# Find the degree of the splitting field of a polynomial over $Q$ [closed]

I want to know how to determine the degree of the extension $K/\mathbb{Q}$, where $K$ is the splitting field of the polynomial $x^6+1\in \mathbb{Q}[x]$ over $\mathbb{Q}$. Do I have to get all the roots in order to find the degree?

## closed as off-topic by user26857, Claude Leibovici, Watson, choco_addicted, user91500May 8 '16 at 8:47

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The splitting field of $x^6+1$ is generated by $\;\zeta=\mathrm e^{\tfrac{\mathrm i\pi}6}$, which is a root of $x^4-x^2+1$.

This polynomial is irreducible, because it is irreducible modulo $7$: indeed it has no root in $\mathbf F_7$, since $-1$ is not a square mod $7$. It can't be factored as a product of two quadratic polynomials either, since a factorisation $$x^4-x^2+1=(x^2+ax+b)(x^2+a'x+b')$$ would lead to the system of equations: $$\begin{cases}a+a'=0\\b+b'+aa'=-1\\ab'+a'b=0\\bb'=1\end{cases}$$ whence $\;b=b'=\pm1$, $a'=-a$, $a^2=2b+1=\begin{cases}-1,\\3.\end{cases}$

Now neither $-1$ nor $3$ are squares modulo $7$. This proves the splitting field of $x^6+1$ has degree $4$.

Added: (thanks to @王李远 )

I forgot the case $a=0$. But the computation can be made directly in $\mathbf Q[x]$: $x^4-x^2+1$ has no integer (rational) roots since the only possibility would be $\pm1$. So if it factorises, it is in the form mentioned above. We obtain the same system of equations, and

• either $a=a'=0$. Then $b, b'$ satisfy the system of equations $b+b'=-1,\enspace bb'=1$. This implies $b,b'$ are roots of the quadratic equation $\;t^2+t+1=0$, which has no real root.
• or $a=-a\ne0$. Then $b=b'$ (by the 3rd eq.) $=\pm1$ by the last equation, and we deduce $a^2=-1$ or $3$, which has no rational root.
• I don't know from the system of equations,how you get b=b'=±1.maybe,it is easier to prove the polynomial is irreducible directly in Q[x] – 王李远 Apr 28 '16 at 8:45
• You're right. I forgot the case $a=0$. Sorry for that :o( – Bernard Apr 28 '16 at 9:13