# Find a normal extension over $\mathbb{Q}$ of degree 3

I've had trouble coming up with one.

I know that if I could find

an irreducible poly $p(x)$ over $\mathbb{Q}$ which has roots $\alpha, \beta, \gamma\in Q(\alpha)$,

then $|\mathbb{Q}(\alpha) : \mathbb{Q}|$ = 3 and would be a normal extension, as $\mathbb{Q}(\alpha)=\mathbb{Q}(\alpha,\beta,\gamma)$ would be a splitting field of $f$ over $\mathbb{Q}$.

However, this is a lot of conditions to find by luck...

Any help appreciated!

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Well, it's not that bad. The splitting field of an irreducible degree-3 polynomial is degree either 3 or 6. – Ben Millwood Dec 10 '12 at 22:20

## 3 Answers

Try to find a polynomial with discriminant $D$ that satisfies $\sqrt{D}\in\mathbb{Q}$.

Why does this help?

First, the only possibilities for the Galois group $G$ are $S_3$ and $A_3$, as Ben Millwood remarked.

Second, every element of $G$ must fix $\sqrt{D}\in\mathbb{Q}$. But you can check that every transposition of two roots of $f$ does not fix $\sqrt{D}$. Therefore $G$ cannot contain any transposition and must be isomorphic to $A_3$.

Spoiler:

Use $f(x) = x^3 -3x -1$

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You may know that the Galois group of $x^n-1$ over the rationals is cyclic of order $\phi(n)$ (that's the Euler phi-function). If $\phi(n)$ is a multiple of $3$ (and it's not hard to find such $n$), then you can find a normal extension of the rationals of degree $3$ as a subfield of the splitting field of $x^n-1$.

This is easiest to do if you don't choose $n$ any bigger than necessary.

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The splitting field of $x^3-7x+7$ is of degree 3 over the rationals, since the discriminant is a rational square and the polynomial is irreducible.

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