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Can anyone help me understand the following proof that if $p|ab$ then $p|a$ or $p|b$? This proof is on a separate question.

Suppose there were a counterexample, with $pa=bc$, $p$ a prime, but neither $b$ nor $c$ divisible by $p$. Then there would be a counterexample with $p$ as small as possible and, for that $p$, $b$ as small as possible. Note that $b>1$, since otherwise we would have $pa=c$, which means $p$ divides $c$.

We first note that $b<p$, since otherwise $pa′=p(a−c)=(b−p)c=b′c$ would be a smaller counterexample. But now $b>1$ implies $b$ is divisible by some prime $q$, which means we have $q$ dividing pa with $q≤b<p$. By the minimality of $p$ as a counterexample, we conclude that $q$ divides $a$ (since it can't divide $p$). If we now write $a=a′q$ and $b=b′q$ and note that $b′<b<p$ implies $p$ doesn't divide $b′$ either, we find that $pa′=b′c$ is a smaller counterexample, which is a contradiction. Thus there can be no counterexample.

I am having trouble understanding how this proves anything. Especially this part:

$pa′=p(a−c)=(b−p)c=b′c$

What is the reasoning behind subtracting $c$ and $p$ from the factors? Would someone be willing to go through this proof step by step and explain why it works?

Question: Proof of Euclid's Lemma

Bill Dubuque
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Michael Munta
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    $p\mid ab\Longrightarrow (p\mid a)\vee (p\mid b)$ is the definition of prime. – Jack D'Aurizio Oct 27 '18 at 14:12
  • Since $pa=bc$ we have $p(a-c)=pa-pc=bc-pc=(b-p)c$. If $b>p$ then this would be a smaller counterexample. Is that what you were asking? – lulu Oct 27 '18 at 14:12
  • (a) If there is a counterexample, there is a smallest; (b) if there is a counterexample there is a smaller one - therefore there is no smallest. Therefore there is no counterexample. – Mark Bennet Oct 27 '18 at 14:31
  • @JackD'Aurizio I agree. But "prime" is used in a lot of elementary work to mean "irreducible", and only later are the two properties properly distinguished. So it is not surprising to see the term "prime" being used loosely in this context. – Mark Bennet Oct 27 '18 at 14:34
  • @JackD'Aurizio Yes it is a definition, but it still requires proof. Usually it is proven by Bezout's theorem, but I am more interested in this particular proof. – Michael Munta Oct 27 '18 at 14:36
  • @lulu Yes, I am having troubles with that. How does that contradict the fact that $b<p$? – Michael Munta Oct 27 '18 at 14:38
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    Well, if $b>p$ then $b>b-p>0$. we see that $p,|,(b-p)c$ so either $p,|,c$ or $p,|,b-p$. But $p$ can't divide $c$ by the original assumption on $bc$, so $p$ must divide $b-p$. But in that case it is easy to see that $p,|,b$, which again contradicts the assumption. – lulu Oct 27 '18 at 14:40
  • @MichaelMunta: no, definitions do not require proofs. If you take $d(p)=2$ as the definition of "$p$ is a prime" then $p\mid ab\Longrightarrow (p\mid a)\vee(p\mid b)$ has to be proved, but if you take $(p>1)\wedge(p\mid ab\Longrightarrow (p\mid a)\vee(p\mid b))$ as the definition of prime there is nothing to prove. – Jack D'Aurizio Oct 27 '18 at 14:45
  • @JackD'Aurizio But this is Euclid's Lemma and it requires a proof, doesn't it? – Michael Munta Oct 27 '18 at 14:59
  • @MichaelMunta: If your definition of prime is $d(p)=2$, but that is not stated anywhere. – Jack D'Aurizio Oct 27 '18 at 15:14
  • @JackD'Aurizio Don't know what $d(p) = 2$ means, but I gave a link to the original question so I was not assuming anything in my question. – Michael Munta Oct 27 '18 at 15:52
  • $d(p)=2$ means that $p$ has exactly $2$ positive integer divisors. In other terms: what is your definition of prime? – Jack D'Aurizio Oct 27 '18 at 16:04
  • @JackD'Aurizio I'm quite puzzled by your comments. Surely you must know that "prime" means "irreducible" for naturals in most elementary number theory courses. Further there can be no doubt that this is the definition used in the question else there would be nothing to prove! – Bill Dubuque Oct 27 '18 at 16:58
  • @Michael When one studies more general number systems it is convenient to distinguish between the properties of being irreducible and the prime divisor property given by Euclid's Lemma, because they are generally no longer equivalent. Non-invertible elements satisfying the latter property are called primes. Primes are always irreducible, but the converse is true iff factorizations into irreducibles are unique, so it is a rather strong assumption.(above I presume our "numbers" are a commutative integral domain) – Bill Dubuque Oct 27 '18 at 17:20
  • @lulu So this $a-c$ part is just a consequence of subtracting $p$ from $b$ looking at this equation $pa=bc$? I was wondering what would be the reasoning to subtract $c$ from $a$. – Michael Munta Oct 29 '18 at 13:21

