Modular Traces and Brauer-Character Independence

Every submodule $N$ of a finite sum $W$ of simple modules has a complement that is a sum of simple modules of the listed types. Here is the induction proving this assertion. For $W=0$, take complement $0$. Write $W=S\oplus T$, with $S$ simple, and suppose the assertion proved for $T$. If $S\subseteq N$, then $N=S\oplus (N\cap T)$; a complement of $N\cap T$ in $T$ complements $N$. Otherwise $N\cap S=0$, and projection to $T$ identifies $N$ with a submodule $N^{\mathrm{\prime }}$ of $T$. Write $T=N^{\mathrm{\prime }}\oplus C$ by the induction hypothesis. There is an $A$-linear map $f:N^{\mathrm{\prime }}\to S$ with $N=\{(f(t),t):t\in N^{\mathrm{\prime }}\}$. Every $(s,t+c)$ equals $(f(t),t)+(s-f(t),c)$, uniquely, so $W=N\oplus (S\oplus C)$. These cases exhaust the possibilities since $N\cap S$ is a submodule of $S$.

For each member of a finite selection of the $S_i$, choose $0\mathrm{\ne }{s}_i\in S_i$. Simplicity gives $A{s}_i=S_i$. Images of a finite basis of $A$ span $S_i$. Scanning that list and retaining a vector precisely when it is outside the previous span gives an independent spanning list $e_{i1},\dots ,e_{i{d}_i}$, with $1\le d_i\le {\dim }_{k}A$. Only finitely many selections have been made.

For this finite selection put $W={⨁}_i{S}_i^{d_i}$ and $v=(e_{ij})$. If $Av\mathrm{\ne }W$, step 1.1 supplies a nonzero simple summand of a complement, isomorphic to some $S_l$, and projection onto it gives a nonzero $A$-linear $F:W\to S_l$ with $F(Av)=0$. Its restrictions to copies of $S_i$ vanish for $i\mathrm{\ne }l$ by Schur, and on the copies of $S_l$ are scalars $c_j$. Since $1v=v$, we have $0=F(v)={\sum }_j{c}_j{e}_{lj}$. Independence gives $c_j=0$ for all $j$, so $F=0$, a contradiction. Therefore $Av=W$.

Given any tuple of $k$-endomorphisms $u_i$, step 2.1 supplies $a\in A$ with $av=(u_i(e_{ij}))$. Then $a{e}_{ij}=u_i(e_{ij})$ for every basis vector, so $a$ induces $u_i$ on all of $S_i$. Thus the action map for every finite selection is surjective. Lifting a vector-space basis of the target gives an independent list in $A$ (apply the map to any relation). Consequently ${\sum }_i{d}_i^2\le {\dim }_{k}A$, in particular the size of the selection is at most ${\dim }_{k}A$.

If $I$ had more than ${\dim }_{k}A$ members, finite induction would select ${\dim }_{k}A+1$ distinct indices, contradicting step 3.1. Thus $I$ is finite and that step proves the asserted surjectivity. If $I$ is empty, the unique map onto zero is surjective; if $A=0$, no nonzero unital simple module exists and this is the only case.

In a basis of the nonzero $S_i$, the matrix $E_{11}$ has trace $1$, including when the characteristic divides $\dim S_i$. By surjectivity choose $a_i$ acting as $E_{11}$ on $S_i$ and as zero on every other $S_j$. Hence ${\tau }_j(a_i)={\delta }_{ij}$.

If ${\sum }_j{c}_j{\tau }_j=0$, evaluation at $a_i$ gives $c_i=0$ for every $i$. For $kG$, a relation vanishing on $G$ vanishes on $a={\sum }_g{a}_{g}g$ since ${\sum }_j{c}_j{\tau }_j(a)={\sum }_g{a}_g{\sum }_j{c}_j{\tau }_j(g)=0$. It is therefore the zero relation by the first assertion. An empty family has only the empty relation.

Write $\mathrm{\mid }g\mathrm{\mid }=m{p}^a$ with $p\nmid m$. Choose $e,f\in \mathbb{Z}$ with $em+f{p}^a=1$, and set $u=g^{em}$, $s=g^{f{p}^a}$. Their product is $g$ and they commute; $u^{p^a}=s^m=1$. This also covers $g=1$ and $a=0$.

Put $N=\rho (u)-I$. The commuting binomial identity in characteristic $p$, iterated $a$ times, gives $N^{p^a}=\rho (u{)}^{p^a}-I=0$. Since $\rho (s)N=N\rho (s)$, the map $T=\rho (s)N$ satisfies $T^{p^a}=\rho (s{)}^{p^a}{N}^{p^a}=0$. Moreover $\rho (g)-\rho (s)=T$.

A nilpotent map $T$ has trace zero over $k$ itself. To see this, extend an independent list successively along $0=\ker T^0\subseteq \ker T\subseteq \cdots \subseteq \ker T^{p^a}=V$. At each stage, if the current list does not span that kernel, append a vector outside its span. The dimension bound forces this finite procedure to terminate. Since $T(\ker T^j)\subseteq \ker T^{j-1}$, its matrix in the resulting basis has zero diagonal. The trace is independent of basis: $\mathrm{tr}(AB)={\sum }_{i,j}{A}_{ij}{B}_{ji}=\mathrm{tr}(BA)$, so $\mathrm{tr}(P^{-1}TP)=\mathrm{tr}(T)$.

By additivity of the diagonal sum, $\mathrm{tr}(\rho (g))-\mathrm{tr}(\rho (s))=\mathrm{tr}(T)=0$. For $V=0$ both sums are empty and zero. No assertion that $\rho (u)=I$ is needed.

