The question is Q4.1.17(a) in Hungerford's Algebra:

If $R$ has an identity and $A$ is an $R$-module, then there are submodules $B$ and $C$ of $A$ such thatt $B$ is unitary $RC = 0$ and $A = B \oplus C.$

[Hint: let $B = \{ 1_{R}a | a \in A \}$]

But I can not see why $1_{R}a$ will be equal a and hence $B$ will be unitary or is it because of that $R$ has identity (a very trivial justification ), could anyone clarify this for me please?

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    $\begingroup$ I think $1_Ra=a$ is an axiom $\endgroup$ – JoseCruz Nov 30 '17 at 23:31
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    $\begingroup$ It is not an axiom in Hungerford. $\endgroup$ – Randall Dec 1 '17 at 0:25
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    $\begingroup$ yes it is not an axiom in Hungerford Algebra @JoseCruz $\endgroup$ – user426277 Dec 1 '17 at 1:59

Even if

$1_R a \ne a, \tag 1$

we certainly have

$1_R (1_R a) = (1_R 1_R) a = 1_R a, \tag 2$

which shows that $B = 1_R A$ is unitary.

Now set

$C = \{ a - 1_R a \mid a \in A\}; \tag 3$

then for $r \in R, a \in A$ we have

$r(a - 1_R a) = ra - r 1_R a = ra - ra = 0, \tag 4$

which shows that

$RC = \{ 0 \}; \tag 5$

note that any $a \in A$ satisfies

$a = 1_R a + (a - 1_R a) \in B + C; \tag 6$

if $b \in B \cap C$, then

$b = 1_Ra \tag 7$

for some $a \in A$, since $b \in B$; since $b \in C$,

$1_R b = 0; \tag 8$

from (6) and (7) we have

$b = 1_R a = (1_R 1_R)a = 1_R (1_R a) = 1_R b = 0. \tag 9$

(6) and (9) together show that

$A = B \oplus C. \tag{10}$

Of course, one should technically show that $B$ and $C$ are submodules of $A$, but I leave this easy and obvious verification to my readers.

  • $\begingroup$ Why you set $C$ in a different form than what is mentioned in the hint? $\endgroup$ – user426277 Dec 1 '17 at 6:49
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    $\begingroup$ All the hint says is that there exists a direct summand $C$ of $A$ such that $RC = \{0\}$. I chose $C$ the way I did to take those parts of $a \in A$ which are not in $B = 1_RA$. $\endgroup$ – Robert Lewis Dec 1 '17 at 6:54
  • $\begingroup$ In identity (4) from where did you get that $r1_{R}a = ra$? $\endgroup$ – user426277 Dec 1 '17 at 9:32
  • $\begingroup$ Identity(7) is not clear for me, could you please clarify it for me? $\endgroup$ – user426277 Dec 1 '17 at 10:33
  • $\begingroup$ Since we assumed right above (7), in the text, that $b \in B \cap C$, we have $b \in B$. But we have previously defined $B$ to be $1_R A$, so $b \in B \Longrightarrow \exists a \in A \mid b = 1_R a$. (7) comes from the definition of $B$. $\endgroup$ – Robert Lewis Dec 1 '17 at 17:04

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