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I'm a bit confused on Artin's proof of Corollary 2.3.9. After defining the $\gcd$ of integers $d = \gcd(a,b)$ by $\mathbb{Z} d = \mathbb{Z} a + \mathbb{Z} b$ and $m = \text{lcm}(a,b)$ by $\mathbb{Z} m = \mathbb{Z} a \cap \mathbb{Z} b$, he proves:

Let $d = \gcd(a,b)$ and $m = \text{lcm}(a,b)$ be the greatest common divisor and least common multiple of a pair $a,b$ of positive integers, respectively. Then $ab = dm$.

I'm going to try to paraphrase his proof in my own words. What's causing me the greatest confusion is that he seems to make use of division, which isn't defined on the integers. If $a \mid b$, it makes sense that $\frac{b}{a}$ is an integer, but I don't think I can say $\frac{a}{b} = ab^{-1}$ because $b^{-1}$ only exists in $\mathbb{Z}$ if $b = \pm 1$. b

By definition, we have $d \mid a$ and $d \mid b$, so $\frac{a}{d}$ and $\frac{b}{d}$ are integers. So we can say, in $\mathbb{Z}$, that $a$ divides the integer $\frac{ab}{d}$ and $b$ divides the integer $\frac{ab}{d}$. As $\frac{ab}{d}$ is divisible by $a$ and $b$, it is divisible by $m$, so there exists $k \in \mathbb{Z}$ such that $mk = \frac{ab}{d}$. Multiplying both sides by $d$ gives $(dm)k = ab$, so $dm \mid ab$. We will now show that $ab \mid dm$. As a consequence of the fact that $\mathbb{Z} d = \mathbb{Z} a + \mathbb{Z} b$, there exist $r,s \in \mathbb{Z}$ such that $d = ra + sb$. Multiplying by $m$, we obtain $dm = ram + sbm$. As $m$ is a multiple of $b$, we have $\frac{m}{b} \in \mathbb{Z}$, hence $\frac{rm}{b} \in \mathbb{Z}$, so $(ab) \cdot \frac{rm}{b} = ram$, so $ram$ is divisible by $ab$. Analogously, $m$ is a multiple of $a$, so $\frac{m}{a} \in \mathbb{Z}$, hence $\frac{sm}{a} \in \mathbb{Z}$, so $(ab) \cdot \frac{sm}{a} = sbm$, so $ab$ divides $sbm$. As $ab$ divides $ram$ and $sbm$, it divides their sum, so $ab$ divides $dm$. As $ab \mid dm$ and $dm \mid ab$, and $ab,dm$ are both positive, we conclude that $ab = dm$.

I think I'm ok with the mechanics, but there are several points where I use division, and even though I'm ultimately still working with integers, I'm not certain this is allowed.

Anne Bauval
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Brad G.
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    Squint your eyes and see a new variable wherever there is a fraction in the argument. – NDB Jul 14 '23 at 21:53
  • Nowhere in the proof is there any mention of $b^{-1}$ nor have I seen any number here that is outside of $\mathbb{Z}$. So what is the confusion? – bb_823 Jul 14 '23 at 21:54
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    I am surprised that Artin uses Bezout. Simpler proof. – Anne Bauval Jul 14 '23 at 22:22
  • That is proof only looks simpler by hiding things under the rug. In particular, the role of division in proving the equivalence in the middle. Euclidean division is one of the most important algorithms there are. It is good to make it explicit. – NDB Jul 14 '23 at 22:53
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    @NDB Nothing is "hidden" - the linked proof uses only very basic properties of division that are expected known at this point. It does not use "Euclidean division" (with remainder). In fact the proof works in any gcd domain (which need not be Euclidean, e.g. $,\Bbb Q[x,y]).,$ The genesis of these divisibilities is much clearer when one learns the beautiful (cofactor) duality that governs the proof (a divisibilty analog of DeMorgan's Law). Follow the links there for more on that. – Bill Dubuque Jul 15 '23 at 00:19
  • @bb_823 My understanding is that the definition of $\frac{a}{b}$ is $a \cdot b^{-1}$. I understand that we may well have $\frac{a}{b} \in \mathbb{Z}$ based on the division algorithm, but is that definition not valid in $\mathbb{Z}$? – Brad G. Jul 15 '23 at 02:25
  • @BillDubuque The double implication is not proven. Why is it a gcd domain? Explicit is better than gaslighting. – NDB Jul 15 '23 at 03:02
  • @NDB It's not clear what you claim is "not proven". Please be precise. The linked proof is correct. A gcd domain in an integral domain where all gcds exist (as you could find by searching). – Bill Dubuque Jul 15 '23 at 03:07
  • @BillDubuque It is not clear what you are claiming. Saying $\mathbb{Z}$ is a gcd domain, is not a proof. Prove it. – NDB Jul 15 '23 at 03:07
  • @NDB Comments are not the correct place to ask new (dupe) questions. – Bill Dubuque Jul 15 '23 at 03:11
  • @BillDubuque It is a rhetorical question. Nobody is asking you to prove anything. I know how to do it. – NDB Jul 15 '23 at 03:11
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    @BradG If $b\neq 0,$ then by definition $,b\mid a,$ means there is a (unique) integer $c$ such that $,bc = a.,$ When so we define the quotient $,a/b := c.,$ There is no need to introduce fractions, inverses, or (Euclidean) division-with-remainder to define such (exact) quotients. – Bill Dubuque Jul 15 '23 at 03:22

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The division here is fine considering that it is kept within $\mathbb{Z}$. As you can see every time division was performed you get an integer as a result. For example $\frac{ab}{d}$ is an integer, because $d|a$ and $d|b$.

Alternate proof

You might find this unnecessary, but here is a different and in my opinion simpler way of proving this.

Let's define two integers $a,b \in \mathbb{Z}$ as $a=p_{1}^{k_1}p_{2}^{k_2}\dots p_{n}^{k_n}$ and $b=p_{1}^{l_1}p_{2}^{l_2}\dots p_{n}^{l_n}$, where $p_1,p_2,\dots,p_n$ are primes and $k_1,\dots,k_n,l_1,\dots,l_n\in \mathbb{N}$.
We know that $$d=p_{1}^{\min(k_1,l_1)}p_{2}^{\min(k_2,l_2)}\dots p_{n}^{\min(k_n,l_n)}$$ $$m=p_{1}^{\max(k_1,l_1)}p_{2}^{\max(k_2,l_2)}\dots p_{n}^{\max(k_n,l_n)}$$ From here we get $$dm=p_{1}^{\min(k_1,l_1)}p_{2}^{\min(k_2,l_2)}\dots p_{n}^{\min(k_n,l_n)}p_{1}^{\max(k_1,l_1)}p_{2}^{\max(k_2,l_2)}\dots p_{n}^{\max(k_n,l_n)}=\\=p_{1}^{k_1+l_1}p_{2}^{k_2+l_2}\dots p_{n}^{k_n+l_n}=p_{1}^{k_1}p_{2}^{k_2}\dots p_{n}^{k_n}p_{1}^{l_1}p_{2}^{l_2}\dots p_{n}^{l_n}=ab$$ $\blacksquare$

bb_823
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