Ytukyg

I am working through Christian Kassel's textbook on Quantum Groups. The Proposition states that 5 statemens are equivalent. The two I am having trouble with are as follows.

1.For any pair $V'subset V$ of finite-dimensional $A$-modules, there exists an $A$-module $V''$ such that $Vcong V'oplus V''$.

3.For any pair $V'subset V$ of finite-dimensional $A$-modules, there exists an $A$-linear map $p:Vrightarrow V'$ with $p^2=p$.

In the proof it has the following for 3 implies 1: Let $V''=operatorname{Ker}(p)$; it is a submodule of $V$. The relations $v=p(v)+big(v-p(v)big)$ and $p^2=p$ prove that $V$ is the direct sum $V''$ and $V'$.

I am completely lost as to how those relations prove the direct sum. Any help would be greatly appreciated.

In this answer, $V'$ and $V$ are not necessarily finite dim. In fact $A$ can be any ring (i.e., not necessarily an algebra over any field, and not necessarily unital). That is the equivalence of the two statements is as follow.

Theorem. Let $V$ be a module over a ring $A$ with a submodule $V'$. Then, $V'$ is a direct summand of $V$ (that is, $V=V'oplus V''$ for some submodule $V''$ of $V$) if and only if there exists a projection $p:Vto V$ such that $operatorname{im}p=V'$. (A projection on $V$ is an $A$-module endomorphism $p:Vto V$ such that $p^2=p$.)

Remark. The OP's version of this theorem is false. If $p$ is not required to be surjective, we have a counterexample, e.g., when $V'$ is a non-zero submodule of $V$ with $p=0$.)

Proof. Suppose that $V'$ and $V$ are $A$-modules such that $V'subseteq V$ and there exists a projection $p:Vto V$ with $V'=operatorname{im}p$. We will show that $V=V'oplus V''$ where $V''=ker p$.

First, $V=V'+V''$. To see this, we note that $v=p(v)+big(v-p(v)big)$. Clearly, $p(v)in V'$. We must prove that $v-p(v)in V''$, which is equivalent to $pbig(v-p(v)big)=0$. But $$pbig(v-p(v)big)=p(v)-p^2(v)=p(v)-p(v)=0$$ by the hypothesis that $p^2=p$. So $v-p(v)in ker p = V''$.

To finish the prove we must show that the sum $V=V'+V''$ is direct. That is $V'cap V''={0}$. Suppose that $win V'cap V''$. Since $win V'=operatorname{im}p$, we must have $w=p(v)$ for some $vin V$. Since $win V''=ker p$, $p(w)=0$. Therefore,
$$0=p(w)=pbig(p(v)big)=p^2(v)=p(v)=w,$$
where we use again the hypothesis that $p^2=p$. This proves that $V'cap V''={0}$. Therefore, $V$ is the internal direct sum $Voplus V''$. $square$

Corollary. For a ring $A$, the following statements are equivalent.

(a) For any pair $(V,V')$ of $A$-modules such that $V'subseteq V$, there exists an $A$-submodule $V''$ of $V$ s.t. $V$ is the internal direct sum $V'oplus V''$.

(b) For any pair $(V,V')$ of $A$-modules such that $V'subseteq V$, there exists a surjective $A$-linear map $p:Vto V$ with $operatorname{im} p =V'$ and $p^2=p$.

If $A$ is unital, then we have another equivalent statement.

(c) Any $A$-module is semisimple.

If you are curious about the whole statement in Kassel's book, I put it below. However, I would like to note that the proposition is not true. My correction will be in bold italic font. In this setting, $A$ is an algebra over a field $k$, and an $A$-module means a left $A$ module.

Proposition I.1.2. The following statements are equivalent.

(i) For any pair $V'subseteq V$ of finite dim $A$-modules, there exists an $A$-module $V''$ s.t. $Vcong V'oplus V''$.

(ii) For any pair $V'subseteq V$ of finite dim $A$-modules where $V'$ is simple, there exists an $A$-module $V''$ such that $Vcong V'oplus V''$.

(iii) For any pair $V'subseteq V$ of finite dim $A$-modules, there exists a surjective $A$-linear map $p:Vto V'$ with $p^2=p$.

(iv) For any pair $V'subseteq V$ of finite dim $A$-modules where $V'$ is simple, there exists a surjective $A$-linear map $p:Vto V'$ with $p^2=p$.

(v) Any finite dim $A$-module is semisimple.

Note that (i) and (iii) are equivalent in the general setting as well. See the corollary after the proof of the theorem above.