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Use the standard defining properties of Category O: finite generation over $U(\mathfrak{g})$, a direct sum of finite-dimensional weight spaces, and locally finite positive root action. In particular the weights lie in a finite union of downward root cones. Submodules and quotients retain the representation weight bounds and finite weight spaces; finite generation of submodules follows from the Poincaré-Birkhoff-Witt theorem and the noetherian property of $U(\mathfrak{g})$.

Choose finitely many weight vector generators. The sum $T$ of their full weight spaces is finite-dimensional and preserved by the centre. The kernel of its central action has finite codimension in $Z(U(\mathfrak{g}))$ and kills the generators, hence, by centrality, all of $M$. Thus only finitely many central characters ${\theta }_1,\dots ,{\theta }_r$ can occur in any subquotient: they are the characters of a fixed finite-dimensional commutative quotient algebra of the centre.

By the Harish-Chandra isomorphism each ${\theta }_i$ determines one finite Weyl orbit of shifted weights. Let $S$ be the finite union of these orbits after subtracting $\rho$. Every nonzero subquotient has a nonzero highest-weight vector, using the bounded-above weights. Within its finite-dimensional highest weight space, the commuting central operators have a common eigenvector. Part (e) and the central-character restriction show that its highest weight belongs to $S$.

Define for any subquotient $N$It is finite, additive in short exact sequences, and positive whenever $N\ne 0$. Hence each strict step of a submodule chain consumes at least one unit of $d(M)$. Refining a nonsimple factor into a proper submodule and quotient cannot continue more than $d(M)$ times. It ends with simple factors, provingThe argument proves the assertion for every object of $\mathcal{O}$, rather than just for a cyclic highest weight object.

The permutation module $M_F^{\mu }$ is the $F$-vector space with the $\mu$-tabloids as basis, with $S_n$ acting by permuting their entries. For a tableau $t$ with column stabilizer $C_t$, its polytabloid isThe Specht module $S_F^{\mu }$ is the submodule of $M_F^{\mu }$ spanned by these polytabloids, equivalently generated by any one $e_t$ under the group action.

The standard basis theorem states thatis an $F$-basis of $S_F^{\mu }$ in every characteristic. Its dimension is the number of standard tableaux of shape $\mu$; in the hook shape used below the basis is exactly the displayed $e_{1ij}$ with $2\le i<j\le n$.

Identify $M=M_F^{(n-1,1)}$ with $F^n$, whose basis vectors record the singleton row. PutIf the characteristic does not divide $n$, then $M=T\oplus U$, and $U$ is simple. If it divides $n$, then $0<T<U<M$ is the submodule chain, with $T$ and $M/U$ trivial and $U/T$ simple of dimension $n-2$. In both cases there are no additional submodules: any vector with unequal coordinates yields a nonzero multiple of $e_i-e_j$ by subtracting its image under $(i,j)$, and the orbit of such a difference spans $U$. A submodule containing no nonconstant vector is contained in $T$.

Now suppose $p\nmid n$. When $p\ne 2$, part (b)'s valid vanishing ${\mathrm{Hom}}_{S_n}(M,W)=0$ implies ${\mathrm{Hom}}_{S_n}(U,W)=0$: extend a putative map using the projection $M\to U$.

When $p=2$, the assumption $p\nmid n$ makes $n$ odd. The corrected calculation in part (b) says every map $M\to W$ sends its singleton-row basis vectors to the same multiple of the globally invariant vector $v$. Hence it kills the augmentation submodule $U$. Extending a map $U\to W$ along the direct-sum projection again makes it zero. Thus, despite the characteristic-two error in the earlier claim, the requested conclusion remains valid:

The structural input is the submodule lattice of the standard symmetric-group permutation module.

If $p\nmid n$, the preceding part says $U=S_F^{(n-1,1)}$ is simple and has no nonzero homomorphism into $W$. Its only nonzero simple quotient is itself, so $W$ cannot contain a submodule isomorphic to such a quotient.

If $p\mid n$, use the given nonzero homomorphism $\varphi :U\to W$. The submodule chain shows that $T$ is the unique maximal proper submodule of $U$ and $U/T$ is simple. Moreover ${\mathrm{Hom}}_{S_n}(T,W)=0$: for odd characteristic this follows from part (b), and in characteristic two the divisibility condition makes $n$ even, where the corrected fixed-space calculation also gives zero. Thus $\varphi$ kills $T$ and descends to a nonzero map $U/T\to W$. Simplicity makes this map injective.

Its image is the required simple submodule, provingIn the divisible case the quotient is explicitly $U/T$, of dimension $n-2$. The argument uses the supplied existence of $\varphi$ and only the fixed-space vanishing that remains valid in that case.

