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 $$a - r = 0, \\<br /> b - ar = 0, \\<br /> c - br = -7, \\<br /> cr = 0.$$ 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 $$a - r = 0, \\<br /> b - ar = 0, \\<br /> c - br = -7, \\<br /> cr = 0.$$ 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