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For a finite-dimensional algebra over an algebraically closed field, basic algebra means $A/J(A)\cong k^n$, so its simple modules are one-dimensional and occur once in the semisimple quotient of the regular module. A hereditary ring has every submodule of a projective module projective; equivalently its global dimension is at most one.

Choose representatives $S_1,\dots ,S_n$ of the simple modules. Our Ext quiver convention gives one vertex $i$ for $S_i$ andWith primitive orthogonal idempotents $e_i$, put $P_i=A{e}_i$ and $J=J(A)$. Applying ${\mathrm{Hom}}_A(-,S_j)$ to $0\to J{P}_i\to P_i\to S_i\to 0$ givesThe restriction map from $\mathrm{Hom}(P_i,S_j)$ is zero since every map to a simple kills $J{P}_i$, and $P_i$ is projective. Thus the arrow multiplicity is $\dim e_j(J/J^2)e_i$. This specifies the orientation rather than silently replacing the quiver by its opposite.

Because $A$ is hereditary, $J{P}_i$ is projective. Its semisimple top has $a_{ij}$ copies of $S_j$, where $a_{ij}$ is the arrow multiplicity. Its projective cover therefore supplies an epimorphismThis splits by projectivity of $J{P}_i$. Its kernel is a summand contained in the radical of the source; projection onto that summand and Nakayama lemma force the kernel to be zero. HenceEvery arrow $i\to j$ forces $\dim P_i>\dim P_j$, so $Q$ has no directed cycles.

Choose lifts in $e_{j}J{e}_i$ of bases of the arrow spaces. Send the vertices and arrows of the path algebra to these idempotents and lifts, multiplying paths in the order $\beta \alpha$ for consecutive arrows $\alpha :i\to j$, $\beta :j\to l$. This defines an algebra homomorphism $kQ\to A$. It is surjective: vertices span $A/J$, arrows span $J/J^2$, and products show that paths of length $r$ span $J^r/J^{r+1}$. Nilpotence of $J$ then gives all of $A$.

Let $N_i$ be the number of directed paths starting at $i$, including the length-zero path. Acyclicity gives the same recursion $N_i=1+{\sum }_j{a}_{ij}{N}_j$, with value one at a sink. Induction from sinks gives $N_i=\dim P_i$. ThereforeThe surjective homomorphism is an isomorphism. This proves the basic hereditary algebra path-algebra theorem:

For each connected component $C$ of the underlying unoriented graph, $e_C={\sum }_{i\in C}{e}_i$ is a central idempotent. Paths never connect distinct components, so $kQ={⨁}_C{e}_{C}kQ$ as a direct sum of two-sided ideals. To prove each is a block, let $z$ be any central idempotent. Modulo the arrow ideal it has coordinates $c_i\in \{0,1\}$. Commuting $z$ with an arrow $i\to j$ and reducing modulo the square of that ideal gives $c_i=c_j$. Thus these coordinates are constant on each component. If they are zero on a component, the restriction of $z$ there is a nilpotent idempotent, hence zero; if they are one, apply that argument to $1-z$. Consequently

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 $L^2\ne 0$: otherwise $LAL=(LA)L=L$ would be zero, since each $la{l}^{\prime }$ 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 $fAe\ne 0$; 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 $\Delta ={\mathrm{End}}_A(L)$, a division algebra by the Schur lemma. Right multiplication givesTaking opposites, with matrix transposition reversing products, provesThe 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 areThe tensor product is central simple, and $A\otimes A^{\mathrm{op}}\cong {\mathrm{End}}_k(A)$ 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 $a\ne 0$, letThis is a quaternion algebra. It is central simple: the center is $\mathbb{Q}$, and after adjoining $i$ its generators have the faithful matrix modelThese generate $M_2(E)$; consequently a nonzero proper two-sided ideal over $\mathbb{Q}$ is impossible. The quaternion conjugation givesA 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 $p\equiv 3(\mathrm{mod}4)$, 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 byare $A_{ab}$ and the split quaternion algebra $(1,b{)}_{\mathbb{Q}}\cong M_2(\mathbb{Q})$. The latter has the matrix model $\mathrm{diag}(1,-1)$ and $\left(\begin{array}{cc}0& b\\ 1& 0\end{array}\right)$. Together the two subalgebras generate the whole tensor product, since they recover $i_2$ and $j_1$. Dimensions are sixteen on both sides, givingQuaternion conjugation identifies $A_a$ with its opposite, so these classes have order at most two. For distinct primes $p,q\equiv 3(\mathrm{mod}4)$, 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 $p_1,\dots ,p_m$, the integer $4p_1\cdots p_m-1$ would have a prime factor congruent to three modulo four outside that list. We concludeIndeed 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 $j\ne 0$ in $D_{-}$. For any $x\in D_{-}$, $x{j}^{-1}$ commutes with $i$, so $D_{-}=\mathbb{C}j$. 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 $j^2=-1$. ConsequentlyConversely 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, soThis uses the finite-dimensional real division-algebra classification directly, not as an unexplained theorem citation.

Finally let $k_s$ be a separable closure and $\Gamma =\mathrm{Gal}(k_s/k)$. The cohomological description isHere $k_s^{\times }$ has the discrete topology and its natural Galois action. For a finite Galois splitting field $L$, identify $A{\otimes }_{k}L$ with a matrix algebra. Its descent action is implemented by matrices $u_{\sigma }$, unique up to scalars, soThe 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 $H^2(\mathrm{Gal}(L/k),L^{\times })\cong \ker (\mathrm{Br}(k)\to \mathrm{Br}(L))$. Every central simple algebra has a finite separable, hence finite Galois, splitting field, yielding the displayed limit.

For a two-sided ideal $I$ of a ring $R$, the quotient $R/I$ has addition and multiplication induced from $R$.