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If $(a,b)=1$ then if $a\mid bx$, why is it necessary that $a\mid x$?

I have seen such claims and I can't find the answer because it is usually used as a little remark, but I am having a hard time understanding why.

I would appreciate your help.

Aaron Maroja
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Meitar Abarbanel
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  • This is known as Gauß's lemma (although it was known before Gauß — actually, he proved an analog for the coefficients of aproduct of polynomials). – Bernard Mar 23 '15 at 19:49
  • @Bernard In English is almost always called Euclid's Lemma. – Bill Dubuque Mar 23 '15 at 19:59
  • Euclid's lemma, strictly speaking, is for prime divisors. – Bernard Mar 23 '15 at 20:03
  • @Bernard Whether or not it is historically precise, the name "Euclid's Lemma" is widely applied to the non-prime case too. Indeed, the linked Wikipedia article states this "generalization is also called Euclid's lemma". My experience as a number theorist agrees with that remark. Only very rarely have I seen it called Gauss's Lemma (that name is already far too overloaded). – Bill Dubuque Mar 23 '15 at 20:27
  • It's used in France… That's the way I learnt it in high school. – Bernard Mar 23 '15 at 20:29

1 Answers1

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If $a,b$ are relatively prime, there are integers $u,v$ such that $au+bv=1$. (This is called Bézout's identity.)

Multiply by $x$ this means $a(xu) + (bx)v = x$.

Since $a\mid a(xu)$ and $a\mid bx$, we know that $a\mid x$.

Thomas Andrews
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  • @Aaron There do exist non-PID Bezout domains. But Euclid's Lemma holds more generally in any gcd domain, see here. – Bill Dubuque Mar 23 '15 at 20:05
  • I think is a particular case of my answer. But people thought it was not useful, then it's okay, Andrew's answer might be the easier way to go. – Aaron Maroja Mar 23 '15 at 20:07
  • @AaronMaroja As I said I comments, the OPs question is usually a lemma on the way to proving the integers are a UFD, so using that the integers are a UFD was premature and unhelpful. (Unhelpful because most people learning this theorem don't even know what UFD stands for.) And no, this is not a 'particular case' of your answer, at least, not in the case of "special case." – Thomas Andrews Mar 23 '15 at 20:27
  • I didn't see this as a lemma to proof anything, in fact, as a corollary from a theorem involving gcd. As for this construction, first I saw the integers as a principal domain and then using an infinite chain of ideals and induction we proved that a principal domain is a UFD. It all depends on how you approach. Thus it may not be considered "premature", since I've seen this as an exercise holding the hypothesis of $A$ being an UFD. – Aaron Maroja Mar 23 '15 at 20:45
  • Arnold Ross used to call this lemma the "real" Fundamental Theorem of Arithmetic. If you are talking ideals, then you are talking already more advanced than most intro number theory questions. Usually, we talk about the integers first, and properties of them, before generalizing. @AaronMaroja – Thomas Andrews Mar 23 '15 at 20:56
  • @ThomasAndrews okay. – Aaron Maroja Mar 23 '15 at 21:17