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 $m$.

Let $x:E\to {\mathbb{P}}^1$ be the $x$-coordinate map and define the naive logarithmic height byThe duplication formula induces a degree-four morphism on the $x$-line, so part (a) givesuniformly in $P$. The telescoping sequence $4^{-n}h(2^{n}P)$ is therefore Cauchy. Define the canonical height of an elliptic curve bySumming the geometric error series proves that $|h(P)-\hat{h}(P)|$ is bounded. If another function has this bounded-difference property and scales by four under doubling, evaluating the bounded difference at $2^{n}P$ and dividing by $4^n$ proves uniqueness.

Suppose first that $\kappa$ is regular and uncountable in $M$. Part (b) and cardinal preservation by chain-condition forcing show that $\kappa$ remains a cardinal number in $M[G]$, while part (a) makes every infinite ground-model cardinal below $\kappa$ countable. Every ordinal below $\kappa$ has ground-model cardinality below $\kappa$ and is therefore countable in the extension. Thus $\kappa$ is the least uncountable ordinal there:

Conversely, suppose $M[G]\models \kappa ={\aleph }_1$ and $\kappa$ were singular in $M$. Let $\mu ={\mathrm{cf}}^M(\kappa )<\kappa$ and take in $M$ a cofinal function $f:\mu \to \kappa$. Part (a) makes $\mu$ countable in $M[G]$, while the same $f$ remains cofinal there. This would give $\kappa$ countable cofinality, contradicting first uncountable ordinal is regular. Hence $\kappa$ was regular in $M$, and for the intended uncountable $\kappa$,

Assume that $\kappa$ is regular and uncountable in $M$, as required for this claim. If an antichain $A\in M$ had size $\kappa$, apply the delta-system lemma at a regular uncountable cardinal inside $M$ to the finite domains of its conditions. After thinning, their domains form a delta-system with finite root $R$.

There are fewer than $\kappa$ possible restrictions to $R$, because each coordinate $(\alpha ,n)\in R$ has fewer than $\kappa$ possible values. Since $\kappa$ is a regular cardinal, we can thin again to $\kappa$ conditions agreeing on $R$. Any two now have disjoint domains outside $R$ and agree on $R$, so their union is a common extension. This contradicts that $A$ is an antichain. ThereforeIf the word “regular” is allowed to include $\kappa =\omega$, the printed claim needs the additional hypothesis that $\kappa$ is uncountable; the finite-condition order can have an infinite antichain when $\kappa =\omega$.

Let $F=\bigcup G$, the generic union for the Finite-condition Lévy collapse, and fix an infinite cardinal number $\lambda <\kappa$ in $M$. For every $n\in \omega$, the set of conditions defining a value at $(\lambda ,n)$ is dense: extend a condition at that fresh coordinate with any value below $\lambda$. Hence $F_{\lambda }(n)=F(\lambda ,n)$ is a total function $\omega \to \lambda$.

For every $\beta <\lambda$, the set of conditions assigning the value $\beta$ at some fresh coordinate $(\lambda ,n)$ is also dense. Genericity therefore makes $F_{\lambda }$ surjective. Thusso $\lambda$ is a countable set in $M[G]$. Finite cardinals are already countable.