《基础代数学》作业 Week7
Let $M$ be a finite dimensional left module over a finite dimensional algebra $A$. Let $D=\mathrm{Hom}_k(-,k)$. Then $M$ is a projective module if and only if $D(M)$ is an injective right $A$-module.
Functors
are contravariant functors between the category of finite dimensional left $A$-modules and the category of finite dimensional right $A$-modules. It is clear that $D$ is a duality, i.e., $D∘D≅1$.
(⇒) Suppose that $_AM$ is projective. Consider any injective homomorphism $i:N→N'$ of right $A$-modules and any homomorphism $f:N→D(M)$. By applying $D$, we have
Since $_AM$ is projective, there exists a homomorphism $M→ D(N')$ such that the above diagram commutes. By applying $D$ again, we have the following commutative diagram:
Thus, $D(M)$ is an injective right $A$-module.
(⇐) Similar.
Remark. 似乎只能在小模范畴上做.
Let $e$ be an idempotent element of a finite dimensional $k$-algebra $A$. Then $D(eA)$ is an injective left $A$-module, and any injective left $A$-module is of this form.
Since $eA$ is a projective right $A$-module ($A_A≅eA⊕(1-e)A$), by the previous problem, $D(eA)$ is an injective left $A$-module.
For any injective left $A$-module $I$, by the previous problem, $D(I)$ is a projective right $A$-module. Since any projective right $A$-module is a direct summand of a free right $A$-module, say $A^{⊕n}$, there exists an idempotent element $e∈A^{⊕n}$ such that $D(I)≅eA$. Thus, $I≅D(eA)$.
Let $S$ be a simple module over a finite dimensional algebra $A$. Then $S$ can be embedded into an indecomposable injective $A$-module.
We have that $D(S)$ is a finite dimensional right $A$-module. There exists a projective right $A$-module $P$ such that $D(S)$ is a quotient module of $P$. By applying $D$ again, we have an injective homomorphism $S→D(P)$. Since $P$ is a projective right $A$-module, $D(P)$ is an injective left $A$-module. Decomposing $D(P)$ into a direct sum of indecomposable injective left $A$-modules (fintie direct sum is the same as finite product), we see that $S$ can be embedded into one of them (since $S$ is simple).
Let $A$ be a finite dimensional algebra. Then TFAE:
(1) $A$ is semisimple;
(2) Every left $A$-module is projective;
(3) Any short exact sequence of $A$-modules splits;
(4) Every left $A$-module is injective.
(1) ⇒ (2) Since $A$ is semisimple, every left $A$-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 left $A$-module is projective (by an earlier problem).
(2) ⇒ (3) By the property of split short exact sequences in abelian categories.
(3) ⇒ (4) For any left $A$-module $I$, any short exact sequence $0→I→M→N→0$ splits, so $I$ is injective.
(4) ⇒ (1) We want to show that $_AA$ is semisimple. For any submodule $M$ of $_AA$, consider the short exact sequence $0→M→A→A/M→0$. Since $M$ is injective, the above short exact sequence splits, so that $A≅M⊕A/M$. Thus, $_AA$ is semisimple.
Solve the indecomposable injective modules of $A=k[x]/⟨x^n⟩ (n≥2)$.
We have shown that the only finite dimensional indecomposable projective module of $A$ is $A$ itself. By an earlier problem, the only indecomposable injective module of $A$ is $D(A)$.
Prove that $\mathbb{Z}_m$ is an injective $\mathbb{Z}_m$-module by using Baer’s criterion. If $d ∣ m$ and $d$ and $m/d$ has common prime factor, then $\mathbb{Z}_d$ is not an injective $\mathbb{Z}_m$-module.
(1) We first show that $\mathbb{Z}_m$ is an injective $\mathbb{Z}_m$-module. By Baer’s criterion, we need to show that for any ideal $I$ of $\mathbb{Z}_m$ and any homomorphism $h:I→\mathbb{Z}_m$, it can be extended to a homomorphism $h':\mathbb{Z}_m→\mathbb{Z}_m$. Since $\mathbb{Z}_m$ is a principal ideal ring, we have $I=⟨\bar{d}⟩$ for some $d∣m$. Let $h(\bar{d})=\bar{c}$. Since $\bar{c}⋅\overline{m/d}=h(\bar{m})=\bar{0}$, $\overline{c/d}$ is well defined in $\mathbb{Z}_m$.
We define a homomorphism $h':\mathbb{Z}_m→\mathbb{Z}_m$ by $h'(\bar{1})=\overline{c/d}$. It is clear that $h'$ is an extension of $h$. Thus, by Baer’s criterion, $\mathbb{Z}_m$ is an injective $\mathbb{Z}_m$-module.
(2) Now we show that if $d ∣ m$ and $d$ and $m/d$ has common prime factor, then $\mathbb{Z}_d$ is not an injective $\mathbb{Z}_m$-module. We only need to find an ideal $I$ of $\mathbb{Z}_m$ and a homomorphism $h:I→\mathbb{Z}_d$ such that it can not be extended to a homomorphism $h':\mathbb{Z}_m→\mathbb{Z}_d$. Since $d$ and $m/d$ has common prime factor, there exists a prime number $p$ such that $p∣d$ and $p∣m/d$. Let $I=⟨\overline{m/d}⟩$. We define a homomorphism $h:I→\mathbb{Z}_d$ by $h(\overline{m/d})=\bar{1}$. If there exists an extension homomorphism $h':\mathbb{Z}_m→\mathbb{Z}_d$, then we have $$\begin{equation} \bar{1}=h(\overline{m/d})=h'(\overline{m/d})=p h'(\overline{\frac{m}{dp}}). \end{equation}$$ So $$\begin{equation} 0≠ \overline{d/p} = \overline{d/p}h(\overline{m/d})= \overline{d/p}h'(\overline{m/d})= \overline{d} h'(\overline{\frac{m}{dp}}) = \bar{0}. \end{equation}$$ Thus, such extension homomorphism $h'$ does not exist, and by Baer’s criterion, $\mathbb{Z}_d$ is not an injective $\mathbb{Z}_m$-module.
Prove that $R$-module $M$ is an injective module if and only if for any left ideal $L$ of $R$ and any homomorphism $h:L→M$, there exists $m∈M$ such that $h(l)=lm,∀ l∈ L$.
(⇒) Since $M$ is injective, the homomorphism $h:L→M$ can be extended to a homomorphism $h':R→M$. Let $m=h'(1)$. Then for any $l∈L$, we have $h(l)=h'(l)=h'(l⋅1)=l⋅h'(1)=lm$.
(⇐) We use Baer’s criterion to prove this. For any left ideal $L$ of $R$ and any homomorphism $h:L→M$, by the assumption, there exists $m∈M$ such that $h(l)=lm,∀ l∈ L$. Now we can define a homomorphism $h':R→M$ by $h'(r)=rm$. It is clear that $h'$ is an extension of $h$. Thus, by Baer’s criterion, $M$ is injective.