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This is from Artin Algebra:

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

Silent
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  • please don't post pictures instead of text. – miracle173 Jan 21 '18 at 16:57
  • I think this is meant in the more general context of rings. Using only the ring-axioms we don't have the implication. In $\mathbb{Z}$ they are equivalent. –  Jan 21 '18 at 17:10
  • The point is that in doing the proof they did $c$ is given; $c \implies a$ and $c\implies b$. They are noting don't attempt to do $d$ generates S implies b) and b) implies c). Because b) doesn't imply c). – fleablood Jan 21 '18 at 17:16
  • Ah, crap... You're right. I had it backwards. And Paul K is correct. I think. – fleablood Jan 21 '18 at 17:17

2 Answers2

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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.

Dietrich Burde
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    $1 = -1 \cdot 7 + 1 \cdot 8$? –  Jan 21 '18 at 16:50
  • @PaulK Yes, but how did $(b)$ help you there? – Dietrich Burde Jan 21 '18 at 16:53
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    $1 = 8 \cdot 7 + (-5) \cdot 11$

    You won't find integers $a,b,d$ which satisfy $e \mid a,b \implies e \mid d$ for all $e \in \mathbb{Z}$ and simultaneously not being represented as an integral linear comnination of $a$ and $b$.

     –  Jan 21 '18 at 17:08
  • Yes, sure, I know. This was not my point. – Dietrich Burde Jan 21 '18 at 17:15
  • @PaulK, so you are saying that in $\Bbb Z$, (b) and (c) are equivalent, even when we take $d$ is not to be $\gcd(a,b)$? – Silent Jan 21 '18 at 17:29
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    Oh, I see! if $d$ satisfies (b), then $e \mid a,b \implies e \mid d$ for all $e\in \Bbb Z$, so $d=k\gcd(a,b)$, hence if $\gcd(a,b)=ra+sb$, then $d=kra+ksb$. Am I right @PaulK? – Silent Jan 21 '18 at 17:35
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    Yes, i just wrote it as an answer. –  Jan 21 '18 at 17:36
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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 $1 = rX + sY$.

  • Your interpretation of $d$ in para one seems fishy, Artin uses $d$ in both b and c as $\gcd(a,b)$, but, you use that for multiple of gcd as well! – Silent Jan 24 '19 at 17:42