Abstract Algebra- Finding the Minimal Polynomial

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
3 replies · 3K views
corky23
Messages
2
Reaction score
0

Homework Statement


Given field extension C of Q, Find the minimal polynomial of a=sqrt( 5 + sqrt(23) ) (element of C).

Homework Equations


The Attempt at a Solution


I may be complicating things, but let me know if you see something missing.

Doing the appropriate algebra, I manipulated the above expression into (a^2 - 5)^2=23

Expanding the left side, we get a^4 - 10*a^2 + 25 = 23 , i.e. a^4 - 10*a^2 + 2 = 0

So I plan to use f(x)=x^4 - 10*x^2 + 2

From here, I just need to show that it's irreducible.

If it is reducible, there will be either a linear factor or a quadratic factor.

My last step was simply to just use brute force to find a contradiction when comparing the following expressions with my polynomial above:

(ax+b)(cx^3 + dx^2 + ex + f) and

(ax^2+bx+c)(dx^2+ex+f)
 
Physics news on Phys.org
corky23 said:

Homework Statement


Given field extension C of Q, Find the minimal polynomial of a=sqrt( 5 + sqrt(23) ) (element of C).


Homework Equations





The Attempt at a Solution


I may be complicating things, but let me know if you see something missing.

Doing the appropriate algebra, I manipulated the above expression into (a^2 - 5)^2=23

Expanding the left side, we get a^4 - 10*a^2 + 25 = 23 , i.e. a^4 - 10*a^2 + 2 = 0

So I plan to use f(x)=x^4 - 10*x^2 + 2

From here, I just need to show that it's irreducible.

If it is reducible, there will be either a linear factor or a quadratic factor.

My last step was simply to just use brute force to find a contradiction when comparing the following expressions with my polynomial above:

(ax+b)(cx^3 + dx^2 + ex + f) and

(ax^2+bx+c)(dx^2+ex+f)

Isn't f(x) a quadratic in ##y = x^2?##
 
corky23 said:

Homework Statement


Given field extension C of Q, Find the minimal polynomial of a=sqrt( 5 + sqrt(23) ) (element of C).

Homework Equations


The Attempt at a Solution


I may be complicating things, but let me know if you see something missing.

Doing the appropriate algebra, I manipulated the above expression into (a^2 - 5)^2=23

Expanding the left side, we get a^4 - 10*a^2 + 25 = 23 , i.e. a^4 - 10*a^2 + 2 = 0

So I plan to use f(x)=x^4 - 10*x^2 + 2

From here, I just need to show that it's irreducible.

If it is reducible, there will be either a linear factor or a quadratic factor.

My last step was simply to just use brute force to find a contradiction when comparing the following expressions with my polynomial above:

(ax+b)(cx^3 + dx^2 + ex + f) and

(ax^2+bx+c)(dx^2+ex+f)

Your minimal polynomial has to be monic, let's call it f(x).

f(x) has to have ##\sqrt{ 5 + \sqrt{23} }## as a root.

You also have to show there are no monic polynomials of smaller degree which have ##\sqrt{ 5 + \sqrt{23} }## as a root.

Your polynomial f(x) is indeed monic and has ##\sqrt{ 5 + \sqrt{23} }## as a root. The last part is not about showing that f(x) is irreducible ( Because it is, it has 4 linear factors ), but you want to show there is no g(x) with deg(g(x)) ≤ 3 such that g(x) is the minimal polynomial.
 
Last edited:
You could use the rational roots theorem to exclude the case of f(x) having a linear rational factor. Or you could use Ray Vickson's hint that you can easily find all of the roots. Once you done that any quadratic factor must have the form (x-r1)(x-r2) where r1 and r2 are two of the roots. Show none of them are rational. Trying to reason from your a,b,c,d,e,f is likely to be a lot messier.