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For a nonzero group representation $V$, there is a $G$-equivariant identificationwhere $G$ acts on endomorphisms by conjugation. The identity endomorphism is a nonzero fixed vector, so its span is a trivial submodule. It follows that the trivial simple module is a Jordan–Hölder factor of $V\otimes V^{\ast }$, and character additivity gives $\boxed{1_G\text{ is a constituent of }\chi \overline{\chi }}$.

Here “constituent” means a composition factor; the trivial submodule need not be a direct summand. Also the assertion needs $\chi \ne 0$: the zero-dimensional module has the zero Brauer character, whose product with its conjugate has no constituent. This is an implicit nonzero-character qualification in the printed statement.

An integral form of a group representation $V$ is a $G$-stable finite free $\mathcal{O}$-submodule $W\subset V$ with $K{\otimes }_{\mathcal{O}}W\cong V$. Such a form exists: take a basis lattice $L$ and replace it by ${\sum }_{g\in G}gL$, which is finite and torsion-free and therefore free over the discrete valuation ring $\mathcal{O}$.

Choose ordinary simple modules $V_i$ with forms $W_i$, and modular simple modules $S_j$ with projective covers $P_j$. With $\pi$ a uniformizer, put ${\overline{W}}_i=W_i/\pi W_i$. The decomposition matrix and Cartan matrix of a group algebra have entriesThe first numbers do not depend on the integral form: the Brauer character of its reduction is the ordinary character restricted to p-regular elements, and part 2(a) determines all composition multiplicities from this restriction.

For the assertion in (i), set $I={\mathrm{Hom}}_{\mathcal{O}G}(W,W^{\prime })$. It is a submodule of the finite free module ${\mathrm{Hom}}_{\mathcal{O}}(W,W^{\prime })$, hence is finite free over $\mathcal{O}$. There is a natural injective mapFor any homomorphism $f$ in the target, clearing the finitely many denominators of its values on an $\mathcal{O}$-basis of $W$ gives ${\pi }^{N}f(W)\subset W^{\prime }$. Thus ${\pi }^{N}f\in I$, and the map is surjective. Therefore $\boxed{I\text{ is an }\mathcal{O}\text{-form of }{\mathrm{Hom}}_{KG}(M,M^{\prime })}$, regarding this Hom space as a vector space. The original PDF supplies part (i), which is missing from the supplied TeX.

For the preliminary definitions, a block is $Re$ for a primitive central idempotent $e$, and a module belongs to it when $eM=M$. For a semisimple algebra, the Artin–Wedderburn theorem identifies its blocks with the factors $M_{n_i}(D_i)$ in its product decomposition. Each factor is a block of a finite-dimensional algebra.

We prove Ext separation of finite-length modules. The extension hypothesis is ${\mathrm{Ext}}^1(S,T)={\mathrm{Ext}}^1(T,S)=0$ for simples in opposite sets. First, for a simple $S\in {\mathcal{C}}_1$ and a module $N$ of finite composition length with factors in ${\mathcal{C}}_2$, we have ${\mathrm{Ext}}^1(S,N)=0$. Induct on the length of $N$: for $0\to N^{\prime }\to N\to T\to 0$ with $T$ simple, the long exact sequence of the Ext functor containswhose outer groups are zero. The same argument works with the two sets interchanged.

Now induct on the length of $M$, with $M=0$ immediate. Take a simple quotient in $0\to N\to M\to S\to 0$. By induction, $N=N_1\oplus N_2$ with factors in the respective sets. Suppose $S\in {\mathcal{C}}_1$; the other case is symmetric. Quotienting by $N_1$ givesThis splits by the preceding Ext vanishing. Let $U_1$ be the inverse image in $M$ of the chosen complementary copy of $S$. Then $U_1\cap N_2=0$, $U_1+N_2=M$, and $0\to N_1\to U_1\to S\to 0$. Thus $U_1$ has only factors in ${\mathcal{C}}_1$, while $U_2=N_2$ has only factors in ${\mathcal{C}}_2$.

If two finite-length modules have factors in disjoint sets, any homomorphism between them is zero: a nonzero image would, by the Jordan–Hölder theorem, have a simple factor belonging to both sets. For any submodule $V$ of $M$ with factors in ${\mathcal{C}}_1$, projection onto $U_2$ is consequently zero, so $V\subset U_1$. The analogous argument applies to $U_2$. HenceIn particular both summands are preserved by every endomorphism of $M$.

We prove $(ii)\Rightarrow (i)$. The central block idempotents split every finite-dimensional module into its block components. A nonzero indecomposable representation, in particular an indecomposable projective module, has only one such component; otherwise these give a nontrivial direct-sum decomposition.

Every Jordan–Hölder factor of a module in a block belongs to that block, since its central idempotent acts as identity on submodules and quotients. Thus any two simple factors of the same projective indecomposable are in the same block. Applying this along the proposed chain yields $\boxed{S,T\text{ belong to the same block}}$.

