Suppose $a\notin \mathrm{acl}(A)$. Choose a small model $M_0$ containing $A$. The complete type $\mathrm{tp}(a/A)$ is nonalgebraic, so choose a realization $a^{\prime }\notin M_0$. Strong homogeneity of the monster model gives an automorphism $\sigma$ fixing $A$ with $\sigma (a)=a^{\prime }$. Then $M={\sigma }^{-1}(M_0)$ contains $A$, but $a\in M$ would imply $a^{\prime }=\sigma (a)\in M_0$. Taking the contrapositive proves

Suppose $a\in \mathrm{acl}(A)$ in the sense of model-theoretic algebraic closure. Some formula $\phi (x,\bar{c})$ over $A$ has exactly $n<\infty$ realizations in the monster model and includes $a$. Every model $M$ containing $A$ is an elementary substructure of the monster and therefore contains $n$ distinct realizations. Since there are only $n$ in the monster, it contains all of them, including $a$. Hence

Put $N=N_G(D)$. For an algebra on which a group acts by conjugation, writefor its transfer ideal of conjugation-fixed elements. In the diagram, $\tau ={\mathrm{Tr}}_D^G$, ${\tau }^{\prime }={\mathrm{Tr}}_D^N$, the upper map is $\beta ={\mathrm{Br}}_D$, and ${\beta }^{\prime }$ is the restriction of ${\mathrm{Br}}_D$ from $(kG{)}^G$ to $(kG{)}_D^G$. Since $N$ normalizes both $D$ and $C_G(D)$, the lower map lands in $(k{C}_G(D){)}_D^N$.

For $a\in (kG{)}^D$, let $D$ act on the left cosets $G/D$. A coset $gD$ is fixed precisely when $g^{-1}Dg\le D$, hence, because the two groups have the same order, precisely when $g\in N$. Every nonfixed orbit has size divisible by $p$. Moreover, after applying ${\mathrm{Br}}_D$, all summands indexed by one $D$-orbit are equal: conjugation by an element of $D$ acts trivially on $k{C}_G(D)$. Those orbits therefore contribute zero in characteristic $p$, while the fixed cosets contribute the trace over $N/D$. ConsequentlyThis is the Brauer morphism and relative trace identity, so the diagram commutes.

Let $D$ act on the group algebra $kG$ by conjugation. Its fixed-point algebra isand the centralizer $C_G(D)$ consists of the elements of $G$ commuting with every element of $D$. The Brauer morphism isThus $\beta$ deletes the coefficients of basis elements outside $C_G(D)$. The nonfixed $D$-orbits have cardinality divisible by $p$, so the usual orbit argument shows that this projection is a unital ring homomorphism. Hence

Assume (v). If the generic fibre $M=K{\otimes }_{\mathcal{O}}W$ had a nonzero proper $KG$-submodule, intersecting it with $W$ and rescaling to obtain a saturated lattice would give a nonzero proper $kG$-submodule of $\overline{W}$. Thus $M$ is simple.

Because $\overline{W}$ is projective, its restriction to a Sylow p-subgroup $S$ is projective. The local algebra $kS$ has only free finitely generated projectives, so $|S|=p^d$ dividesConsequently

Under (ii), tensor the direct-product decomposition with the residue field $k$. Since $W$ is free over $\mathcal{O}$,Henceand the reduced action map is the projection onto the first factor. This proves

Assume (iii). Reduction of the projective lattice $W$ is projective, because a direct-summand decomposition of a free $\mathcal{O}G$-module remains one after tensoring with $k$. Completeness of $\mathcal{O}$ and idempotent lifting show that $\overline{W}$ is indecomposable: otherwise a nontrivial idempotent of ${\mathrm{End}}_{kG}(\overline{W})$ would lift and split $W$, hence split the simple $KG$-module $M$.

