A question regarding greatest common divisor

Question

This is from Artin Algebra:

I can't understand here what b does not imply c means. I will appreciate a counterexample.

Answer 1

In $\mathbb{Z}$ these are equivalent, the implication $c) \implies b)$ was done. For the reverse, consider $d \in \mathbb{Z}$ such that for all $e \mid a,b$ we have $e \mid d$. Let $e = \gcd(a, b)$. Then $e \mid a,b$ and thus $e \mid d$, i.e. $d = ce$ for some $c$. Now, by the given proposition we have $r,s$ with $e = ra +sb$ and thus $$d = ce = cra + csb = r'a + s'b$$ with $r' = cr, s' = cs$.

Now, what does he actually mean? I think he is talking about a more general context. Take a polynomial ring in two variables, i.e. $R = \mathbb{C}[X, Y]$. Let $a = X, b = Y$. Then $d = 1$ satisfies the property in b), since every polynomial dividing $a$ and $b$ is constant. But you won't find polynomials $r,s \in \mathbb{C}[X, Y]$ with $1 = rX + sY$.

Answer 2

If $d=gcd(a,b)$, then by the extended Euclidean algorithm, there exists integers $r,s$ with $d=ra+sb$. So there cannot be a counterexample for $(c)$, and all three properties are equivalent. Perhaps it is meant that (b) alone does not give such numbers $r,s$. Take $a=7$ and $b=11$. Then (b) only leaves $e=1$, but this does not help us to find $r,s$ with $1=7r+11s$. On the other hand, dropping the assumption $d=gcd(a,b)$ from the Proposition, the three properties are of course no longer equivalent.