Schur's lemma

Theorem

Schur's lemma

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Statement

Let \(k\) be a field, let \(G\) be a group (or more generally let \(A\) be any \(k\)-algebra), and let \(V\) and \(W\) be irreducible (simple) \(k[G]\)-modules of finite \(k\)-dimension, i.e. representations \(\rho_V:G\to GL(V)\), \(\rho_W:G\to GL(W)\) with no proper nonzero \(G\)-invariant subspace. Let \(f:V\to W\) be a \(k[G]\)-module homomorphism, i.e. a \(k\)-linear map satisfying \(f(g\cdot v)=g\cdot f(v)\) for all \(g\in G\), \(v\in V\). Then either \(f=0\) or \(f\) is an isomorphism. Moreover, if \(k=\mathbb{C}\) (or any algebraically closed field) and \(V=W\), then every such \(f\) is a scalar multiple of the identity: \(f=\lambda\,\mathrm{id}_V\) for some \(\lambda\in\mathbb{C}\).

Why it matters

Schur's lemma is the single load-bearing fact underneath representation theory: it converts the abstract statement "irreducible representations are the atoms of the category" into concrete linear-algebra control over intertwiners. Every downstream structural theorem — the orthogonality relations for characters, the decomposition of the group algebra \(k[G]\cong\bigoplus_i \mathrm{Mat}_{n_i}(k)\) (Wedderburn/Artin–Wedderburn for \(k[G]\)), and the fact that isotypic components of a module are canonically determined — is proved by applying Schur's lemma to a cleverly chosen intertwiner.

Its second, sharper form (the scalar version over \(\mathbb{C}\)) is what makes representation theory computationally tractable: it says the "endomorphism algebra" of an irreducible complex representation is as small as it can possibly be, namely \(\mathbb{C}\) itself, and this single fact is the engine behind computing character tables, proving the row/column orthogonality relations, and diagonalizing group actions.

Hypotheses

\(V\) and \(W\) are irreducible.If \(V=k^2\) with trivial \(G\)-action, \(V\) has the invariant line spanned by \(e_1\), and the projection \(\pi:V\to V\), \(\pi(e_1)=e_1,\pi(e_2)=0\) is a nonzero, non-invertible \(G\)-module map — Schur's conclusion fails outright once \(V\) is reducible. \(f\) is a \(G\)-module homomorphism (intertwiner), not merely \(k\)-linear.Take \(V=W=\mathbb{C}^2\) irreducible under some \(G\); an arbitrary linear map that does not commute with the \(G\)-action (e.g. a map sending one weight vector to another with mismatched eigenvalues) need be neither zero nor invertible in any representation-theoretic sense compatible with the action — the whole force of the lemma comes from equivariance. \(k\) is algebraically closed, for the scalar conclusion.Over \(k=\mathbb{R}\), take \(G=\mathbb{Z}/4\) acting on \(V=\mathbb{R}^2\) by rotation through \(90^\circ\); \(V\) is irreducible over \(\mathbb{R}\) (no real invariant line), yet \(\mathrm{End}_{\mathbb{R}G}(V)\cong\mathbb{C}\) is two-dimensional, not one-dimensional — an intertwiner \(V\to V\) need not be a real scalar multiple of the identity (it can be any element of that copy of \(\mathbb{C}\), i.e. any rotation-and-scaling). \(V\) is finite-dimensional (for the scalar form's eigenvalue argument).The vanishing/isomorphism dichotomy holds for arbitrary (Jacobson-)simple modules over any algebra with no finiteness assumption, but the "\(f\) has an eigenvalue" step used to force \(f=\lambda\,\mathrm{id}\) requires a nonzero eigenvalue to exist, which is guaranteed over an algebraically closed field only when \(\dim_k V\lt\infty\) (a nonzero endomorphism of an infinite-dimensional space need not have any eigenvalue at all).

Proof

\(\ker f=\{v\in V: f(v)=0\}\) is a \(G\)-submodule of \(V\).

