GCD is same in a field and its superfield.

  • Context: MHB 
  • Thread starter Thread starter caffeinemachine
  • Start date Start date
  • Tags Tags
    Field Gcd
Click For Summary
SUMMARY

The monic greatest common divisor (GCD) of two polynomials $p(t)$ and $q(t)$ in a field $F[t]$ is identical to the monic GCD of the same polynomials in an extension field $K$. This conclusion is reached by assuming the existence of a non-trivial common factor in $K[t]$ while $p(t)$ and $q(t)$ lack such a factor in $F[t]$. The contradiction arising from the existence of coefficients $a, b \in F[t]$ satisfying $pa + qb = 1$ confirms that the GCD remains unchanged across the fields.

PREREQUISITES
  • Understanding of polynomial rings, specifically $F[t]$ and $K[t]$.
  • Knowledge of the concept of greatest common divisors in the context of polynomials.
  • Familiarity with field extensions and their properties.
  • Basic algebraic manipulation skills involving polynomials.
NEXT STEPS
  • Study the properties of polynomial rings over fields, focusing on $F[t]$ and $K[t]$.
  • Learn about the Euclidean algorithm for finding GCDs of polynomials.
  • Explore the implications of field extensions on algebraic structures.
  • Investigate examples of non-trivial common factors in polynomial rings.
USEFUL FOR

Mathematicians, algebraists, and students studying abstract algebra, particularly those interested in polynomial theory and field extensions.

caffeinemachine
Gold Member
MHB
Messages
799
Reaction score
15
Let $K$ be an extension field of a field $F$ and let $p(t),q(t)\in F[t]$. Show that the monic greatest common divisors of $p(t)$ and $q(t)$ in $F[t]$ is same as the monic greatest common divisor of $p(t)$ and $q(t)$ in $K$.
 
Physics news on Phys.org
Hint:

Let $p(t)$ and $q(t)$ have a non trivial common factor in $K[t]$. Assume that $p$ and $q$ don't have a non-trivial common factor in
$F[t]$. Then there exist $a,b\in F[t]$ such that $pa+qb=1$. But this contradicts the fact that $p$ and $q$ have a non-trivial common factor in $K[t]$.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 19 ·
Replies
19
Views
5K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 2 ·
Replies
2
Views
677
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 24 ·
Replies
24
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K