2 Answers2

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These "direct" proofs of Euclid's Lemma achieve descent via (Euclidean) division with remainder, i.e. we use division to reduce to a smaller instance of the claim, then apply (complete) induction.

The first reduction step replaces any $\,b> p\,$ by a smaller $\,b'\equiv b\pmod{\!p},\,$ which doesn't alter the truth of the statement since $\,p\mid bc\iff p\mid b'c,\,$ and we still have$\,(b',p) = (b,p) = 1$. The OP chooses $\,b' = b-p,\,$ but we could also choose $\,b' = b\bmod p < p\,$ as in the equivalent proof you posted a few days ago.

By the above step(s) we reduce to the case $1 < b < p.\,$ We don't need prime factorizations for descent in this second step. Instead it is more constructive is to replace $\,b\,$ by its smaller remainder $\,p\bmod b = p - qb.\,$

Combining the above two descent steps yield the following variant of the Euclidean algorithm, which applies when one argument is a prime $p\,$ (and $\,p\nmid b)$

$$\begin{align} &(b,p) = (b\bmod p,\,p)\ \ {\rm if}\ \ b > p\ \ \ \ \ \ [\![1]\!]\\[.3em] &(b,p) = (p\bmod b,\,p)\ \ {\rm if}\ \ b < p\ \ \ \ \ \ [\![2]\!]\end{align}$$

This form of the proof essentially uses $\,p\mid bc\,\Rightarrow\, p\mid(p,b)c = c\,$ by $\,(p,b) = 1,\,$ while using the above two descent steps to iteratively calculate the gcd $(p,b) = 1.\,$ Here is a simple example.

$$\begin{align} &31\mid 38c\\ \Rightarrow\ &\color{#c00}{31\mid 7}c\ \ \ {\rm by}\ \ \ 7 \,=\, 38\bmod 31\ \ \&\ \ \ [\![1]\!]\\ \Rightarrow\ &\color{#c00}{31\mid 3}c\ \ \ {\rm by}\ \ \ 3 \,=\, 31\bmod 7\ \ \ \ \&\ \ \ [\![2]\!]\\ \Rightarrow\ &31\mid 1c\ \ \ {\rm by}\ \ \ 1 \,=\, 31\bmod 3\ \ \ \ \&\ \ \ [\![2]\!] \end{align}\quad\ \ $$

which essentially inlines the following gcd calculation using $[\![1]\!]$ and $[\![2]\!]$

$$(31,38) \,\overset{[\![1]\!]}=\, \color{#c00}{(31,7)\,\overset{[\![2]\!]}=\, (31,3)}\,\overset{[\![2]\!]}=\, (31,1)=1$$

the relation being: $\,31\mid 7c\!\iff\! 31\mid \color{#c00}{(31,7)}c = \color{#c00}{(31,3)}c\!\iff\! 31\mid 3c$

Eliminating the (unneeded) contradictive form and viewing it positively leads to Gauss's algorithm for computing inverses and fractions $\!\bmod p$

See also this closely related proof.