If \(f(v)=0\) then for any \(g\in G\), \(f(g\cdot v)=g\cdot f(v)=g\cdot 0=0\), using the intertwining property \(f(g\cdot v)=g\cdot f(v)\) which holds by hypothesis. Also \(\ker f\) is a \(k\)-subspace since \(f\) is \(k\)-linear. A

\(\operatorname{im} f=\{f(v):v\in V\}\subseteq W\) is a \(G\)-submodule of \(W\).

For \(w=f(v)\in\operatorname{im}f\) and \(g\in G\): \(g\cdot w = g\cdot f(v)=f(g\cdot v)\in\operatorname{im}f\), again by equivariance of \(f\); closure under \(k\)-linear combinations is immediate since \(f\) is linear. A

Since \(V\) is irreducible, \(\ker f\in\{0,V\}\); since \(W\) is irreducible, \(\operatorname{im}f\in\{0,W\}\).

This is exactly the definition of irreducibility applied to the submodules found in Steps 1–2: an irreducible module has no submodules other than \(0\) and itself. A

Case \(\ker f=V\): then \(f=0\).

\(\ker f=V\) means \(f(v)=0\) for every \(v\in V\), which is precisely the statement \(f=0\) as a map. A

Case \(\ker f=0\): then \(f\) is injective, so \(\operatorname{im}f\neq 0\) (assuming \(V\neq 0\)), forcing \(\operatorname{im}f=W\) by Step 3; hence \(f\) is bijective.

Injectivity from \(\ker f=0\) is the standard linear-algebra fact that a linear map is injective iff its kernel is trivial. If \(V\neq 0\) then injectivity gives \(\operatorname{im}f\neq 0\), and the dichotomy of Step 3 rules out \(\operatorname{im}f=0\), leaving \(\operatorname{im}f=W\), i.e. surjectivity. A linear bijection is automatically an isomorphism of underlying vector spaces; its inverse \(f^{-1}:W\to V\) is again a \(G\)-module map because for \(w=f(v)\), \(g\cdot f^{-1}(w)=g\cdot v\) and \(f^{-1}(g\cdot w)=f^{-1}(f(g\cdot v))=g\cdot v\) agree, using equivariance of \(f\) once more. B

Combining Steps 4–5: every \(G\)-module homomorphism \(f:V\to W\) between irreducibles is either the zero map or a \(G\)-module isomorphism. This proves the first assertion.

Case split is exhaustive because Step 1 forces \(\ker f\) to be exactly \(0\) or \(V\), with no third option, by irreducibility of \(V\). A

Now suppose \(k=\mathbb{C}\), \(V=W\), \(\dim_{\mathbb{C}}V=n\lt\infty\), and \(f:V\to V\) is a nonzero \(G\)-module endomorphism. Since \(\mathbb{C}\) is algebraically closed, the characteristic polynomial \(\det(f-\lambda\,\mathrm{id}_V)\) has a root \(\lambda_0\in\mathbb{C}\), so \(f\) has an eigenvalue \(\lambda_0\) with a nonzero eigenvector.

This uses the Fundamental Theorem of Algebra (every nonconstant polynomial over \(\mathbb{C}\) has a root) applied to the characteristic polynomial of the linear operator \(f\) on the finite-dimensional space \(V\); this is exactly where algebraic closedness and finite-dimensionality are used. C

Set \(h:=f-\lambda_0\,\mathrm{id}_V\). Then \(h\) is again a \(G\)-module endomorphism of \(V\), and \(h\) is not injective (it kills the \(\lambda_0\)-eigenvector found in Step 7).

