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$F$ is a field.
$a(x),b(x)\in F[X]- F$.
$\gcd(a(x),b(x)) = d(x)$ and $u(x)a(x)+v(x)b(x)=d(x)$.

I need to prove that: $\deg(u)<\deg(b)-\deg(d)$.
I use the fact that degree is replacing the absolute value, but I still stuck...

Can you give a hint please? Or what is the direction of the proof?

Thank you!!

CS1
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1 Answers1

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The proof is essentially the same as for integers - descent via (euclidean) division with remainder. To compute the Bezout identity for $\,\gcd(f,g,h)\,$ note that the set $I$ of polynomials of the form $\, a f + b g +ch$ is closed under addition and scaling so it is closed under remainder = mod, since that is a composition of such operations: $f_i\bmod g_i = f_i - q\, g_i. $ It follows that the $\rm\color{#c00}{least}$ degree $d\in I$ divides every $e\in I\,$ (else $0\neq e\bmod d\,$ is in $\, I\,$ but has $\rm\color{#c00}{smaller}$ degree than $d).\,$

So $\,f,g,h\in I\Rightarrow d\,$ is a common divisor of $\,f,g,h\,$ necessarily a greatest (degree) common divisor by $\, d'\mid f,g,h\,\Rightarrow\, d'\mid d\!=\! \bar a f + \bar b g+\bar ch,\,$ so $\,\deg d'\le \deg d.\,$ To force the gcd $\,d\,$ to be unique the common convention is scale it to be monic (lead coef $=1).\,$

The extended Euclidean algorithm is an efficient way to search $I$ for a polynomial of minimal degree, while keeping track of each element's representation as a linear combination of $\,f\,$ and $\,g.$

The same idea works for any Euclidean domain (i.e. enjoying division with (smaller) remainder).

We show there exists a Bezout identity with the sought $\rm\color{#c00}{degree\ bound}$ on the coefficient $\,\color{#c00}u\,$ of $\,a.\,$ By above there is a Bezout identity $\ u' a + v'b = d.\,$ Dividing $\,u'\,$ by $\,b/d\,$ yields $\,u' = q\, b/d + u\,$ with quotient $\,q\,$ and remainder $\, u\,$ satisyfing the sought $\,\color{#c00}{\deg(u) <} \deg(b/d) = \deg(b)-\deg(d)$

Substituting $\,u' = q\, b/d + u\,$ into $\ d = u'a + v'b \ $ yields the sought Bezout identity, i.e.

$$ d = u'a+v'b = (q\, b/d + u)\,a + v'b = u\,a + (\color{#0a0}{v' + q\,a/d})\, b = \color{#c00}u\,a+\color{#0a0}v\,b \qquad\qquad $$

It's clearer via mod: $\ 1 = u' a/d + v' b/d\iff u'\equiv (a/d)^{-1}\!\pmod {\!b/d}$ and any such inverse $u'$ remains an inverse when reduced $\!\bmod b/d\,$ to $\, u:= u' \bmod b/d = $ the remainder above. So the algebra above amounts to modular reducing an inverse to force its degree less than the modulus.

