Abstract Algebra- Finding the Minimal Polynomial

In summary, the minimal polynomial of a=sqrt(5+sqrt(23)) in the field extension of C over Q is f(x)=x^4 - 10*x^2 + 2, and it can be shown that there are no monic polynomials of smaller degree which have a as a root, making it the minimal polynomial. This can be done by either using the rational roots theorem or finding all of the roots and showing that none of them are rational.
  • #1
corky23
2
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)
 
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  • #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)

Isn't f(x) a quadratic in ##y = x^2?##
 
  • #3
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:
  • #4
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.
 

1. What is the purpose of finding the minimal polynomial in abstract algebra?

The minimal polynomial is used to determine the algebraic structure of a given field or ring. It provides information about the elements that are contained within the field or ring and their relationships with one another.

2. How is the minimal polynomial different from other types of polynomial equations?

The minimal polynomial is unique in that it is the smallest degree polynomial that has a given element as a root. It is also the only polynomial that can be used to generate all other polynomials that have the same root.

3. Can the minimal polynomial be found for any element in a field or ring?

Yes, the minimal polynomial can be found for any element in a field or ring, as long as the field or ring is finite and has a finite number of elements.

4. How do you determine the degree of the minimal polynomial?

The degree of the minimal polynomial is equal to the degree of the field or ring over which it is defined. For example, if the field or ring is of degree n, then the minimal polynomial will also be of degree n.

5. Can the minimal polynomial be used to find the inverse of an element in a field or ring?

Yes, the minimal polynomial can be used to find the inverse of an element in a field or ring. By solving the minimal polynomial for the inverse root, the inverse of the given element can be determined.

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