\(\mathrm{id}_V\) is trivially a \(G\)-module map (\(g\cdot\mathrm{id}_V(v)=g\cdot v=\mathrm{id}_V(g\cdot v)\)), and the space of \(G\)-module maps \(V\to V\) is a \(k\)-vector space (closed under the \(k\)-linear combination \(f-\lambda_0\,\mathrm{id}_V\)) because equivariance is preserved under linear combinations: \((af+bh)(g\cdot v)=a\,g\cdot f(v)+b\,g\cdot h(v)=g\cdot(af+bh)(v)\). Non-injectivity is immediate from \(h(v_0)=f(v_0)-\lambda_0 v_0=0\) for the eigenvector \(v_0\). B

By the dichotomy of Step 6 applied to \(h:V\to V\) (an intertwiner between the irreducible \(V\) and itself), and since \(h\) is not injective hence not an isomorphism, we must have \(h=0\), i.e. \(f=\lambda_0\,\mathrm{id}_V\).

Direct reapplication of the already-established general Schur dichotomy (Steps 1–6) to the specific map \(h\); non-injective rules out the "isomorphism" branch, leaving only "\(h=0\)". B

Hence every \(G\)-module endomorphism of an irreducible finite-dimensional complex representation \(V\) is a scalar: \(\mathrm{End}_{\mathbb{C}G}(V)=\mathbb{C}\cdot\mathrm{id}_V\).

This is exactly the content of Step 9 restated as a statement about the endomorphism algebra; it also uses that the zero map is \(\lambda=0\) times the identity, unifying with the case \(f=0\) from Step 6. B

Result

f:V\to W \text{ a } G\text{-map between irreducibles} \implies f=0 \text{ or } f \text{ iso}; \quad \text{over } \mathbb{C},\ V=W \implies f=\lambda\,\mathrm{id}_V

Reading. An equivariant map between irreducible representations has no room to be "partly" an isomorphism — its kernel and image are forced by irreducibility to be all-or-nothing. When source and target are literally the same irreducible complex representation, the only equivariant maps available are the trivial ones: multiples of the identity.

Scope. The dichotomy (zero-or-isomorphism) holds for irreducible modules over any \(k\)-algebra \(A\) (group algebras \(k[G]\) are the motivating case, but nothing in Steps 1–6 used group structure beyond \(A\)-linearity). The scalar refinement additionally needs \(k\) algebraically closed and \(\dim_k V\lt\infty\); it applies verbatim to Lie algebra representations, associative algebra modules, and more generally to any simple module over an algebra that is finite-dimensional over an algebraically closed field.

Corollaries & converses

If \(V\not\cong W\) as \(G\)-modules (no equivariant isomorphism exists between them) and both are irreducible, then \(\mathrm{Hom}_{kG}(V,W)=0\): the only intertwiner is the zero map, since Step 6 forbids any nonzero non-isomorphism.
If \(V\cong W\) with \(k\) algebraically closed and \(\dim V\lt\infty\), then \(\dim_k \mathrm{Hom}_{kG}(V,W)=1\): the space of intertwiners is exactly the scalar multiples of a fixed isomorphism, by Step 10 transported along that isomorphism.
For \(k=\mathbb{C}\), any two irreducible representations commuting with all of \(G\)-action and agreeing on a nonzero intertwiner are isomorphic — this is the tool used to show characters of non-isomorphic irreducibles are orthogonal.
The converse is false: "\(f\) is zero or an isomorphism for every equivariant \(f:V\to V\)" does not imply \(V\) is irreducible in general algebraic settings without extra finiteness, but for finite-dimensional modules over \(k[G]\) it is in fact equivalent to indecomposability plus the endomorphism ring being a division ring (a form of the Fitting lemma) — the precise converse needs the module to be finite length, illustrating that Schur's lemma is a consequence of irreducibility, not an equivalent restatement of it without further hypotheses.

