Finding the splitting field of x^4-7x in C over Q

[PsychonautQQ]

Homework Statement

Hello PF. I need to find a splitting field of x^4-7x in C over Q

Homework Equations

The Attempt at a Solution

letting r be a root, I did the division and got x^4-7x = (x-r)(x^3+r*x^2+x*r^2+r^3). I'm a little confused on what to do now, do I just take another root and do the division again?

 

[mfb]

You can find all the roots (in C) and see how many of them are in Q.

 

[PsychonautQQ]

You can find all the roots (in C) and see how many of them are in Q.

Is the best way to find all the roots in C to do what I've been doing? Assume an element is a root and then divide?

 

[pasmith]

Homework Statement

Hello PF. I need to find a splitting field of x^4-7x in C over Q

Homework Equations

The Attempt at a Solution

letting r be a root, I did the division and got x^4-7x = (x-r)(x^3+r*x^2+x*r^2+r^3). I'm a little confused on what to do now, do I just take another root and do the division again?

The right hand side is x^4 - r^4 which for fixed r is not identically equal to the left hand side for every x.

Starting with x^4 - 7x = (x - r)(x^3 + ax^2 + bx + c) and comparing coefficients of powers of x leads to <br /> a - r = 0, \\<br /> b - ar = 0, \\<br /> c - br = -7, \\<br /> cr = 0. This is a system in four unknowns a, b, c and r which has the solution a = r, b = r^2, c = r^3 - 7 and r(r^3 - 7) = 0. This of course gets you no closer to actually finding r.

Instead observe that x^4 - 7x = x(x^3 - 7) and then use the identity x^3 - r^3 \equiv (x - r)(x^2 + rx + r^2). That leaves you to factorize a quadratic.

 

[PsychonautQQ]

The right hand side is x^4 - r^4 which for fixed r is not identically equal to the left hand side for every x.

Starting with x^4 - 7x = (x - r)(x^3 + ax^2 + bx + c) and comparing coefficients of powers of x leads to <br /> a - r = 0, \\<br /> b - ar = 0, \\<br /> c - br = -7, \\<br /> cr = 0. This is a system in four unknowns a, b, c and r which has the solution a = r, b = r^2, c = r^3 - 7 and r(r^3 - 7) = 0. This of course gets you no closer to actually finding r.

Instead observe that x^4 - 7x = x(x^3 - 7) and then use the identity x^3 - r^3 \equiv (x - r)(x^2 + rx + r^2). That leaves you to factorize a quadratic.

Where r = (7)^1/3? Thank you by the way. I feel like this was a really obvious question in retrospect