_{ab}].}
$$
Quaternion conjugation identifies $A_a$ with its opposite, so these classes have order at most two. For distinct primes $p,q\equiv3\pmod4$, the parameter $pq$ has odd valuation at $p$, so $A_{pq}$ is a <division algebra>. Thus $[A_p]\ne[A_q]$. There are infinitely many such primes: if their list were $p_1,\ldots,p_m$, the integer $4p_1\cdots p_m-1$ would have a prime factor congruent to three modulo four outside that list. We conclude
$$
\boxed{\operatorname{Br}(\mathbb Q)\text{ is infinite}.}
$$
Indeed the same valuation argument for any nonempty finite product gives <independent quaternion classes over rational numbers>.
For $\mathbb R$, we prove the required division-algebra classification. Let $D$ be finite-dimensional and central over $\mathbb R$. If $D\ne\mathbb R$, choose a nonreal element; its irreducible real <minimal polynomial> is quadratic, so translating and rescaling it gives $i\in D$ with $i^2=-1$. The centralizer $D_+$ of $i$ is a finite-dimensional <division algebra> over $\mathbb C=\mathbb R(i)$, with that <field> central. Every element has a split <minimal polynomial> over $\mathbb C$, hence belongs to $\mathbb C$, so $D_+=\mathbb C$.
Conjugation by $i$ is an involution, giving $D=D_+\oplus D_-$, where $D_-=\{x:ix=-xi\}$. If $D_-=0$, then $D=\mathbb C$, contrary to its center being $\mathbb R$. Choose $j\ne0$ in $D_-$. For any $x\in D_-$, $xj^{-1}$ commutes with $i$, so $D_-=\mathbb Cj$. Also $j^2\in\mathbb C$ and commutes with $j$, forcing $j^2\in\mathbb R$. It cannot be positive, since $(j-\sqrt{j^2})(j+\sqrt{j^2})=0$, or zero, since $D$ is a <division algebra>. Rescale to $j^2=-1$. Consequently
$$
D=\mathbb R\oplus\mathbb Ri\oplus\mathbb Rj\oplus\mathbb Rij\cong\mathbb H.
$$
Conversely the positive definite quaternion norm makes $\mathbb H$ a <division algebra>. The two possible central division representatives are therefore $\mathbb R$ and $\mathbb H$. They are not equivalent, and conjugation identifies $\mathbb H$ with its opposite, so
$$
\boxed{\operatorname{Br}(\mathbb R)\cong\mathbb Z/2,\quad\text{generated by }[\mathbb H].}
$$
This uses the <finite-dimensional real division-algebra classification> directly, not as an unexplained theorem citation.
Finally let $k_s$ be a separable closure and $\Gamma=\operatorname{Gal}(k_s/k)$. The cohomological description is
$$
\boxed{\operatorname{Br}(k)\cong H^2_{\mathrm{cont}}(\Gamma,k_s^\times)
=\varinjlim_{L/k\ \mathrm{finite\ Galois}}H^2(\operatorname{Gal}(L/k),L^\times).}
$$
Here $k_s^\times$ has the discrete topology and its natural Galois action. For a finite Galois splitting <field> $L$, identify $A\otimes_kL$ with a <matrix algebra>. Its descent action is implemented by matrices $u_\sigma$, unique up to scalars, so
$$
u_\sigma\,\sigma(u_\tau)\,u_{\sigma\tau}^{-1}=c(\sigma,\tau)^{-1}\in L^\times.
$$
The inverse convention here agrees with the crossed-product convention below. The semilinear operators $u_\sigma\sigma$ commute with the descended algebra $A$; their crossed product has class $[A]^{-1}$ by the commuting-factor decomposition of the full endomorphism algebra. Inverting its scalar cocycle therefore gives the class $[A]$. Associativity gives the <two-cocycle> identity, and changing the representatives changes $c$ by a coboundary. Conversely a <two-cocycle> defines the <crossed-product algebra of a Galois extension>, with $u_\sigma a=\sigma(a)u_\sigma$ and $u_\sigma u_\tau=c(\sigma,\tau)u_{\sigma\tau}$. It is central simple and split by $L$. <Galois descent> identifies coboundaries with split matrix-algebra classes and gives $H^2(\operatorname{Gal}(L/k),L^\times)\cong\ker(\operatorname{Br}(k)\to\operatorname{Br}(L))$. Every <central simple algebra> has a finite separable, hence finite Galois, splitting <field>, yielding the displayed limit.
Back to article page