Let the torsion submodule of the module over a ring $M$ beThis is a submodule. Indeed, if $rm=0$ and $sn=0$ with $r,s\ne 0$, then $rs\ne 0$ because the scalar ring is an integral domain, and $rs(m+n)=0$; scalar multiples and additive inverses are handled similarly.

Take the quotient module $M_0=M/T(M)$ and let $q:M\to M_0$ be the quotient R-module homomorphism. It is torsion-free: if $0\ne r\in R$ and $r(m+T(M))=0$, then $rm\in T(M)$, so $srm=0$ for some $s\ne 0$. Since $sr\ne 0$, this puts $m$ in $T(M)$ and therefore $m+T(M)=0$.

Now let $N$ be torsion-free and let $f:M\to N$ be an $R$-module homomorphism. If $m\in T(M)$ and $rm=0$ for $r\ne 0$, then $rf(m)=f(rm)=0$, so torsion-freeness gives $f(m)=0$. Hence $T(M)\subseteq \ker f$, and the universal property of a quotient module gives a unique homomorphismwith $\boxed{f=f_0\circ q}$. Thus $M/T(M)$ is the maximal torsion-free quotient of $M$.

Suppose first that $M$ is finitely generated. Its quotient $M/N$ is generated by the images of any finite generating set of $M$. A principal ideal domain is Noetherian, so every submodule of the finitely generated $R$-module $M$ is finitely generated; in particular, $N$ is finitely generated.

Conversely, supposeand $M/N$ is generated by the cosets ${\overline{m}}_1,\dots ,{\overline{m}}_s$. For any $m\in M$, its coset is an $R$-linear combination of the ${\overline{m}}_j$, soExpressing this remainder in terms of the $n_i$ shows thatThus $M$ is finitely generated if and only if both the submodule $N$ and the quotient module $M/N$ are finitely generated.

An $R$-module $M$ is Noetherian when every submodule is finitely generated, equivalently when every ascending chain of submodules stabilizes.

A free module is an $R$-module with a basis: every element has a unique expression as a finite linear combination of basis elements.

A flat module $M$ is one for which the tensor functor $-{\otimes }_{R}M$ is exact. Since tensor products are always right exact, it is equivalent to require that tensoring with $M$ preserve injections.

A projective module $P$ has the lifting property: for every surjection $E\twoheadrightarrow F$ and every map $P\to F$, there is a map $P\to E$ making the resulting triangle commute. Equivalently, $P$ is a direct summand of a free module.

No. Give $R=k[x]$ the x-adic filtration $R_n=(x^n)$, take $M=R$, and let $N=(x)$. At $n=1$ the intersection filtration giveswhereas the induced filtration of $N$ givesThus intersection with a submodule need not equal the induced filtration.

The Rees module of an $R$-module $M$ with respect to an ideal $I$ is ${\mathcal{R}}_I(M)={⨁}_{j\ge 0}{I}^{j}M{t}^j$, a graded module over the Rees ring $\mathcal{R}(I)$. Generators of $M$ in degree zero generate this module. For $N\subseteq M$, the graded submodule ${⨁}_j(I^{j}M\cap N)t^j$ is finitely generated when $R$ is a Noetherian ring; a bound on its generators' degrees yields the Artin-Rees lemma.

A right ideal $I$ in a unital associative algebra $A$ is a vector subspace with $IA\subseteq I$, equivalently a submodule of the right regular module. It need not be a left ideal. Its finite generation refers to right multiplication by elements of $A$.

For a module $M$ over an integral domain $R$, its torsion submodule isThe domain condition makes this a submodule: products of nonzero annihilators remain nonzero and annihilate sums.

A nonzero module is uniform if every pair of nonzero submodules has nonzero intersection. Equivalently, every nonzero submodule is an essential submodule. A right Noetherian domain is uniform as a right module: if $aA\cap bA=0$ for nonzero $a,b$, the right ideals $b^{i}aA$, $i\ge 0$, form an infinite direct sum, contradicting the ascending chain condition.

If an R-module homomorphism $f:M\to P$ vanishes on a submodule $N$, there is a unique homomorphism $\bar{f}:M/N\to P$ satisfying $f=\bar{f}\circ q$, where $q:M\to M/N$ is the quotient map. It is given by $\bar{f}(m+N)=f(m)$.

The standard-tableau right-ideal decomposition of $\mathbb{C}{S}_n$, tensored over that algebra with $V^{\otimes n}$, gives this direct sum of general linear group modules. With $e_t=h_t/H_{\lambda }$, the map $e_t\mathbb{C}{S}_n\otimes T\to e_{t}T$ is an isomorphism. Individual summands generally need not be symmetric group submodules; instead each is a Schur module for the commuting action.