Bill Dubuque
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  • What is $c$ at your answer? Thank you! – CS1 Jun 25 '19 at 16:54
  • @CS1 There $\ c\mid f,g,,$ i.e. $,c,$ is any common divisor of $,f,g,$ (we are aiming to show that every common divisor $c$ divides $d,,$ hence $d$ is a greatest common divisor). $\ \ $ – Bill Dubuque Jun 25 '19 at 16:59
  • What is $f$ here? – CS1 Jun 25 '19 at 17:33
  • OK, I'm waiting... :-) – CS1 Jun 25 '19 at 18:16
  • You right! I mix the letters, sorry: this is the right one: $\deg(u) < \deg(b) - \deg(d)$ – CS1 Jun 25 '19 at 18:23
  • I deleted them! Thank you so much! You help me a lot!! – CS1 Jun 25 '19 at 19:05
  • @CS1 Ok, I'm back and I added a simpler approach to my answer (it's equivalent to what I wrote before but doesn't require knowing the prior-linked result). – Bill Dubuque Jun 25 '19 at 19:37
  • Thank you!! Can you explain me please the first line of your edit? I don't understand it... – CS1 Jun 25 '19 at 19:41
  • @CS1 Please say more precisely where you are stuck. The idea is simply that if $,u,v,$ are Bezout coef's then so too are $\bar u,, \bar v$ where $\bar u$ has the sought minimal degree. – Bill Dubuque Jun 25 '19 at 19:44
  • I stuck here: $,\bar u = u\bmod b/d = u!-! q,b/d,,$ can explain this? how did you get it? – CS1 Jun 25 '19 at 19:44
  • Divide $,u,$ by $,b/d,$ to get $\ u = q, b/d + \bar u,\ $ with remainder $, \bar u = u\bmod b/d,$ of degree $< \deg(b/d)\ \ $ – Bill Dubuque Jun 25 '19 at 19:47
  • So $\bar u$ is the reminder? (the same $u$ at $ua+vb=d$?) – CS1 Jun 25 '19 at 20:46
  • @CS1 It's not clear what you mean by "same as". – Bill Dubuque Jun 25 '19 at 21:00
  • Sorry, I'll try to explain better: We know that $ua+vb=d$ so the $u$ is equal to: $\ u = q, b/d + \bar u$? And if yes, how exactly? – CS1 Jun 25 '19 at 21:37
  • @CS1 No, as said above we divide $,u,$ by $,b/d,$ to get the remainder $,\bar u,,$ i.e. $, \bar u = u\bmod (b/d)\ \ $ – Bill Dubuque Jun 25 '19 at 21:41
  • OK! I think I understand... But how you get it from: $ua+vb=d$? you divide it by $d$ not by $\frac{d}{b}$ (I think I miss something here...) – CS1 Jun 25 '19 at 22:18
  • @CS1 We don't use the Bezout identity to do the division. We use division with remainder to write $, u = \bar u + q, b/d,$ then use that to substitute for $,u,$ in the Bezout identity to reformulate it in terms of $\bar u\ \ $ – Bill Dubuque Jun 25 '19 at 22:29
  • So - how do you get this $u = \bar u + q, b/d$ and $v$ is the same thing but with $a$? i.e. $u = \bar u + q, b/d$ is by definition or something else? Thank you!! – CS1 Jun 25 '19 at 22:57
  • @CS1 Please point to a specific location in the answer where you are stuck. – Bill Dubuque Jun 25 '19 at 22:59
  • OK - first of all $\bar u = u\bmod b/d = u!-! q,b/d$ - you define it? because when I try to find the connection to $ua+vb=d$ and I don't understand how to connect it... – CS1 Jun 25 '19 at 23:02
  • And other thing: about - $,\deg(\bar u) < \deg(b/d) = \deg b - \deg d$, it should be $,\deg(u) < \deg(b/d) = \deg b - \deg d$, so how I convert the $\deg(\bar u)$ to $u$? – CS1 Jun 25 '19 at 23:06
  • The proof shows that given any Bezout identity we can transform it into one where the coef of $,a,$ has the sought degree (via $, u\to \bar u,$ as above). – Bill Dubuque Jun 25 '19 at 23:26
  • So I think that what I missed is how do you make this transform... This is what I don't understand... – CS1 Jun 25 '19 at 23:46
  • Maybe be problem is due to the fact that the problem is misstated. It should say that there exists some $u$ satisfying the degree bound. Is that clear? – Bill Dubuque Jun 25 '19 at 23:50
  • I'm not sure because this is what I need to prove...But you right - I think that " there exists some u satisfying the degree bound." as you write describes it well... But I still don't understand the connection between $\bar u$ and what I need to prove.. :-( When I what you wrote - I see the proof, but it's feels like the are some steps that are "missing" for me... – CS1 Jun 26 '19 at 00:10
  • @CS1 What I call $\bar u,$ is the minimal $,u,$ that you seek. What I call $,u,$ you can call something else, e.g. $,u_0,$ to get the notation you seek. – Bill Dubuque Jun 26 '19 at 00:16
  • OK, I'll try to understand if I will have any questions I'll ask you here (I think there is some mess at my head :)) – CS1 Jun 26 '19 at 07:31
  • I still don't understand this: $\bar u = u\bmod b/d = u!-! q,b/d$ it's something you define? how do you get this? If you can explain more it will be excellent!! – CS1 Jun 26 '19 at 07:44
  • @CS1 By the polynomial division algorithm we can divide $,u,$ by $,b/d,$ to get $, u = q, b/d + r,$ with quotient $,q,$ and remainder $,r,\ $ where $, \deg r < \deg b/d.\ $ We define $,\bar u,$ as the remainder $,r,\ $ which can also be written as $, \bar u = u\bmod b/d\ \ $ – Bill Dubuque Jun 26 '19 at 13:52
  • But this is the same $u$ as at the $ua$? – CS1 Jun 26 '19 at 15:27
  • @CS1 The proof in the answer shows that there exists some Bezout identity $, u,a + v,b = d.,$ We transform that into $, \bar u ,a + \bar v, b = d,$ to get a Bezout identity with the required degree bound on the coeff of $a\ \ $ – Bill Dubuque Jun 26 '19 at 15:32
  • So if I understand right you are doing "zoom-in" to $u$ at the $ua$ and call it $\bar u$ - and then you bound it? (bound the degree). – CS1 Jun 26 '19 at 15:39
  • @CS1 I rewrote the 2nd half of the answer. Maybe it will be clearer to you now. – Bill Dubuque Jun 26 '19 at 16:15
  • It's much more clear now, thank you! So you decided to to divide it by $b/d$ to get the wanted result? – CS1 Jun 26 '19 at 17:17
  • @CS1 Yes, the above proof shows that if $,u'$ is a Bezout coeff for $,a,$ then so to is $, u := u'+q,b/d,$ for any $,q.,$ We simply chose $q$ so that it leads to the lowest possible degree for $,u,,$ which occurs when $,q,$ is said quotient, so $,u,$ is the remainder $= u'\bmod b/d\ \ $ – Bill Dubuque Jun 26 '19 at 17:49
  • It would be useful to collect the information from the many comments and incorporate them into the answer. Comments are not for extended discussions. – robjohn Jun 26 '19 at 18:27
  • I read it few times and try to understand - but at your last comment you decided to divide it ($\bar u$) by $b/d$ to get the wanted result - I was wondering - you can divide it by ANYTHING and you can get any bound you want... What I miss here? – CS1 Jun 26 '19 at 21:33
  • @CS1 $,b/d,$ works because $(b/d)a = (a/d)b = {\rm lcm}(a,b),$ is a common multiple of $,a,b,$ which allows us to move the summand $,(q,b/d)a,$ from $,u'a,$ into a summand $,(q,a/d)b,$ in $,v,b.,$ Being the least common multiple it's the least degree polynomial with this property. The second approach I gave using $!\bmod,$ yields a more intuitive viewpoint. – Bill Dubuque Jun 26 '19 at 22:06