Let $\chi$ be the ordinary irreducible character of $M$. Since $W$ is projective, $\chi$ vanishes on p-singular elements and restricts to the projective character $\Phi$ of $\overline{W}$. ThusAn indecomposable projective module that is not simple has, besides its identity, a nonzero noninvertible endomorphism obtained by projecting onto its simple head and embedding the isomorphic simple socle. Hence its endomorphism algebra has dimension at least two. Therefore $\overline{W}$ is simple as well as projective, proving

Assume (ii), and write the ring decomposition asThe natural column module $W\cong {\mathcal{O}}^n$ is projective over $M_n(\mathcal{O})$ by Morita equivalence; extending it by zero across $B$ makes it projective over $\mathcal{O}G$. After extending scalars to $K$, the action factors through $M_n(K)$ on its natural module $K^n$, which is simple. Therefore

The divisibility in (i) is the standard dimension test for a projective modular representation. If $P$ is a projective $kG$-module and $S$ is a Sylow p-subgroup of order $p^d$, then ${\mathrm{Res}}_S^{G}P$ is projective over $kS$. The group algebra of a p-group in characteristic p is local, so every finitely generated projective $kS$-module is free. ConsequentlyThis will apply to $P=\overline{W}$ in the implication (v)$\Rightarrow$(i).

The eight-dimensional simple module is projective, so its tensor product of group representations with the natural three-dimensional module is projective. Its Brauer character isFrom part (iii),The duality of simple and projective Brauer characters makes the decomposition multiplicities unique. Hence, writing $P_3$ and $P_8$ for the corresponding projective covers,This completes the explicit 2-modular representation theory of GL3 of F2 calculation.

The rows of the Cartan matrix of a group algebra express the projective characters in the simple Brauer-character basis. ThusEvaluating gives

On the 2-regular classes the six ordinary characters decompose asTherefore the decomposition matrix, with columns ordered $1,3,\overline{3},8$, isThe Cartan matrix of a group algebra is

Only the classes $1,3,7A,7B$ are 2-regular. The trivial module and the natural three-dimensional module are simple; the dual natural module gives the conjugate three-dimensional character. Restricting the ordinary characters to the odd-order classes and using $\alpha +\overline{\alpha }=-1$ produces the fourth simple character of degree eight. The Brauer character table isThe ordinary degree-eight character restricts exactly to ${\varphi }_8$. Since its degree contains the full 2-part $8$ of $|G|$, this simple module is projective and its singleton block has defect zero. Thus

If $M$ is projective over $RG$, then its restriction is projective over $RH$ because $RG$ is a free right $RH$-module and a free $RG$-module restricts to a free $RH$-module.

Conversely, suppose ${\mathrm{Res}}_H^{G}M$ is projective. Thenis projective over $RG$. Part (i) says that $M$ is a direct summand of this induced module, so $M$ is projective. Hence projectivity detected on a subgroup of invertible index gives

SetEvery conjugate of $\alpha$ equals $\alpha$, soThe D. Higman criterion therefore gives

Order the simple modules as in part (i), so that $S_r$ records the $r$th diagonal character. The quotient $N^i/N^{i+1}$ has basisFor $a\in A$, multiplication givesbecause every term below row $r$ lies on a lower subdiagonal. Hence the line generated by $e_{r,r-i}$ is $S_r$, and the lines are independent. Therefore the radical layers of the lower triangular matrix algebra are

For the left regular module, the radical series of a module is obtained by multiplying by powers of the Jacobson radical:Products of matrix units show thatwith $N^0=A$ and $N^n=0$. Thus each step removes one subdiagonal, and the Loewy length of ${}_{A}A$ is $n$.

For $1\le r\le n$, let $S_r=k$ with a lower triangular matrix $a=(a_{ij})$ acting by the scalar $a_{rr}$. These are pairwise nonisomorphic simple modules, since the diagonal matrix units $e_{ii}$ distinguish them.

Let $N$ be the ideal of strictly lower triangular matrices. It is nilpotent, andis a semisimple algebra. Hence $N$ is the Jacobson radical; explicitly,

Let $T$ have order $m>0$. Property (ii) gives $\hat{h}(T)=0$, sinceIt also givesThereforefor every rational torsion point of an elliptic curve. The construction in part (i), quadraticity in part (ii), and this identity establish existence; the bounded-difference and doubling argument in part (i) establishes uniqueness.

The addition law and part (a) give the approximate parallelogram identityReplace $P,Q$ by $2^{n}P,2^{n}Q$, divide by $4^n$, and let $n\to \infty$. The error disappears, leaving the exact parallelogram lawTogether with $\hat{h}(2P)=4\hat{h}(P)$, induction on $|m|$ yieldsfor every integer $m$.