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Let $F$ be a field and suppose that $d(x)$ is a greatest common divisor of two polynomials $p(x)$ and $q(x)$ in $F[x]$. Then there exist polynomials $r(x)$ and $s(x)$ such that $d(x)=r(x)p(x)+s(x)q(x).$ Furthermore, the greatest common divisor of two polynomials is unique.

🔗 Let $d(x)$ be the monic polynomial of smallest degree in the set

$S =\{f(x)p(x)+g(x)q(x):f(x),g(x)∈F\}$. We can write $d(x)=r(x)p(x)+s(x)q(x)$ for two polynomials $r(x)$ and $s(x)$ in $F[x]$. We need to show that $d(x)$ divides both $p(x)$ and $q(x)$. We shall first show that $d(x)$ divides $p(x)$. By the division algorithm, there exist polynomials $a(x)$ and $b(x)$ such that $p(x)=a(x)d(x)+b(x)$, where $b(x)$ is either the zero polynomial or deg $b(x)$ < deg $d(x)$. Therefore,

$b(x)$ $=p(x)−a(x)d(x)$ $=p(x)−a(x)(r(x)p(x)+s(x)q(x))$ $=p(x)−a(x)r(x)p(x)−a(x)s(x)q(x)$ $=p(x)(1−a(x)r(x))+q(x)(−a(x)s(x))$

is a linear combination of $p(x)$ and $q(x)$ and therefore must be in $S$ . However, $b(x)$ must be the zero polynomial, since $d(x)$ was chosen to be of smallest degree; consequently, $d(x)$ divides $p(x)$. A symmetric argument shows that $d(x)$ must also divide $q(x)$; hence, $d(x)$ is a common divisor of $p(x)$ and $q(x)$.

My questions are

  1. Why does $d(x)$ need to be a monic polynomial?

  2. The theorem says that gcd is monic but that is not the case with all polynomials!?

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    $\begingroup$ Factorization of integers is unique up to factors of units $\pm1$, so we choose positive integers as representatives; analogously, in the ring of polynomials over a field, all non-zero constant polynomials are units (they are invertible, as elements of the field), so we choose monic polynomials as representatives $\endgroup$ Aug 2, 2020 at 18:38
  • $\begingroup$ For example, over $\mathbb Q$, we could have $x-2,$ $2x-4$, $\frac12x-1$, etc.; cf. this paragraph in Wikipedia $\endgroup$ Aug 2, 2020 at 18:46

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If $d(x)$ were not constrained to be monic, then the gcd of $p(x),q(x)\in F[x]$ would not be unique; we could multiply $d(x)$ by any non-zero element of $F$, and it would still be a greatest common divisor of $p(x)$ and $g(x)$. On the other hand, given any greatest common divisor of $p(x)$ and $g(x)$, we can multiply it by the inverse of the leading coefficient -- note that all non-zero elements of a field are invertible -- to get a monic greatest common divisor. For this reason, the monic polynomial g.c.d. is taken as the greatest common divisor, just as for integers the positive greatest common divisor is taken as the g.c.d., even though its opposite also divides both integers and all common divisors of both integers also divide its opposite.

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    $\begingroup$ Thank you very much Mr. Tanner that did clear up alot. $\endgroup$ Aug 2, 2020 at 19:08
  • $\begingroup$ I just have one more question consider the two polynomials (2x^2+5)(x^4+4) and (2x^2+5)(x^5+2) have gcd (2x^2+5) which is not monic but according to the theorem they should be monic....am i missing out something over here $\endgroup$ Aug 2, 2020 at 19:27
  • $\begingroup$ I would call it $x^2+\frac52$ ;-) $\endgroup$ Aug 2, 2020 at 19:39

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