HW7

1. Let $A$ be an Artinian ring. Then there are only a finite number of maximal ideals, and the intersection of them coincides with $J(A)$.

\begin{proof} Since $A/J(A)$ is semisimple, it is a direct sum of finitely many simple Artinian rings. Thus, $A/J(A)$ has finitely many two-sided ideals. Recall that the ideals of $A/J(A)$ is $1$-$1$ corresponds to the ideals of $A$ containing $J(A)$. Thus, to show $A$ has finitely many two-sided maximal ideal, it is enough to show each maximal two-sided ideal contains $J(A)$. By definition, $J(A)$ is the intersection of all left maximal ideals of $A$.

Assume that there exists a two-sided maximal ideal $I\subseteq A$ such that $I⊉J(A)$. Then $I+J(A)=A$. Since $I$ is a proper two-sided ideal, $I$ is also a proper left ideal and so it is contained in some maximal left ideal $J$. Remark that $J\supseteq J(A)$ and $I+J(A)\subseteq J+J(A)=J$, which is a contradiction. Therefore, all two-sided maximal ideals of $A$ contain $J(A)$. We have proved that $A$ has finitely many maximal ideals.

Define $M$ as the intersection of all two-sided maximal ideal. By the argument above, we have $M\supseteq J(A)$. On the other hand, each two-sided maximal ideal is contained in a left maximal ideal, which deduces that $M\subseteq J(A)$.

Therefore, $M=J(A)$, i.e., the intersection of all two-sided maximal ideals coincides with $J(A)$. \end{proof}

2. If $\phi :A\to A^{\prime }$ is an epimorphism of rings, then $\phi (J(A))\subset J\left(A^{\prime }\right)$.

\begin{proof} For any $a\in J(A)$, $1-xa$ has left inverse for all $x\in A$, i.e. there exists $b\in A$ such that $b(1-xa)=1$. It deduces that $1-\phi (x)\phi (a)$ has left inverse of all $a\in A$. Since $\phi$ is epimorphism, $1-x^{\prime }\phi (a)$ has left inverse for all $x^{\prime }\in A^{\prime }$. Therefore, $\phi (a)\in J(A^{\prime })$ and so $\phi (J(A))\subset J\left(A^{\prime }\right)$. \end{proof}

3. Let $A$ be a finite-dimensional algebra over a field. Suppose that $V$ is an $A$-module, and that a certain simple $A$-module $S$ occurs as a composition factor of $V$ with multiplicity $1$. Suppose that there exist nonzero homomorphisms $S\to V$ and $V\to S$. Prove that $S$ is a direct summand of $V$.

\begin{proof} Define $\phi :S\to V$ and $\psi :V\to S$, then $\ker \phi ,\mathrm{im}\psi$ are $A$-submodules of $S$. Since $S$ is simple, $\ker \phi =0$ and $\mathrm{im}\psi =S$. It deduces that $V/\ker \psi \simeq S$ and $\mathrm{im}\phi \simeq S$ is an $A$-submodule of $V$. We aim to show $V=\ker \psi \oplus \mathrm{im}\phi$. If $\ker \psi \cap \mathrm{im}\phi \ne \{0\}$, then $\ker \psi$ has an $A$-submodule isomorphic to $S$. It is impossible because $V/\ker \psi \simeq S$ and $S$ occurs as a composition factor of $V$ with multiplicity $1$. Therefore, $\ker \psi \cap \mathrm{im}\phi =\{0\}$. Furthermore, $V/\ker \psi \simeq \mathrm{im}\phi$ yields that $V=\ker \psi \oplus \mathrm{im}\phi =\ker \psi \oplus S$. Now we finish the proof. \end{proof}

4. Suppose that $U,V$, and $W$ are modules for a finite-dimensional algebra $A$ over a field and $P_W$ is the projective cover of $W$. Assume that these modules are finite-dimensional.

• (a) Show that $U\to W$ is an essential epimorphism if and only if there is a surjective homomorphism $P_W\to U$ so that the composite $P_W\to$ $U\to W$ is a projective cover of $W$. In this situation, show that $P_W\to U$ must be a projective cover of $U$.
• (b) Prove the following “extension and converse” to Nakayama’s Lemma: let $V$ be any submodule of $U$. Then $U\to U/V$ is an essential epimorphism $\iff V\subseteq \mathrm{Rad}U$.

\begin{proof} a) Assume that $f$ is an essential epimorphism. Since $P_W$ is a projective cover, there exists $g$ such that $f\circ g=\phi$. Then by ^q7sonb, $g$ is also essential. If $g$ is not surjective, then $g(P_W)\ne U$ and $f(g(P_W))=W$, which contradicts with $f$ being an essential epimorphism.

Conversely, assume that there is a surjective homomorphism $g:P_W\to U$ so that the composite $\phi :P_W\stackrel{g}{\to }U\stackrel{f}{\to }W$ is a projective cover of $W$. We aim to show $g$ is essential. If there exists $K⊊P_W$ such that $g(K)=U$, then $f(g(K))=f(U)=W$. Since $f(g(K))=\phi (K)$ and $\phi$ is a projective cover, we have $K=P_W$, which is a contradiction. Thus, $f$ is an essential epimorphism.

b) Assume that $\psi :U\to U/V$ is an essential epimorphism. If there exists $v\in V\setminus \mathrm{Rad}U$, then there exists a maximal submodule $M$ such that $v\notin M$. It deduces that $U=M+Av\subseteq M+V$. Since $\psi$ is an essential epimorphism and $M+V=U$, one have $M=U$, which is impossible.

Conversely, assume that $V\subseteq \mathrm{Rad}U$. We aim to show $\psi :U\to U/V$ is an essential epimorphism, that is, for any $L\subseteq U$, if $L+V=U$, then $L=U$. Otherwise, there exists $L⊊U$ with $L+V=U$. Then there is a maximal submodule $M^{\prime }\supseteq L$, and $M=M+V\supseteq L+V=U$ by $V\subseteq \mathrm{Rad}U\subseteq M$, which is a contradiction. Thus, if $L$ is a submodule of $U$ such that $L+V=U$, then $L=U$. Now we finish the proof. \end{proof}

Remark. The propositions used in exercise 4 is as follows.

• Let $f_i:U_i\to V_i$ be homomorphisms of Noetherian modules for $i=1,\cdots ,n$, then $f_i$ are all essential epimorphisms iff $\oplus f_i:\oplus U_i\to \oplus V_i$ is an essential epimorphism.

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