What is the relationship between $[K:F]$ and the degree of the polynomial $f$?

  • Thread starter Thread starter Chris L T521
  • Start date Start date
Click For Summary
The relationship between the field extension degree $[K:F]$ and the polynomial degree $n$ is established through the properties of Galois extensions. Since $K$ is the splitting field of the polynomial $f \in F[x]$, it forms a Galois extension over $F$. The Galois group $G$ acts by permuting the roots of $f$, making $|G| = [K:F]$ a divisor of the order of the symmetric group $S_n$, which is $n!$. Therefore, it is proven that $[K:F]$ divides $n!$. This conclusion highlights the connection between field extensions and the factorial of the polynomial's degree.
Chris L T521
Gold Member
MHB
Messages
913
Reaction score
0
Here's this week's problem!

-----

Problem
: Let $F$ be a field, $f\in F[x]$ be a polynomial of degree $n$, and let $K$ be a splitting field of $f$. Prove that $[K:F]$ divides $n!$.

-----

Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
Physics news on Phys.org
No one answered this week's problem. You can find the solution below.

[sp]Since $K$ is a splitting field of $f\in F[x]$, $K/F$ is a Galois extension, and a Galois automorphism is determined by its action on the roots of $f$. This action can only permute the roots (since it must be an automorphism, and it must fix $f$); therefore, the Galois group $G$ is a subgroup of $S_n$ and thus $|G| = [K:F]$ divides $|S_n|=n!$.$\hspace{.25in}\blacksquare$[/sp]
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
5K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K