Fails without

Drop irreducibility of \(V\): with \(G=\{e\}\) trivial, \(V=k^2\), \(W=k^2\), the map \(f(x,y)=(x,0)\) is trivially \(G\)-equivariant (any linear map is, since \(G\) acts trivially), yet \(f\) is neither \(0\) nor invertible — it is a nontrivial idempotent projection, exactly the failure mode the hypothesis rules out.
Drop algebraic closedness (for the scalar form): with \(G=\mathbb{Z}/4=\langle r\rangle\) acting on \(V=\mathbb{R}^2\) by \(r\mapsto \left(\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right)\), \(V\) is \(\mathbb{R}\)-irreducible (no invariant real line) yet \(\mathrm{End}_{\mathbb{R}G}(V)\) is the full quaternion-free division algebra \(\mathbb{C}\) (2-dimensional over \(\mathbb{R}\)): rotation-scalings other than real multiples of the identity are legitimate equivariant endomorphisms, so the "scalar" conclusion strictly fails, though the zero-or-isomorphism dichotomy still holds.
Drop equivariance (allow arbitrary linear \(f\)): for any irreducible \(V,W\) of dimension \(\geq 2\), a generic \(k\)-linear map \(V\to W\) (chosen with no relation to the \(G\)-actions) will typically have kernel of intermediate dimension, giving neither \(0\) nor an isomorphism — equivariance is what collapses \(\ker f\) to an actual submodule and hence to one of only two possibilities.

Common errors

Applying the scalar form ("\(f=\lambda\,\mathrm{id}\)") over \(\mathbb{R}\) or another non-algebraically-closed field, forgetting the two-dimensional real representations that only become reducible after tensoring up to \(\mathbb{C}\).
Forgetting that the scalar conclusion needs \(V=W\) (same module on both sides); students sometimes claim an intertwiner between two different irreducibles is "a scalar times some fixed map", which is only correct when an isomorphism \(V\cong W\) has been fixed to identify them, and is meaningless (there is no such intertwiner at all) when \(V\not\cong W\).
Treating "irreducible" and "indecomposable" as synonyms; Schur's lemma genuinely needs irreducibility (no proper nonzero submodule), and can fail for merely indecomposable modules over non-semisimple algebras (e.g. modular representations in characteristic \(p\mid |G|\)).
Misapplying Step 1–2 by forgetting to check that \(\ker f\) and \(\operatorname{im}f\) are \(G\)-stable, not just vector subspaces — this equivariance check is the entire content of the proof and is not automatic for a general linear map.
Assuming the finite-dimensionality hypothesis is decorative: over infinite-dimensional spaces a nonzero equivariant endomorphism can fail to have any eigenvalue at all (e.g. the shift operator on \(\ell^2\)), so the "isomorphism-or-zero" dichotomy from Steps 1–6 still holds, but the eigenvalue argument of Steps 7–10 producing a scalar breaks down completely.

Discussion

Issai Schur proved this lemma in the early 1900s in his work on group representations, and its two-line proof — irreducibility forces kernel and image to be trivial or everything — belies its structural weight. The lemma is the representation-theoretic incarnation of a much older algebraic fact: the endomorphism ring of a simple module over any ring is a division ring (this is sometimes stated as "Schur's lemma" in full generality, with the group-theoretic and character-theoretic corollaries being specializations to \(k[G]\)-modules).

The scalar refinement over \(\mathbb{C}\) is what makes the character table machinery work: because \(\mathrm{End}_{\mathbb{C}G}(V)=\mathbb{C}\) for irreducible \(V\), the projection operators \(e_i=\frac{\dim V_i}{|G|}\sum_{g\in G}\overline{\chi_i(g)}\,\rho(g)\) built from characters act as scalars on each irreducible constituent, which is precisely the mechanism behind the orthogonality relations \(\frac{1}{|G|}\sum_g \chi_i(g)\overline{\chi_j(g)}=\delta_{ij}\).

There is a useful reformulation via the group algebra: irreducible \(k[G]\)-modules correspond to simple modules over the (semisimple, when \(\mathrm{char}\,k\nmid |G|\), by Maschke's theorem) algebra \(k[G]\), and Schur's lemma combined with the Artin–Wedderburn theorem is exactly what identifies \(\mathbb{C}[G]\cong\bigoplus_i \mathrm{Mat}_{n_i}(\mathbb{C})\), with each matrix block corresponding to one irreducible of dimension \(n_i\) acting on itself by left multiplication.

