I'm trying to answer the following question from Lang's Undergraduate Algebra:
I first attempted to prove this directly, however I failed at proving uniqueness. After my first attempted, I used an alternative, equivalent definition for the uniqueness of the sums $a= b+c$, i.e. $A_r \cap A_s = \{0\}$:
Assume $A_r \cap A_s \ne \{0\}$ then $\exists d \ne 0$ s.t. $d \in A_r \cap A_s \implies d \in A_r \: \text{and} \: d \in A_r$. We then have that $$dr = 0 \: \text{and} \: ds = 0 \implies dr + ds=0 \\ \implies dr = -(ds) \: \text{and} \: ds=-(ds)$$
Now, since inverses are unique we have $s=-r$, meaning at least one of $r,s$ is negative. This is a contradiction to our assumption that $r,s$ are positive. Thus $A_r \cap A_s = \{0\}$ and so uniqueness is satisfied.
That was the proof I came up with, but for some reason it just doesn't feel correct; I feel like I've assumed something in the "inverses are unique" step that isn't actually valid.
I was wondering if anyone can guide me as to where I have gone wrong, if I have gone wrong at all.
Let $n=rs$. Then for each element $a\in A$, $0=na = (rs)a= r(sa) = s(ra)$ and so $ra\in A_s$ and $sa\in A_r$. The sets $A_r,A_s$ are nonempty.
Suppose $a\in A_r\cap A_s$. By hypothesis, $ur+vs=1$ and so $a=(ur)a + (vs)a= u(ra)+v(sa) = 0$ as required.
Finally, $A = A_r + A_s$. For this, note that $A_r,A_s\subseteq A$ and so $A_r+A_s\subseteq A$. Conversely, write $ur+vs=1$. Then for each $a\in A$, $a=u(ra)+v(sa)$. Since $s(ra)=0$ and $r(sa)=0$, $(ur)a\in A_s$ and $(vs)a\in A_r$ as required.