《基础代数学》作业 Week6-2
Prove that finite dimensional algebras are IBN rings.
Just consider the dimension.
The following statements about a module $P$ are equivalent:
(1) $P$ is projective;
(2) For any surjective homomorphism $\pi:M→P$, $\pi$ is split surjective;
(3) For any surjective homomorphism $\pi:M→P$, $P$ is a direct summand of $M$.
(1) ⇒ (2) Given a surjective homomorphism $\pi:M→P$, consider the identity homomorphism $1_P:P→P$. By the definition of projective module, there exists a homomorphism $f:P→M$ such that $\pi∘f=1_P$. Thus, $\pi$ is split surjective.
(2) ⇒ (3) Given a surjective homomorphism $\pi:M→P$, by (2), there exists a homomorphism $f:P→M$ such that $\pi∘f=1_P$. It is clear that $M= \ker \pi ⊕ \operatorname{im} f$, so that $P≅\operatorname{im} f$ is a direct summand of $M$.
(3) ⇒ (1) Consider a surjective homomorphism $\pi:F→P$ where $F$ is a free module. By (3), $P$ is a direct summand of $F$. Since free modules are projective, $P$ is also projective (direct summands of a projective module are also projective).
Let $e$ be an idempotent element of a ring $R$. Prove that the left $R$-module $Re$ is projective.
We have $$\begin{equation} Re⊕R(1-e)=R(e+1-e)=R. \end{equation}$$ Since $_RR$ is a free module, it is projective. Thus, its direct summand $Re$ is also projective.
Prove that a finite dimensional algebra $A$ is semisimple if and only if every finite dimensional $A$-module is projective.
(⇒) Since $A$ is semisimple, every finite dimensional $A$-module is semisimple. Then all surjective homomorphisms are split surjective (this is because the homomorphic image of $\pi$ will be isomorphic to the direct complement of $\ker \pi$), so every finite dimensional $A$-module is projective (by the previous problem).
(⇐) We want to show that $_AA$ is semisimple. For any submodule $M$ of $_AA$, consider the surjective homomorphism $\pi:A→A/M$. Since $A/M$ is projective, $\pi$ is split surjective, so that $A≅M⊕\operatorname{im} f$ for some homomorphism $f:A/M→A$. Thus, $_AA$ is semisimple.
Find all the finite dimensional indecomposable projective modules of $A=k[x]/⟨x^n⟩ (n≥2)$.
The only finite dimensional indecomposable projective module of $A$ is $A$ itself. Now we prove this.
We have shown in class that all the finite dimensional indecomposable modules of $A$ are $M_i = ⟨x^{n-i}⟩/⟨x^n⟩, i = 1, 2, \cdots, n$. It is clear that $M_n = A$ is projective. Now we show that for any $1 ≤ i < n$, $M_i$ is not projective. Consider the surjective homomorphism $\pi:A→M_i$ defined by $\pi(1 + ⟨x^n⟩) = x^{n-i} + ⟨x^n⟩$. If $M_i$ is projective, then $\pi$ is split surjective, and $M_i$ will be a direct summand of $A$. However, since $A$ is indecomposable, this is impossible. Thus, $M_i$ is not projective for any $1 ≤ i < n$.