A common misconception is that Schur's lemma is somehow a "finiteness" or "compactness" phenomenon specific to finite groups; in fact the zero-or-isomorphism dichotomy (Steps 1–6) holds for irreducible representations of arbitrary groups, Lie algebras, and associative algebras with no cardinality restriction whatsoever — the only place finiteness and topology genuinely enter is in guaranteeing enough eigenvalues exist for the scalar refinement, which is why the analogous statement for unitary representations of compact (possibly infinite) groups still holds via a version using compactness of the group to secure complete reducibility, while genuinely infinite-dimensional non-compact settings (e.g. Banach space representations) can break the scalar conclusion even over \(\mathbb{C}\).

Worked examples

Let \(G=S_3\) act on \(V=\mathbb{C}^2\) via its standard 2-dimensional irreducible representation (the reflection representation), and let \(f:V\to V\) be a linear map known to satisfy \(f(\rho(g)v)=\rho(g)f(v)\) for all \(g\in S_3\), with \(f\) known to send the vector \(v_0=(1,0)\) (a non-fixed vector) to \((2,0)\).

Setup: \(V\) is irreducible (a standard fact — the reflection representation of \(S_3\) has no invariant line, since the three reflections have distinct axes), so Steps 6 and 7–10 (\(k=\mathbb{C}\), \(\dim V=2\lt\infty\)) apply directly to \(f\). B

By the scalar form of Schur's lemma, \(f=\lambda\,\mathrm{id}_V\) for some \(\lambda\in\mathbb{C}\).

Direct application of the Result box (scalar case), since \(V\) is irreducible, finite-dimensional, and \(k=\mathbb{C}\) is algebraically closed. A

Since \(f(v_0)=(2,0)=2\cdot(1,0)=2v_0\), matching \(\lambda v_0\) forces \(\lambda=2\).

Substitute the known value into \(f=\lambda\,\mathrm{id}_V\) and compare coordinates; a single nonzero vector's image determines \(\lambda\) uniquely because \(\mathrm{id}_V\) is injective. A

f = 2\cdot \mathrm{id}_V \text{ on all of } V, \text{ e.g. } f(0,1)=(0,2)

Reading. Knowing that an equivariant map on an irreducible complex representation acts as a scalar on one vector pins it down completely, everywhere — an enormous shortcut compared to checking the map coordinatewise.

Let \(G=\mathbb{Z}/3=\{e,r,r^2\}\), and let \(V\) be the irreducible complex representation with \(\rho(r)=\omega=e^{2\pi i/3}\) acting as a scalar on \(V=\mathbb{C}\), and \(W\) the irreducible representation with \(\rho(r)=\omega^2\) acting as a scalar on \(W=\mathbb{C}\). Both are 1-dimensional hence automatically irreducible. Suppose \(f:V\to W\) is \(G\)-equivariant.

Setup: \(\mathbb{Z}/3\) is abelian, so its irreducible complex representations are exactly the three 1-dimensional characters \(1,\omega,\omega^2\) — a standard fact from the theory of characters of finite abelian groups (each irreducible representation of an abelian group is 1-dimensional, since Steps 7–10 applied to \(\rho(g)\) itself for each \(g\) force every \(\rho(g)\) to be a scalar, hence every subspace is invariant, forcing \(\dim=1\) for irreducibility). C

Equivariance forces \(f(r\cdot v)=r\cdot f(v)\), i.e. \(\omega f(v) = \omega^2 f(v)\) for all \(v\in V=\mathbb{C}\).

Direct unwinding of the definitions \(\rho_V(r)v=\omega v\) and \(\rho_W(r)f(v)=\omega^2 f(v)\), combined with the equivariance condition \(f(\rho_V(r)v)=\rho_W(r)f(v)\) (here \(f\) is a scalar multiplication map \(f(v)=c v\), so \(f(\omega v)=\omega c v\), giving \(\omega c v=\omega^2 c v\)). A

Since \(\omega\neq\omega^2\) (as \(\omega=e^{2\pi i/3}\) has order 3, so \(\omega\neq\omega^2\)), the equation \((\omega-\omega^2)f(v)=0\) forces \(f(v)=0\) for all \(v\).