Let $n_1,\dots ,n_t$ generate the submodule $N$, and let $m_1+N,\dots ,m_s+N$ generate the quotient module $M/N$. Choose each $m_i\in M$ as a representative of its class. For any $m\in M$, there are $r_1,\dots ,r_s\in R$ such thatTherefore $m-{\sum }_i{r}_i{m}_i\in N$. Expressing that difference using the generating set of a module $n_1,\dots ,n_t$ givesHenceis a finitely generated module. This proves that finite generation is preserved by module extensions.

Put $K=\ker \varphi$, $H=\ker \psi$, and $Q=\mathrm{Im}q$. The kernel of a module homomorphism $H$ is a submodule, so its image $L=q(H)$ is a submodule. For $x\in H$, $\varphi (q(x))=\psi (x)=0$, showing $L\subseteq K$. Thus the quotient module $K/L$ exists. Also $L\subseteq Q$.

Consider the restriction to $K$ of the natural module homomorphism $R^m\to R^m/Q$:Its kernel of a module homomorphism is $K\cap Q$. We have $L\subseteq K\cap Q$. Conversely, if $y\in K\cap Q$, choose $x\in R^n$ with $y=q(x)$. Thenso $x\in H$ and $y\in L$. Hence $\boxed{K\cap Q=L}$.

The map $\alpha$ is surjective. Given $v+Q\in R^m/Q$, use the surjectivity of $\psi$ to choose $x\in R^n$ with $\psi (x)=\varphi (v)$. Set $y=v-q(x)$. ThenThus $y\in K$ is a preimage of the arbitrary class $v+Q$.

By the first isomorphism theorem for modules, $\alpha$ induces the module isomorphismExplicitly, replacing $y$ by $y+\ell$ with $\ell \in L$ leaves its class modulo $Q$ unchanged, proving well-definedness. Its formula preserves addition and scalar multiplication. If its value is zero, $y\in K\cap Q=L$, so $y+L=0$, proving injectivity; surjectivity was proved above. These checks give all the requested properties of the induced map.

Assume $H=\ker \psi$ is a finitely generated module. The restriction $q:H\to L$ is a surjective module homomorphism, so part (a)(i) makes $L$ a finitely generated module.

The free module $R^m$ has its finite standard basis of a module. Its quotient module $R^m/\mathrm{Im}q$ is therefore a finitely generated module by part (a)(i). Part (c) transfers this property to the isomorphic quotient module $\ker \varphi /L$. Finally apply part (a)(ii) to the submodule $L\subseteq \ker \varphi$. It follows thatMore concretely, take the images under $q$ of finitely many generators of $H$, and adjoin lifts in $\ker \varphi$ of finitely many generators of $\ker \varphi /L$. This finite list generates $\ker \varphi$. The argument shows why finite presentation of a module is independent of the chosen finite generating set of a module.

Put $N=⟨E^{\prime }⟩$ and $P=⟨E^{\prime \prime }⟩$, the submodules generated by the two parts of the basis of a module. Every finite expression in the basis of a module splits into its $E^{\prime }$ terms and its $E^{\prime \prime }$ terms. Thus $M=N+P$.

If $m\in N\cap P$, there are finite expressionsSubtracting gives a finite relation among distinct members of $E$, because $E^{\prime }\cap E^{\prime \prime }=\varnothing$. Their linear independence in a module forces every coefficient to vanish, so $m=0$. HenceThis proves the basis partition gives a module direct sum lemma, also when one part is empty.

Under the temporary assumption that $E$ is a basis of a module for $B$, part (a)(v) applied to the countably generated submodule $F$ gives a countable set $E^{\prime }\subseteq E$ such that $F\subseteq C=⟨E^{\prime }⟩$. By part (a)(iv),Use part (a)(iii) to project onto the second direct summand. This gives a module homomorphism $\pi :B\to B$ withIt remains to check the required size of this kernel. Enumerate $E^{\prime }$ as $b_1,b_2,\dots$, with the finite case treated the same way. Every element of $C$ is a finite sum of these elements with integer coefficients. For each $m$, the set $⟨b_1,\dots ,b_m⟩$ is the image of the countable set ${\mathbb{Z}}^m$, so it is countable. Taking their countable union provesThis is the countability of countably generated abelian groups argument; it uses the countability of the scalar ring $\mathbb{Z}$.

For $j\in {\mathbb{N}}_0$, let $e_j\in B$ have $e_j(j)=1$ and $e_j(k)=0$ for $k\ne j$. If $f,g\in F$ and $z\in \mathbb{Z}$, then $f+g$ can be nonzero only where $f$ or $g$ is nonzero, and $zf$ can be nonzero only where $f$ is nonzero. These are finite sets. Since $0\in F$, this proves that $F$ is a submodule.

Every $f\in F$ has the finite expressionEvaluating at coordinate $k$ recovers the coefficient of $e_k$, so the expression is unique. Thus $\{e_j:j\in {\mathbb{N}}_0\}$ is a basis of a module for $F$, andis a free module. By contrast, the ambient module of integer sequences $B={\prod }_{j\in {\mathbb{N}}_0}\mathbb{Z}$ allows arbitrary integer sequences.

