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The injective hull of a uniform module is uniform. An injective endomorphism of a uniform injective module splits its image, so its complementary summand must vanish and it is an automorphism. Consequently all nonunits are precisely the noninjective endomorphisms. Their nonzero kernels are essential; intersection of two such kernels is nonzero, making their sum noninjective. Composition on either side preserves noninjectivity, unless the other map is a unit, which merely transports the kernel. Thus the nonunits form the unique maximal ideal, and the quotient is a division ring.

A uniform module is a nonzero module in which any two nonzero submodules intersect nontrivially. An essential extension $V\subseteq E$ means that $V$ is an essential submodule of $E$: every nonzero submodule of $E$ meets $V$. Since $R$ is a left Noetherian ring, the finitely generated $M$ is a Noetherian module.

Every nonzero Noetherian module contains a uniform submodule. Otherwise split a nonuniform submodule into two disjoint nonzero submodules, retain one for further splitting and set the other aside. If no uniform submodule is reached, the pieces set aside form an infinite direct sum, whose partial sums contradict the ascending chain condition. Now successively choose uniform submodules disjoint from the current direct sum. The process again must stop by Noetherianity, giving $V={⨁}_{i=1}^r{V}_i\subseteq M$. At termination no nonzero submodule is disjoint from $V$, soThis proves the finite uniform decomposition of a Noetherian module.

An injective hull $E(M)$ is an injective module containing $M$ essentially. It is equivalently a minimal injective extension; we use the general existence theorem for injective hulls. To prove uniqueness, let $E_1,E_2$ be two hulls of $M$. Injectivity of $E_2$ extends the identity on $M$ to $f:E_1\to E_2$. Its kernel is disjoint from $M$ and therefore zero by essentiality. Its image is injective, so its inclusion splits in $E_2$. The complementary summand is disjoint from $M$, hence zero. Thus $f$ is an isomorphism fixing $M$.

A finite direct sum of injective modules is injective, and a finite direct sum of essential inclusions is essential. For the latter assertion, start with a nonzero vector and process its coordinates in turn: if a coordinate lies outside its specified essential submodule, multiply by a ring element taking that coordinate nontrivially into that submodule. Already processed coordinates stay inside their submodules; the chosen coordinate keeps the vector nonzero. Essentiality is also transitive. Consequently both $E(M)$ and ${⨁}_{i}E(V_i)$ are injective hulls of $V$, so uniqueness givesThe essential extension of a uniform module is uniform, since intersecting any two nonzero submodules with the original uniform module already gives a nonzero common intersection.

Write $E=E(V_i)$ and $S={\mathrm{End}}_R(E)$. If $f\in S$ is injective, its image is isomorphic to the injective module $E$, hence is a direct summand. Uniformity leaves no nonzero complementary summand, so $f$ is an automorphism. Therefore the nonunits are exactly the noninjective endomorphisms. Let $J$ be their set. Two such maps have nonzero kernels, whose intersection is nonzero by uniformity; their sum kills this intersection. For a product, composition on the left preserves the original kernel. Composition on the right either has a noninjective right factor, or that factor is an automorphism and transports the original nonzero kernel. Thus $J$ is a proper two-sided ideal. Every element outside $J$ is a unit, so every nonzero residue class is invertible and every proper left or right ideal is contained in $J$. This proves the local endomorphism ring of a uniform injective module assertion:For the whole hull, let $E_i=E(V_i)$. Its endomorphism ring is the ring of finite block matriceswith multiplication given by summing compositions. The diagonal entries have the local rings just proved. Off-diagonal entries are homomorphism groups, not necessarily zero; in particular this ring is not generally the product of its diagonal rings. Repeated isomorphic summands give matrix blocks over their common local endomorphism ring.

For the left regular module of $A_1(\mathbb{C})$, the Bernstein filtration has graded ring $\mathbb{C}[x,\xi ]$, so finite leading-term cancellation proves that $A_1$ is left and right Noetherian and a domain. Let $Q$ be its classical division ring of fractions. It is an essential extension of $A_1$: a nonzero fraction $q=s^{-1}a$ satisfies $sq=a\ne 0$ in $A_1$.