Cancellation in the field \(\mathbb{C}\): \(\omega-\omega^2\neq 0\) is a nonzero scalar, and \(\mathbb{C}\) has no zero divisors, so the equation forces the other factor to vanish. This matches the "\(f=0\)" branch of Schur's dichotomy (Step 6), consistent with \(V\not\cong W\) as the Corollary predicts. B

V\not\cong W \implies \mathrm{Hom}_{\mathbb{C}[\mathbb{Z}/3]}(V,W)=0

Reading. Two non-isomorphic irreducibles of an abelian group admit no nonzero equivariant map at all — this is the algebraic root of why distinct characters are "independent" and underlies orthogonality of characters.

Problems

Let \(V\) be an irreducible \(\mathbb{C}[G]\)-module of dimension \(n\), and suppose \(f:V\to V\) is \(G\)-equivariant with \(f^2=f\) (idempotent). Show \(f=0\) or \(f=\mathrm{id}_V\).
Solution
By the scalar form of Schur's lemma (Result box), \(f=\lambda\,\mathrm{id}_V\) for some \(\lambda\in\mathbb{C}\), since \(V\) is irreducible, finite-dimensional, and \(\mathbb{C}\) is algebraically closed. Then \(f^2=f\) gives \(\lambda^2\,\mathrm{id}_V=\lambda\,\mathrm{id}_V\), so \(\lambda^2=\lambda\) (comparing the scalar coefficients, since \(\mathrm{id}_V\neq 0\) for \(V\neq 0\)), i.e. \(\lambda(\lambda-1)=0\), giving \(\lambda=0\) or \(\lambda=1\). Hence \(f=0\) or \(f=\mathrm{id}_V\).
Let \(V,W\) be irreducible \(\mathbb{C}[G]\)-modules with \(\dim V\neq \dim W\). Show directly, without invoking character orthogonality, that \(\mathrm{Hom}_{\mathbb{C}G}(V,W)=0\).
Solution
Let \(f:V\to W\) be any \(G\)-equivariant map. By Schur's lemma (Steps 1–6), \(f=0\) or \(f\) is an isomorphism. If \(f\) were an isomorphism, it would in particular be a linear isomorphism of underlying vector spaces, forcing \(\dim V=\dim W\), contradicting the hypothesis. Hence the isomorphism branch is impossible, so \(f=0\). Since \(f\) was arbitrary, \(\mathrm{Hom}_{\mathbb{C}G}(V,W)=0\).
Let \(G\) be a finite group and \(V\) an irreducible \(\mathbb{C}[G]\)-module. Let \(z\) be a central element of \(\mathbb{C}[G]\) (i.e. \(z\) commutes with every element of the group algebra). Show that \(\rho(z)\) acts as a scalar on \(V\).
Solution
Since \(z\) is central in \(\mathbb{C}[G]\), for every \(g\in G\) we have \(zg=gz\) in \(\mathbb{C}[G]\), hence \(\rho(z)\rho(g)=\rho(g)\rho(z)\) as operators on \(V\) (applying the representation \(\rho\), which is an algebra homomorphism \(\mathbb{C}[G]\to \mathrm{End}(V)\)). This says exactly that \(\rho(z):V\to V\) is a \(G\)-module homomorphism, since \(\rho(z)(g\cdot v)=\rho(z)\rho(g)v=\rho(g)\rho(z)v=g\cdot(\rho(z)v)\) for all \(g,v\). \(V\) is irreducible and finite-dimensional over the algebraically closed field \(\mathbb{C}\), so by the scalar form of Schur's lemma (Result box, applied to \(f=\rho(z)\)), \(\rho(z)=\lambda\,\mathrm{id}_V\) for some \(\lambda\in\mathbb{C}\).
Give an explicit example (as in the "Fails without" section) of a group \(G\), a field \(k\), and an irreducible \(k[G]\)-module \(V\) with \(\mathrm{End}_{kG}(V)\neq k\cdot\mathrm{id}_V\), and identify exactly which hypothesis of the scalar form of Schur's lemma fails.
