Algebra by Michael Artin Prop $2.3.5$
$1.$ Please check my understanding of the $2$ proofs of $(b)$ (If an integer $e$ divides both $a$ and $b$, it also divides $d$.) presented
Proof $1$: (I believe this is the original proof in Prop 2.3.5. This uses $(c)$ and shows that $\mathbb Z d=\mathbb Z a + \mathbb Z b$ implies $\mathbb Z (ra+sb) \subseteq \mathbb Z e$.)
By $(c)$, there are integers $r, s$ s.t. $d=ra+sb$. Then $\frac d e = r \frac a e + s \frac b e$. Now suppose that $e$ divides both $a$ and $b$. By definition $\frac a e$ and $\frac b e$, are integers i.e. $a,b \in \mathbb Z e$. Then $\frac d e = r \frac a e + s \frac b e$ is an integer, i.e. $ra+sb = d \in \mathbb Z e$. Finally, $\mathbb Z d$ is the subgroup of $\mathbb Z$ generated by $d$, so $\mathbb Z d = \mathbb Z (ra+sb) \subseteq \mathbb Z e$.
QED
Proof $2$: (I believe this is the proof after Prop $2.3.5$. This uses $(c)$ and the note after Prop 2.3.5, shows that $\mathbb Z d=\mathbb Z a + \mathbb Z b$ implies $\mathbb Z (ma+nb) \subseteq \mathbb Z e$ and shows another way how $(c)$ implies $(b)$.)
Suppose $e$ divides both $a$ and $b$. We must show $e$ divides $d$, i.e. $d \in \mathbb Z e$ i.e. $\mathbb Z d \subseteq \mathbb Z e$.
Therefore, by the note, $e$ divides both $ma+nb$ for any integers $m, n$ including the integers $r, s$ s.t. $d=ra+sb$ by $(c)$. Finally, for any $m, n$, we have that $\mathbb Z d = \mathbb Z (ra+sb) \subseteq \mathbb Z (ma+nb) \subseteq \mathbb Z e$.
QED
$2.$ Is there a way we can prove $\mathbb Z d=\mathbb Z a + \mathbb Z b \implies (b)$ without using both $(c)$ and the note besides Proof 1, preferably without using either?
Best Answer
Proof 3: (This uses the note after Prop 2.3.5 but not $(c)$.)
$a,b \in \mathbb Z e$
$\implies \mathbb Z a + \mathbb Z b \subseteq \mathbb Z e$
$\implies \mathbb Z d = \mathbb Z a + \mathbb Z b \subseteq \mathbb Z e$
QED
The first implication is actually the note.