Bill Dubuque
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  • Why were you able to replace $b$ by $p$ $mod$ $b$? – Michael Munta Oct 30 '18 at 21:21
  • @Michael That's explained in the 2nd sentence "which doesn't alter...". Ditto for reducing $b$ to $b-p$ (note $b\bmod p = b-kp$ can be viewed as an iteration of that till we reach the remainder $r = b\bmod $p).\ \ $ – Bill Dubuque Oct 30 '18 at 21:55
  • can you please help me with understanding your version of this proof. I understand now how it works with prime factorization, but I am not sure why your proof works. Via email or edit to your answer? – Michael Munta Dec 25 '18 at 16:20
  • @Michael Please tell me which parts are still not clear to you. – Bill Dubuque Dec 25 '18 at 16:24
  • If it is possible I would like that you go through everything in more detail. You say that it achieves descent via division algorithm so could you please go through it in full. Thank you – Michael Munta Dec 25 '18 at 16:59
  • @Michael The key idea is that $,p\mid bc\Rightarrow p\mid jbc!+!kpc = (\color{#c00}{jb!+!kp})c.,$ If $,b>1,$ can choose $,j,k,$ to make $ \color{#c00}{jb!+!kp} < b,$ so reducing to a smaller problem, e.g. doing it quickly in one-step by Bezout we can choose $,j,k,$ so $,\color{#c00}{jb!+!kp} = \gcd(b,p) = \color{#c00}1,$ so $,p\mid \color{#c00}1 c$ and we're done. Alternatively we can do it slower in a few steps by choosing $,j,k,$ to be $,b\bmod p,$ or $,p\bmod b,$ analogously to the $\bmod$ reduction steps in the Euclidean algorithm for $,\gcd(b,p).$ – Bill Dubuque Dec 25 '18 at 17:00
  • @Michael I added a concrete example to the answer. – Bill Dubuque Dec 25 '18 at 17:09
  • Can you use the same variable letters as the proof in my question just so I can connect everything more easily? – Michael Munta Dec 25 '18 at 17:09
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    @Michael Your quoted proof uses the above Euclidean descent only in the first half - to reduce to the case $,b < p, $ (essentially by repeatedly subtracting $,p,$ from $b$, i.e. by replacing $,b,$ by $,b\bmod p).,$ After that it uses prime factorization. But as I show above we can continue to use the Euclidean descent in the case $,b < p.\ $ – Bill Dubuque Dec 25 '18 at 17:20
  • can you please also add a concrete example for your comment with $j$ and $k$ variables? Just so I have all bases covered. Thank you – Michael Munta Dec 28 '18 at 15:46
  • @Michael That was already done in my 3rd last comment (it can be more concrete with symbolic $j,k)\ \ $ – Bill Dubuque Dec 28 '18 at 18:15
  • I don't know how $jbc + kpc$ got there so I want to understand that part – Michael Munta Dec 28 '18 at 19:17
  • @MichaelMunta There $,j,k,$ are arbitrary integers. Since $p$ divides $bc$ it also divides its multiple $jbc$ etc. – Bill Dubuque Dec 28 '18 at 19:24
  • So then essentially your version is nothing different from the proof that uses Bezout identity to see that $kp + jb = 1$ and then $kpc + jbc = (kp + jb)c = c$. You just use the longer method of descent to get to $1$. – Michael Munta Dec 28 '18 at 21:12
  • @MichaelMunta Right, that's how these "direct" proofs work for Euclid's Lemma, unique factorization etc. They eliminate higher-level conceptual structure by replacing it with lower-level equivalents (akin to compiling a high-level programming language into lower-level assembly language). Follow the links I gave here on Zermelo's proof for further discussion on that. – Bill Dubuque Dec 28 '18 at 21:30
  • I have one other question. You say that this is a 'variant' of Euclidean algorithm. In finding $gcd(p,b)$ you keep $p$ the same and only replace $b$ by the remainders, you don't actually swap places of the numbers like in the normal case of Euclid algorithm. Is this something that can be generalized? Does it have a name? – Michael Munta Dec 30 '18 at 19:29
  • @Michael It works because $p$ is prime, see this answer on Gauss's algorithm. – Bill Dubuque Dec 30 '18 at 19:46
  • Ok, I checked the link and I would like to add that $(a,p)->(p$ mod $a, p)$ iterations only work when $p > a$. Am I right? $38$ and $31$ are an example of this. $38$ needs to be less than $31$ to start iterating in the way you have shown. – Michael Munta Dec 31 '18 at 08:29
