An integer cyclic presentation descends the simple-head generator line

We prove the regular-trace radical criterion directly over ANY characteristic-zero field $F$, without a splitting field or algebraic closure. Let $B$ be a finite-dimensional $F$-algebra and ${\beta }_B(a,b)={\mathrm{Tr}}_F(L_{ab}:B\to B)$. Every element of $J(B)$ is orthogonal to $B$, because its product with any element lies in the nilpotent radical and hence has nilpotent left multiplication and zero trace. Choose a finite composition series of the left regular module. F1–F2 give $B\mathrm{/}J(B)\cong {\prod }_i{M}_{n_i}(D_i)$, where each $D_i$ is a finite-dimensional division algebra over $F$; every simple composition factor is its column module $D_i^{n_i}$. Additivity of trace along the composition series shows that on this quotient the pairing is ${\sum }_i{m}_i{\mathrm{Tr}}_F({\rho }_i(X_i{Y}_i):D_i^{n_i}\to D_i^{n_i})$, where $m_i$ is the positive integer multiplicity of the column module in the regular composition series. Every $m_i$ is positive because the regular quotient already contains the corresponding column module. For a nonzero entry $x_{ab}$ of a block $X_i$, choose $Y_i$ with its only nonzero entry $(b,a)$ equal to $x_{ab}^{-1}$, and set all other blocks of $Y$ to zero. The product $X_i{Y}_i$ has exactly one nonzero diagonal entry, namely $1$ at position $a$. On the $F$-space $D_i^{n_i}$, its action therefore has trace ${\dim }_F{D}_i$; off-diagonal blocks contribute no trace. The full pairing on this $Y$ is $m_i{\dim }_F{D}_i\mathrm{\ne }0$ in characteristic zero. Thus the quotient pairing is nondegenerate and $\ker {\beta }_B=J(B)$.

Apply step 1.1 over $F=\mathbb{Q}$ and $F=K$. The regular multiplication and trace matrices for $A_K$ are scalar extensions of those for $A_{\mathbb{Q}}$. Row reduction of a finite matrix commutes with field extension, so the kernels satisfy $\ker {\beta }_{A_K}=K\otimes \ker {\beta }_{A_{\mathbb{Q}}}$. The radical criterion gives $J(A_K)=K\otimes J(A_{\mathbb{Q}})$. This argument applies equally to nonsplit division factors and needs no separable splitting theorem.

Put $S_{\mathbb{Q}}=\mathbb{Q}\otimes S_{\mathbb{Z}}$ and $D_{\mathbb{Q}}=S_{\mathbb{Q}}\mathrm{/}J(A_{\mathbb{Q}})S_{\mathbb{Q}}$. Tensoring quotient presentations is right exact and $K\mathrm{/}\mathbb{Q}$ is flat, so $K\otimes D_{\mathbb{Q}}=S_K\mathrm{/}J(A_K)S_K$. This last quotient is precisely the semisimple head: $A_K\mathrm{/}J(A_K)$ is semisimple, so the quotient is semisimple, while every simple quotient kills $J(A_K)S_K$. Equivalently, intersections of maximal submodules equal $J(A_K)S_K$, since the intersection of the maximal submodules of a finite semisimple module is zero. The assumed head is simple, so this quotient is $D_K$. Its rational precursor is nonzero and simple: a proper nonzero rational submodule would remain proper and nonzero after faithful scalar extension.

The radical of a finite-dimensional $\mathbb{Q}$-graded algebra is homogeneous. Indeed all scaling automorphisms $a_d\mapsto t^d{a}_d$ preserve its characteristic ideal $J$, for $t\in {\mathbb{Q}}^{\times }$. If $a={\sum }_{d\in E}{a}_d\in J$ with finite degree set $E$, take $t=2^j$ for $j=0,\dots ,\mathrm{\mid }E\mathrm{\mid }-1$. The resulting coefficient matrix $((2^d{)}^j{)}_{j,d}$ is Vandermonde in the pairwise distinct rational numbers $2^d$, hence invertible; each $a_d$ is a rational linear combination of elements of $J$. Thus $J(A_{\mathbb{Q}})S_{\mathbb{Q}}$ is homogeneous and $D_{\mathbb{Q}}$ has the required quotient grading. Put $v_{\mathbb{Q}}=z_{\mathbb{Q}}+J(A_{\mathbb{Q}})S_{\mathbb{Q}}$. It is nonzero: otherwise cyclicity would force $D_{\mathbb{Q}}=0$. The scalar-extension quotient isomorphism preserves the generator, and hence its image is exactly $v$, including its degree. Any prescribed normalization shift is imposed simultaneously on both modules.

The orbit $A_{\mathbb{Z}}{v}_{\mathbb{Q}}$ is generated as an abelian group by the images of a finite integer basis of $A_{\mathbb{Z}}$. It is torsion-free because it is a subgroup of a rational vector space. Its rational span is $D_{\mathbb{Q}}$, by cyclicity, so its rank equals ${\dim }_{\mathbb{Q}}{D}_{\mathbb{Q}}={\dim }_K{D}_K$. The generator-preserving scalar-extension isomorphism identifies this orbit with $A_{\mathbb{Z}}v$. To prove freeness here, choose rational coordinates on $D_{\mathbb{Q}}$ and clear the denominators of the finitely many generators, embedding this orbit in ${\mathbb{Z}}^n$. Every subgroup $H\le {\mathbb{Z}}^n$ is free by induction on $n$: its projection to the last coordinate is either zero or $m\mathbb{Z}$ with $m>0$; in the latter case a lift $h$ of $m$ gives $H=(H\cap {\mathbb{Z}}^{n-1})\oplus \mathbb{Z}h$, and the zero case reduces directly to $n-1$. This proves the lattice assertion.

If a fixed normalized $D_K$ is absolutely simple, a nonzero map $S_K\to D_K$ is surjective and kills the radical. Since the head is already the simple $D_K$, such maps factor through its canonical quotient and differ by an invertible endomorphism of $D_K$. This endomorphism $f$ is scalar using a finite root-field construction. Its nonconstant characteristic polynomial has a monic irreducible factor; by F4 the corresponding quotient field $L\mathrm{/}K$ supplies a root $\alpha$. The matrix $f_L-\alpha I$ has zero determinant and hence a nonzero kernel. Absolute simplicity makes $D_L$ simple, so this commuting endomorphism must be zero. Thus $f_L=\alpha I$, and any diagonal entry of the original $K$-matrix shows $\alpha \in K$. Thus every allowed generator image is a scalar multiple of the constructed image, and scaling the lattice preserves freeness and rank. No claim concerning an arbitrary vector in $D_K$ was used.