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Over a splitting field for finite group representations of characteristic $p$, the irreducible Brauer characters form a complex basis of the class functions on the p-regular elements. Independence is the character form of the Brauer–Nesbitt theorem. To obtain spanning, extend such a class function by zero on the p-singular classes. Ordinary irreducible characters form a basis of all class functions by character orthogonality. Restricting them to p-regular elements yields Brauer characters of reductions of an integral form of a group representation in a compatible splitting p-modular system; if needed, first extend scalars, which does not change the simple-module list under the splitting hypothesis. Each restriction is a nonnegative integral sum of simple Brauer characters by exact-sequence additivity and the Jordan–Hölder theorem. Thus these restrictions span, proving the assertion. Consequently the number of simple modules equals the number of p-regular conjugacy classes, and evaluation identifies the complexified modular representation ring with the product of one copy of $\mathbb{C}$ for each such class.

Since $S^{\mu }\subset M^{\mu }$, Question 2 shows that every composition factor of $S^{\mu }$ is $D^{\lambda }$ for a regular partition $\lambda ⊵\mu$. If $\mu$ is regular, its radical quotient $D^{\mu }$ occurs once: it occurs as a quotient, and its multiplicity is bounded above by the multiplicity one in $M^{\mu }$. All remaining factors have strictly dominating labels. A nonregular row has only strictly dominating regular labels.

Define $d_{\mu \lambda }=[S^{\mu }:D^{\lambda }]$. Order the row labels, all partitions of an integer of $n$, and the column labels, the regular partitions of an integer, by the same total order refining decreasing dominance. ThenThe matrix is generally rectangular. Its submatrix on regular rows is lower unitriangular in this decreasing convention; placing least dominant shapes first would transpose the visual triangular convention, not the indices.

This is the decomposition matrix, not just an abstract composition factor table. The standard polytabloid basis makes $S_{\mathbb{Z}}^{\mu }$ a free invariant integral lattice whose extension to characteristic zero is the ordinary irreducible $S^{\mu }$. Reducing this same lattice modulo $p$ gives exactly the modular $S^{\mu }$. On p-regular elements, reduction of its ordinary character is the sum of the Brauer characters of its composition factors, with coefficients $d_{\mu \lambda }$. These are the defining decomposition numbers; the resulting class in the modular representation ring is independent of the chosen invariant lattice.

For $S_3$, take rows $(3),(2,1),(1,1,1)$ in that order. In characteristic zero, or $p>3$, all three Specht modules are irreducible by Maschke's theorem, andFor $p=2$ the regular labels are $(3),(2,1)$ andIndeed the sign representation equals the trivial representation. The two-dimensional $S^{(2,1)}$ is the sum-zero subspace of the three-point permutation module. Over an algebraic closure, a $3$-cycle has two distinct nontrivial third-root eigenvalues there, and a transposition exchanges their lines. Neither line is invariant under $S_3$, proving absolute irreducibility.

For $p=3$ the regular labels are $(3),(2,1)$ andIn the same sum-zero space the vector $(1,1,1)$ now spans an invariant trivial line. Its one-dimensional quotient is sign: the determinant of the sum-zero representation is sign, and the determinant of the invariant line is trivial. Therefore $S^{(2,1)}$ has one trivial and one sign factor, with sign its simple head $D^{(2,1)}$. The bottom Specht module is itself sign. This proves every entry of the three possible matrices.

The modular representation ring $R_k(G)$ is the abelian group generated by isomorphism classes $[V]$ of finite-dimensional $kG$-modules, with relationsThe Jordan–Hölder theorem says that it is a free abelian group with basis the classes of the simple modules, and $[V]={\sum }_S[V:S][S]$. To see independence, composition multiplicity for each fixed simple $S$ is an additive map to $\mathbb{Z}$ and reads off its basis coefficient.

Define $[V][W]=[V{\otimes }_{k}W]$, using the diagonal action in the tensor product of group representations. Tensoring over a field is exact in either argument, so multiplication respects the defining relations. Associativity and the symmetry $v\otimes w\mapsto w\otimes v$ give a commutative ring with identity $[k]$, the trivial representation. Thus $\boxed{R_k(G)\text{ is a commutative ring, freely based on simple modules}}$.

Completeness permits idempotent lifting from $kG$ to $\mathcal{O}G$. In particular, each projective cover $P_j$ lifts to a finite projective $\mathcal{O}G$-lattice ${\tilde{P}}_j$; one can lift an idempotent presenting it as a summand of a finite free module. Set $Q_j=K{\otimes }_{\mathcal{O}}{\tilde{P}}_j$.

For a simple $S_{\ell }$, every map $P_j\to S_{\ell }$ factors through its simple head $S_j$. Splitting and Schur lemma give ${\dim }_k{\mathrm{Hom}}_{kG}(P_j,S_{\ell })={\delta }_{j\ell }$. Exactness of this Hom functor along a composition series therefore givesPart (iii) identifies this with ${\dim }_K{\mathrm{Hom}}_{KG}(Q_j,V_i)$. By Maschke's theorem and splitting, $KG$ is split semisimple, soThe reductions of any two integral forms of the same ordinary module have identical Brauer characters, hence identical composition multiplicities. We may thus reduce a direct-sum form for the displayed decomposition instead of ${\tilde{P}}_j$. In the modular representation ring this givesComparing the simple basis coefficients proves $c_{\ell j}={\sum }_i{d}_{i\ell }{d}_{ij}$, or $\boxed{C=D^{T}D}$.