  • @MichaelMunta This is already in the answer above - where the descent steps depend on $,b > p,$ or $,b < p.,$ The first step reduces to $, b < p,$ and subsequent steps decrease $b$ if $,b > 1.,$ It must terminate with $,b = 1,$ (else the final positive $b$ would be a proper divisor of the prime $p)\ $ – Bill Dubuque Dec 31 '18 at 15:39
  • I meant it more in the context on your Gauss's algorithm answer because you have not specified anywhere that requirement is $p > a$. Otherwise after the first step you would end up with $gcd(p, p)$. So to be able to iterate like $(a,p)->(p$ mod $a,p)$ you first need to reduce $a$. Am I right? – Michael Munta Dec 31 '18 at 15:44
  • @Micheal That's because the cited theorem of Gauss (art.13) only concerns the case where $, a < p.,$ The reduction step when $,a > p,$ is in Gauss's next result (art.14) – Bill Dubuque Dec 31 '18 at 16:02
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    @Michael You can read Gauss's Disq. Artth. proof in Google Books. Art.13 p. 5 employs the descent $,p\mid ab,\Rightarrow, p\mid a(p\bmod b)$ to decrease $b$ when $,1<b<p.\ $ – Bill Dubuque Dec 31 '18 at 16:14
  • Thanks for the reference. Can you explain to me what is a descending chain of multiples of $a$ mod $p$? – Michael Munta Dec 31 '18 at 16:19
  • @Michael Gauss's proof shows $p$ divides all $,ab > ab'> ab'' > \ldots > a,,$ which is a descending chain of multiples of $,a.\ \ $ – Bill Dubuque Dec 31 '18 at 16:26
  • What are these numbers in the example you have given to the answer? With 38 and 31? – Michael Munta Dec 31 '18 at 16:27
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    @Michael $\ 31\ $ divides each of $\ 7c > 3c > c\ \ $ – Bill Dubuque Dec 31 '18 at 16:29
  • I think I understand everything now, even though Gauss proves different things and 'algorithm' is very implicit, but what happens underneath is exactly what you have specified in your answer. – Michael Munta Jan 01 '19 at 09:58
  • In your answer with the fraction presentation of Gauss's algorithm you are scaling and reducing fractions until denominator reaches $1$. How does that process connect to the article in Gauss's book? He does nothing with fractions there. – Michael Munta Jan 01 '19 at 19:51
  • @Michael It's a fractional reformulation of the descent used in Gauss's proof. – Bill Dubuque Jan 01 '19 at 20:03
  • Would you be willing to explain how this fraction reformulation really is equivalent to Gauss's proof? Because I can not see the connection. – Michael Munta Jan 01 '19 at 20:32
  • @Michael The fraction form solves $\ bx\equiv a\ $ by continually scaling it in a way that reduces $,b,$ till it is $1$ (assuming $,b\not\equiv 0).,$ Coalescing all these scalings amounts to a single scaling by $,b^{-1}.,$ The proof in Disq. Arith is just the special case when $,a\equiv 0,,$ i.e. $\ bx\equiv 0,\Rightarrow, x\equiv b^{-1}0\equiv 0.\ $ – Bill Dubuque Jan 01 '19 at 21:02
  • Unfortunately I am still not seeing it. I will open another question where you or someone else can go into more detail on this. – Michael Munta Jan 02 '19 at 08:57
  • Bill, I added a new question. Please follow up on it. https://math.stackexchange.com/questions/3059260/gausss-algorithm-fraction-form – Michael Munta Jan 02 '19 at 16:09
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First we want to show that if b is as small as possible, then necessarily b < p. If not, what we can do is subtract c from the factor of a, and get a smaller b$^{\,\prim4} that satisfies out condition.

$$ pa^{\,\prime} = p(a-c) = pa - pc = bc - ac = (b-a)c = b^{\,\prime}c$$.

Thus we can always reduce our value b until b < p. With that in mind, we proceed : We find a prime factor of b, call it q. So q divides bc = pa. q cannot divide p, so p divides a. Since p divides both a and b, we can rewrite our initial equation to get a counter example with values smaller than we had. This is the contradiction that such a prime p exists.

Joel Pereira
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