To prove that $Q$ is injective, use the Baer criterion. For a homomorphism $f:I\to Q$ on a nonzero left ideal, fix $a\in I\setminus \{0\}$. For any nonzero $x\in I$, take common left multiples $sx=ta$ with $s,t\ne 0$. Then $sf(x)=tf(a)$, so in the division ringThus right multiplication by $a^{-1}f(a)$ extends $f$ from $I$ to $A_1$. The zero ideal is harmless. Hence $Q$ is injective and $E({}_{A_1}{A}_1)=Q$. For any $A_1$-linear endomorphism $F$ of $Q$, clearing the left denominator in $q$ gives $F(q)=qF(1)$. These are all right multiplications, and their composition reverses the multiplying elements. ThereforeThe opposite ring in the injective hull of an Ore domain computation is essential for left-module conventions.

Give the dual $M^{\ast }={\mathrm{Hom}}_k(M,k)$ the contragredient action $(g\lambda )(m)=\lambda (g^{-1}m)$. If ${\delta }_h$ is the functional dual to the basis element $h\in G$, thenThus the map $h\mapsto {\delta }_h$ extends to a $kG$-isomorphismThis is also the left-module form of the fact that a group algebra is a symmetric algebra.

If $P$ is finitely generated and projective, it is a direct summand of $(kG{)}^r$. Dualizing makes $P^{\ast }$ a direct summand of $((kG{)}^{\ast }{)}^r\cong (kG{)}^r$, so $P^{\ast }$ is projective. The converse follows by dualizing again and using $P\cong P^{\ast \ast }$.

For any finite-dimensional algebra $A$, the module $A^{\ast }$ is injective becauseis exact. Since $kG\cong (kG{)}^{\ast }$, free $kG$-modules are injective, and so are their projective direct summands. Conversely, duality sends injectives to projectives, so every finite-dimensional injective is projective. Hence projective modules over a finite group algebra are injective.

Finally, let $P$ be indecomposable projective. It is also an indecomposable injective. Its nonzero socle contains a simple module $S$, and the injective hull $E(S)$ is a direct summand of $P$. Indecomposability forces $P=E(S)$. Since $S$ is essential in its injective hull, every simple submodule of $P$ equals $S$. Therefore

Put $N_G={\sum }_{g\in G}g$. Its image on every module lies in the invariant submodule, since $h{N}_G=N_G$. In the regular module,and this line is the socle of the projective cover $P_k$ of the trivial module.

Suppose $N_{G}M\ne 0$. Choose $m\in M$ with $N_{G}m\ne 0$ and consider the homomorphismIts restriction $f{|}_{P_k}$ is nonzero on $\mathrm{Soc}(P_k)=k{N}_G$. Because $P_k$ is the injective hull of its simple socle, that socle is essential: every nonzero submodule meets it. Hence $\ker (f{|}_{P_k})=0$. The resulting embedding $P_k\hookrightarrow M$ splits because $P_k$ is injective. Since $M$ is indecomposable, $M\cong P_k$.

Conversely, on $P_k$ the image of $N_G$ is its one-dimensional socle. Thus the group norm element detects the trivial projective cover:

Because $N$ is essential in $M$, every associated prime of $M$ occurs in $N$. Since $JN=0$, all these primes contain $J$. For a finitely generated module over a commutative Noetherian ring,Hence $J\subseteq \sqrt{\mathrm{ann}M}$; finite generation of the ideal $J$ gives $J^{n}M=0$ for some $n$.

Now $R/J$ is essential in its injective hull. For $x\in E(R/J)$, the finitely generated module $Rx+R/J$ has essential submodule $R/J$, so the result just proved gives $J^{n}x=0$ for some $n$. The reverse inclusion is tautological, and therefore