Solution
Take \(G=\mathbb{Z}/4=\langle r\rangle\), \(k=\mathbb{R}\), and \(V=\mathbb{R}^2\) with \(\rho(r)=\begin{pmatrix}0&-1\\1&0\end{pmatrix}\) (rotation by \(90^\circ\)). \(V\) has no \(G\)-invariant real line (a line spanned by \((a,b)\) would need to be sent to a scalar multiple of itself by a \(90^\circ\) rotation, which never happens for a real line since rotation by \(90^\circ\) has no real eigenvalue, as its characteristic polynomial \(\lambda^2+1\) has no real roots), so \(V\) is irreducible over \(\mathbb{R}\). The dichotomy (Steps 1–6, needing no algebraic closedness) still holds: any nonzero equivariant \(f:V\to V\) is an isomorphism. But the scalar form fails because \(\rho(r)\) itself is a nonzero, non-scalar equivariant endomorphism of \(V\) commuting with \(G\) (it visibly commutes with all powers of itself), giving \(\mathrm{End}_{\mathbb{R}G}(V)\supseteq\mathbb{R}[\rho(r)]\cong\mathbb{C}\), which is 2-dimensional over \(\mathbb{R}\), not the 1-dimensional \(\mathbb{R}\cdot\mathrm{id}_V\). The failing hypothesis is algebraic closedness of the field \(k\) (Steps 7 needs a root of the characteristic polynomial over \(k\), which \(\mathbb{R}\) does not guarantee).
Let \(A\) be a finite-dimensional associative \(\mathbb{C}\)-algebra and let \(S\) be a simple (irreducible) left \(A\)-module of finite \(\mathbb{C}\)-dimension. Prove that \(D:=\mathrm{End}_A(S)\) is a division ring, and then, using finite-dimensionality and algebraic closedness of \(\mathbb{C}\), prove \(D=\mathbb{C}\).
Solution
Division ring part: Let \(f\in D=\mathrm{End}_A(S)\) be nonzero. Exactly as in Steps 1–3 of the proof (with \(G\)-module replaced by \(A\)-module — nothing in that argument used group structure, only \(A\)-linearity), \(\ker f\) and \(\operatorname{im}f\) are \(A\)-submodules of \(S\), and simplicity of \(S\) forces each to be \(0\) or \(S\). Since \(f\neq 0\), \(\ker f\neq S\) so \(\ker f=0\), and then \(\operatorname{im}f\neq 0\) so \(\operatorname{im}f=S\); hence \(f\) is a bijective \(A\)-module map, i.e. invertible in \(D\) with inverse \(f^{-1}\) also in \(D\) (an \(A\)-module map, by the same argument as Step 5). So every nonzero element of \(D\) is invertible, i.e. \(D\) is a division ring. \(D=\mathbb{C}\) part: \(D\) is a \(\mathbb{C}\)-subalgebra of \(\mathrm{End}_{\mathbb{C}}(S)\), which is finite-dimensional since \(\dim_{\mathbb{C}}S\lt\infty\); hence \(D\) is a finite-dimensional \(\mathbb{C}\)-division algebra. Take any \(f\in D\); exactly as in Steps 7–9, since \(\mathbb{C}\) is algebraically closed and \(S\) finite-dimensional, \(f\) has an eigenvalue \(\lambda_0\in\mathbb{C}\) (root of its characteristic polynomial), and \(h=f-\lambda_0\,\mathrm{id}_S\in D\) is non-injective, hence (division ring — no nonzero non-invertible elements) \(h=0\), so \(f=\lambda_0\,\mathrm{id}_S\). Thus every element of \(D\) is a scalar multiple of the identity, i.e. \(D=\mathbb{C}\cdot\mathrm{id}_S\cong\mathbb{C}\).