Exercise 17, Section 4.1 of Hungerford’s Algebra

Question

(a) If $R$ has an identity and $A$ is an $R$-module, then there are submodules $B$ and $C$ of $A$ such that $B$ is unitary, $RC= 0$ and $A=B\oplus C$. [Hint: let $B=\{1_Ra\mid a\in A\}$ and $C=\{a\in A\mid 1_Ra=0\}$ and observe that for all $a\in A$, $a-1_Ra\in C$.]

(b) Let $A_1$ be another $R$-module, with $A_1=B_1\oplus C_1$ ($B_1$ unitary, $RC_1=0$). If $f: A\to A_1$ is an $R$-module homomorphism then $f(B)\subseteq B_1$ and $f(C)\subseteq C_1$.

(c) If the map $f$ of part (b) is an epimorphism [resp. isomorphism], then so are $f|_B :B \to B_1$ and $f|_C : C \to C_1$.

Question: In part (b), are we considering arbitrary $B_1$, $C_1$ which satisfies above three condition or specific $B_1=\{1_Ra_1\mid a_1\in A_1\}$, $C_1=\{a_1\in A_1\mid 1_Ra_1=0\}$? Former case seem pretty hard problem.

In part (c), I can’t prove $f|_C:C\to C_1$ is surjective. Let $c_1\in C_1$. Since $f$ is surjective, $\exists a\in A$ such that $f(a)=c_1$. So $f(1_Ra)=1_Rf(a)=1_Rc_1=0$. How to show $1_Ra=0$?

Answer 1

In b) we are indeed, a priori, considering arbitrary $B_1,C_1$. However, it's not difficult to argue that the 'arbitrary' $B_1,C_1$ must actually be $\{a\in A_1:1a=a\}$ and $\{a\in A_1:1a=0\}$ in order for $A_1=B_1\oplus C_1$ to hold as an internal direct sum. I leave this as an auxiliary exercise, I guess! Hint: there are some clear containments, and to show these are equalities you can express some element $x$ as a sum $b+c$ for $b\in B_1,c\in C_1$.

For $c)$, again I have to hint that $a=b+c$ for $b\in B,c\in C$. You know $f(1a)=0$, but you can say something about $1a$. Also, there is more than one choice of $a$. For any specific $a$, it is possible that $a\notin C$, but you can arrange for $a$ to be in $C$!

I will add more hints if you wish but this should get you going.

Answer 2

In this answer, I will fill in details in hint for (b) and (c) parts provided by user FShrike.

I initially thought $A=B\oplus C$ denote external direct sum, i.e. $A\stackrel{\phi}{\cong}B\oplus C$. For each $a\in A$, $\exists !(b,c)\in B\times C$ such that $\phi (a)=(b,c)=(b,0)+(0,c)$. So $A=\phi^{-1}(B\times \{0\})\oplus \phi^{-1}(\{0\}\times C)$ as internal direct sum. Which makes problem bit messy.

(b) We claim $\exists !$ pair $(B,C)$ of submodule of $A$ such that $B$ is unitary, $RC=0$ and $A=B\oplus C$ as internal direct sum. Let $B’,C’$ be submodule of $A$ such that $B’$ is unitary, $RC’=0$ and $A=B’\oplus C’$. We want to show $B=B’$ and $C=C’$.

Let $b\in B\subseteq A$. Since $A=B’\oplus C’$, we have $b=b’+c’$ for some $b’\in B’$, $c’\in C’$. By unitary of $B$, $B’$ and $RC’=0$, $$b=1_Rb=1_Rb’+1_Rc’=b’+0=b’\in B’.$$ Thus $B\subseteq B’$. Let $b’\in B’\subseteq A$. Since $A=B\oplus C$, we have $b’=b+c$ for some $b\in B$, $c\in C$. By unitary of $B$, $B’$ and $RC=0$, $$b’=1_Rb’=1_Rb+1_Rc=b+0=b\in B.$$ Thus $B\supseteq B’$. Hence $B=B’$.

Let $c\in C\subseteq A$. Since $A=B’\oplus C’$, we have $c=b’+c’$ for some $b’\in B’$, $c’\in C’$. By unitary of $B’$, $RC=0$ and $RC’=0$, $$0=1_Rc=1_Rb’+1_Rc’=1_Rb’+0=b’.$$ So $b’=0$ and $c=c’\in C’$. Thus $C\subseteq C’$. Similarly, $C\supseteq C’$. Hence $C=C’$.

Author have shown the existence of submodule $B$ and $C$ such that $B$ is unitary, $RC=0$ and $A=B\oplus C$. By uniqueness part, $B’=\{1_Ra\mid a\in A\}$ and $C’=\{a\in A\mid 1_Ra=0\}$.

(c) let $c_1\in C_1$. Then $\exists a\in A$ such that $f(a)=c_1$. Since $A=B\oplus C$, we have $a=b+c$ for some $b\in B$, $c\in C$. So $$f(a)=f(b+c)=f(b)+f(c)=c_1\in C_1.$$ By (b) part, $f(b)\in f(B)\subseteq B_1$ and $f(c)\in f(C)\subseteq C_1$. Since $B_1\cap C_1=0$, we have $f(b)=0$. So $f(a)=f(c)=c_1$. Thus $\exists c\in C$ such that $f(c)=c_1$. Hence $f(C)=C_1$.