Normalize the valuation by $v(\pi )=1$. An element is a unit exactly when its value is zero: if $v(x)=0$, then $v(x^{-1})=0$; if both $x,x^{-1}$ are integral, their nonnegative values sum to zero. Thus the maximal ideal is $(\pi )$. For ${\lambda }^m=1$ choose one representative $x_0$; it is a unit, as is $f^{\mathrm{\prime }}(x_0)=m{x}_0^{m-1}$ for $f(X)=X^m-1$.

Define deterministically $x_{n+1}=x_n-f(x_n)\mathrm{/}{f}^{\mathrm{\prime }}(x_n)$. Taylor expansion $f(x+h)=f(x)+f^{\mathrm{\prime }}(x)h+h^{2}R$ with $R\in \mathcal{O}$ shows $v(f(x_{n+1}))\ge 2v(f(x_n))$. All $x_n$ retain residue $\lambda$, so every derivative is a unit. Starting with $v(f(x_0))\ge 1$, we get $v(x_{n+1}-x_n)\ge 2^n$ (a zero error stays zero). The sequence converges by completeness to $x\equiv \lambda$; continuity of the finite polynomial operations and separation give $f(x)=0$. This is the simple-root Newton construction used in the Stacks complete-local-ring lifting proof specialized to a DVR.

If $x^m=y^m=1$ and $\bar{x}=\bar{y}=\lambda$, then $0=(x-y){\sum }_{j=0}^{m-1}{x}^{m-1-j}{y}^j$. The sum has residue $m{\lambda }^{m-1}\mathrm{\ne }0$, hence is a unit by step 1.1. Thus $x=y$, the Stacks simple-root uniqueness argument.

Products and inverses of lifted roots are roots lifting the corresponding products and inverses. Uniqueness therefore gives $\widehat{\lambda \nu }=\widehat{\lambda }\widehat{\nu }$, $\widehat{1}=1$, and $\widehat{{\lambda }^{-1}}=(\widehat{\lambda }{)}^{-1}$. If two exponents occur, both lifts are roots for their least common multiple, still prime to $p$, so uniqueness there identifies them. When $m=1$ both groups are $\{1\}$.

The polynomial $X^m-1$ splits in $k$. Indeed, for any monic irreducible factor $h$, the field $k[X]\mathrm{/}(h)$ is a simple module for $k⟨g⟩$; multiplication by its elements gives module endomorphisms. The scalar-endomorphism condition forces this field to equal $k$, so $\mathrm{deg}h=1$. Since the derivative $m{X}^{m-1}$ has no common root with $X^m-1$, the roots are distinct.

For each root $\lambda$ put $P_{\lambda }(X)={\prod }_{\nu \mathrm{\ne }\lambda }(X-\nu )\mathrm{/}(\lambda -\nu )$. Polynomial interpolation gives ${\sum }_{\lambda }{P}_{\lambda }(X)=1$ and $(X-\lambda )P_{\lambda }(X)=0$ modulo $X^m-1$. Applying these identities to $\rho (g)$ expresses $V$ as the direct sum of its eigenspaces: the sum spans, and application of $P_{\lambda }(\rho (g))$ isolates each summand. Thus all displayed eigenvalues lie in $k$ and have mth power one.

The unique lifts exist in $\mathcal{O}$, and their multiset depends only on the characteristic polynomial. A change of basis or conjugation of $g$ conjugates its matrix and preserves that polynomial, hence preserves the sum. For $V=0$ the sum is zero; at $g=1$ it is $d\cdot 1_{\mathcal{O}}$. These prove the stated well-definedness.

Reduction $\mathcal{O}\to k$ is a ring homomorphism. Therefore $\overline{{\phi }_V(g)}={\sum }_{j=1}^{\dim V}\overline{\widehat{{\lambda }_j}}={\sum }_j{\lambda }_j$. This uses additivity of reduction, not additivity of root lifting.

In an eigenbasis of the p-regular operator the diagonal entries are precisely the ${\lambda }_j$, so their sum is its trace. The definition supplies that eigenbasis decomposition. If $V=0$ both sides are zero; if $g=1$, both reduce to $(\dim V)1_k$.

Consider any finite relation ${\sum }_{i=1}^r{c}_i{\phi }_i=0$ over $K$. If some coefficient is nonzero, let $b={\min }_{c_i\mathrm{\ne }0}v(c_i)$ and replace every $c_i$ by ${\pi }^{-b}{c}_i$. Then all coefficients lie in $\mathcal{O}$ and at least one has valuation zero, hence has nonzero residue.

Reduce the relation at every p-regular element. It becomes ${\sum }_i{\bar{c}}_i\mathrm{tr}(s\mathrm{\mid }{S}_i)=0$. For arbitrary $g$, replace each trace by its trace at the same p-regular part $s$ of $g$. Thus ${\sum }_i{\bar{c}}_i\mathrm{tr}(g\mathrm{\mid }{S}_i)=0$ for every $g\in G$.

Since k splits G, modular trace independence forces every ${\bar{c}}_i=0$, contradicting step 1.1. Hence the original relation has all coefficients zero. This includes the empty family.

For any finite family the matrix of values on the finite p-regular set has independent columns. Successive elimination using nonzero pivots therefore supplies a square minor of full column size with nonzero determinant. That determinant remains nonzero under any field embedding, proving independence after extension. All entries lie in the subfield $E\subseteq K$ generated over $\mathbb{Q}$ by the finitely many lifted roots used here; these are roots of unity, so E is cyclotomic. The same minor lies in E and remains nonzero under a fixed embedding $E\to \mathbb{C}$. No embedding of all of K into C is assumed.