A linear operator $T:V\to V$ on a vector space over a field $k$ makes $V$ a module over the polynomial ring $k[t]$: let $t$ act as $T$, let constants act as the original scalars, and define $p(t)\cdot v=p(T)v$. The identity $(pq)(T)=p(T)q(T)$ for polynomials $p,q$ verifies the multiplication axiom. The submodules are precisely the invariant subspaces of $T$. This construction is described in Keith Conrad's Introductory Notes on Modules, section 6.

For a computed example, take $k=\mathbb{Q}$, $V=k^2$, andThen $T{e}_2=e_1$, $T{e}_1=0$, and $T^2=0$. ConsequentlyEvery $(a,b)$ equals $(b+at)\cdot e_2$. The module homomorphism $k[t]\to V$, $p\mapsto p(T)e_2$, is therefore surjective, and its kernel is the ideal $(t^2)$. It induces a module isomorphism $V\cong k[t]/(t^2)$. This is a cyclic module and a torsion module over $k[t]$, since $t^{2}V=0$. Over $k$ it is a free module with basis $(e_1,e_2)$ and has no nonzero torsion elements of a module. Over $\mathbb{Q}$ its nonzero elements also have infinite additive order of a group element. Changing scalars changes the questions one asks about the same addition law.

For $T=\lambda I+N$ with $N^2=0\ne N$ and a polynomial $p$, the binomial formula gives $p(T)=p(\lambda )I+p^{\prime }(\lambda )N$. Hence its module on the two-dimensional space is indecomposable exactly when $p^{\prime }(\lambda )\ne 0$: in that case the nonzero nilpotent part is one Jordan block; otherwise the operator is scalar and every line is a submodule. For $p(X)=X^2$ over a field of characteristic different from two, the criterion is $\lambda \ne 0$.

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 $A$.

If an algebra acts on a vector space only by scalar multiples of the identity, every vector generates at most a one-dimensional submodule. A higher-dimensional such module is therefore not cyclic. For example upper triangular $n$-by-$n$ matrices act by the first diagonal entry on their first-row ideal. For $n>1$ this ideal is a noncyclic submodule of the cyclic regular module. Thus a submodule of a cyclic module need not be cyclic, even over a finite-dimensional algebra.

In a complex unital Banach algebra, right multiplication by $a$ induces a bounded module endomorphism $T$ of $A/L$ when $La\subseteq L$. Every nonzero module endomorphism of this simple module is algebraically bijective, since its kernel and range are submodules. Its inverse is bounded by the bounded inverse theorem. Choosing $\lambda$ in the nonempty spectrum of an element of $T$ forces $T-\lambda I=0$. Applying this identity to $1+L$ gives $a-\lambda 1\in L$, with uniqueness because $1\notin L$.

A Specht filtration is a finite chain of submodules whose successive quotients are Specht modules. In positive characteristic it need not split and is not a composition series, because its Specht factors may themselves be reducible. Dimensions and ordinary lifted characters can still be added along the filtration.

Suppose the $S_i$ are pairwise nonisomorphic simple modules. Any submodule $W$ of their finite direct sum that projects onto every summand is the entire sum. Induct on $k$. Projection onto the first $k-1$ factors is onto by induction. Its kernel is a submodule of $S_k$, hence zero or all of $S_k$. If zero, $W$ is isomorphic to the first $k-1$ factors but also has quotient $S_k$, contradicting their distinct composition factors and the Jordan–Hölder theorem. Therefore the kernel is $S_k$, proving the assertion. Consequently, if $S_i=A{x}_i$, the map $a\mapsto (a{x}_i{)}_i$ is onto and gives $A/({\bigcap }_i{\mathrm{Ann}}_A(x_i))\cong {⨁}_i{S}_i$ as left modules. The annihilators of these vectors need only be left ideals.

For $n\ge 3$, let $S_n$ permute the coordinates of $M=F^n$. Write $T$ for the constant vectors and $U=S^{(n-1,1)}$ for the augmentation submodule. These and zero and $M$ are all the submodules. Indeed a vector with two unequal coordinates yields a nonzero multiple of $e_i-e_j$ by subtracting its image under $(i,j)$; the orbit of such a difference spans $U$. A submodule without such a vector lies in $T$. If the characteristic does not divide $n$, then $M=T\oplus U$ and $U$ is simple. If it divides $n$, the lattice is the chain $0<T<U<M$, with $U/T$ simple of dimension $n-2$ and trivial end factors. The direct-sum projection in the first case is $x\mapsto x-({\sum }_i{x}_i)1/n$.

A submodule of a filtered module inherits the displayed filtration by intersection. It is also called the induced filtration.