Really struggling with this one. I've tried using Bezout's Identity to solve this, but algebraic manipulation with the definition of divisibility gets me nowhere. I also considered that $gcd(a,b)|b$ and $gcd(b,c)|b$ implies that $gcd(a,b)gcd(b,c)|b^2$, but that didn't seem to help either since there's no good way to get rid of the square without compromising the proof. Any advice/hints for this one?
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From How to ask a good question: Your question should be clear without the title. After the title has drawn someone's attention to the question by giving a good description, its purpose is done. The title is not the first sentence of your question, so make sure that the question body does not rely on specific information in the title. – jjagmath Feb 04 '22 at 16:50
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Using the Fundamental Theorem of Arithmetic this reduces to prove:
$$\min\{\alpha,\beta\}+\min\{\beta,\gamma\} \le \beta + \min\{\alpha,\gamma\}$$
We may assume, by symmetry, that $\alpha \le \gamma$, so we are left with three cases:
- $\beta \le \alpha \le \gamma$
- $\alpha \le \beta \le \gamma$
- $\alpha \le \gamma \le \beta$
And the proof of this cases is straightforward.
jjagmath
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In terms of this notation, this takes $\alpha:=\operatorname{ord}_pa$ etc. for each $p\in\Bbb P$. – J.G. Feb 04 '22 at 17:34
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Proof $1$: $\,\ \begin{align}&\color{#c00}{(a,b)\mid a,b}\\ &\color{#0a0}{(b,c)\:\!\mid b,c}\end{align}\:\!\Rightarrow\: \overbrace{\color{#c00}{(a,b)}\color{#0a0}{(b,c)}}^{\textstyle d}\mid \color{#c00}a\color{#0a0}b,\color{#c00}b\color{#0a0}c\,\Rightarrow\, d\mid(ab,bc)=b(a,c)\ \ $
$\!\!\begin{align} {\rm Proof}\ 2\!:\ (a,b)(b,c) &=(\color{#c00}{ab},\color{#0a0}{bc},\ bb,ac) \mid \color{#c00}{ab},\color{#0a0}{bc}\, \ldots \ \text{ as above.}\\ {\bf or}\ &= (b(a,c),bb,ac)\mid b(a,c) \end{align}$
Proof $3\!:\ (a,b)(b,c) = (\color{#c00}{ab},bb,ac,\color{#0a0}{bc}) \!\supseteq\! (\color{#c00}{ab},\color{#0a0}{bc}) = b(a,c)\ \ $ [prior using principal ideals]
Bill Dubuque
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