# Constructing a splitting field for polynomial over F_5

• PsychonautQQ
Not so complicated. Just remember the formula ##(x-a)(x+a)=x^2-a^2##. This gives you ##(x^2-1)(x^2+1)## in our case. The first factor splits again according to the same formula. To split the second, now write ##x^2+1=x^2-4## and consider ##2\cdot 2=4## even in ##\mathbb{F}_5##.Edit: The fourth roots of unity will work as well, but with the way above, it's easier to see, that ##i=2## in ##\mathbb{F}_5##. In summary, we are trying to show that the
PsychonautQQ

## Homework Statement

f(x) = x^7 + 3x^6 + 3x^5 - x^3 - 3x^2 - 3x where the coefficients are elements of F_5. Show that this polynomial is divisible by x^5-x and construct a splitting field L for f over F_5 and computer [L:F_5]

## The Attempt at a Solution

So the first thing I did was turn all the negative coefficients positive so that f(x) = x^7 + 3x^6 + 3x^5 + 4x^3 + 2x^2 + 2x and we want to divide this polynomial by x^5+4x. Upon doing the long division, I get a quotient x^2 + 3x + 3 with remainder 4x^3... I checked my steps and I got the same answer with the same remainder, can anyone check my work here?

##x^2+3x+3## is correct, without remainder. You don't need to convert the coefficients. The equations keep being true, even outside the standard representation of ##\mathbb{F}_5.## I only used it to simplify the roots of this last polynomial of degree ##2,## e.g. ##\frac{1}{2}=3.##
After that one can write down all ##7## roots which makes the splitting field quite obvious.

PsychonautQQ
fresh_42 said:
##x^2+3x+3## is correct, without remainder. You don't need to convert the coefficients. The equations keep being true, even outside the standard representation of ##\mathbb{F}_5.## I only used it to simplify the roots of this last polynomial of degree ##2,## e.g. ##\frac{1}{2}=3.##
After that one can write down all ##7## roots which makes the splitting field quite obvious.
Yes, excellent. And yet I am still having trouble finding all the roots. So the polynomial reduces to ##(x^2+3x+3)(x^5+4x)=(x^2+3x+3)(x^4+4)x. So the first three roots are the solutions to the quadratic and zero, are the other 4 roots c*i(4)^1/4 where c is a primitive fourth root of unity?

You can factor ##x^4+4##. It would be more obvious if you had kept the sign: ##x^4+4=x^4-1##. Now use ##x^4=(x^2)^2##.
And ##x^2+3x+3=0## can be done by the usual formula. The only critical point in the formula is ##\frac{1}{2}## but this is simply ##3##.

PsychonautQQ
fresh_42 said:
You can factor ##x^4+4##. It would be more obvious if you had kept the sign: ##x^4+4=x^4-1##. Now use ##x^4=(x^2)^2##.
And ##x^2+3x+3=0## can be done by the usual formula. The only critical point in the formula is ##\frac{1}{2}## but this is simply ##3##.

are the solutions to ##x^4-1=0## the powers of a primitive fourth root of unity? i.e. the powers of i? even in ##F_5##?

PsychonautQQ said:
are the solutions to ##x^4-1=0## the powers of a primitive fourth root of unity? i.e. the powers of i? even in ##F_5##?
Not so complicated. Just remember the formula ##(x-a)(x+a)=x^2-a^2##. This gives you ##(x^2-1)(x^2+1)## in our case. The first factor splits again according to the same formula. To split the second, now write ##x^2+1=x^2-4## and consider ##2\cdot 2=4## even in ##\mathbb{F}_5##.

Edit: The fourth roots of unity will work as well, but with the way above, it's easier to see, that ##i=2## in ##\mathbb{F}_5##.

Last edited:
PsychonautQQ

## 1. What is a splitting field for a polynomial over F_5?

A splitting field for a polynomial over F_5 is a field that contains all of the roots of the polynomial over F_5. This means that when the polynomial is factored, all of its factors are linear polynomials with coefficients from F_5.

## 2. How do you construct a splitting field for a polynomial over F_5?

To construct a splitting field for a polynomial over F_5, we first need to find all of the roots of the polynomial over F_5. Then, we can create a field extension by adjoining these roots to F_5. This new field will be the splitting field for the polynomial.

## 3. Why is constructing a splitting field important?

Constructing a splitting field is important because it allows us to find all of the roots of a polynomial over a given field. This is useful for solving equations and understanding the structure of the polynomial.

## 4. Can a splitting field be constructed for any polynomial over F_5?

Yes, a splitting field can be constructed for any polynomial over F_5. However, the degree of the splitting field may vary depending on the degree of the polynomial and the number of distinct roots it has over F_5.

## 5. How is a splitting field related to Galois theory?

A splitting field is closely related to Galois theory, as it is the smallest field extension over which a given polynomial can be fully factored into linear factors. This relates to Galois theory because it involves studying the symmetries and structures of these field extensions.

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