Splitting Field and its Galois Group

Question

Over $\mathbb{Q}$, how can I compute the splitting field for $X^3+2X+2$ and compute its Galois group?

Answer 1

First of all, I would prove that $$p(X)=X^3+3X+2$$ is an irreducible polynomial over $\mathbb{Q}$. This is preatty easy using Eisenstein with $p=2$.

Now, what I would try to do is the following.

Suppose $\alpha$ is one of the roots of $p(X)$ in its splitting field. Try to find another root of $p(X)$ which is of the form $\alpha_2 = \frac{q_1(\alpha_1)}{q_2(\alpha_1)}$ where $q_1$ and $q_2$ are polynomials with coefficients in $\mathbb{Q}$.

• If you find this root, you would be able to conclude that $\mathbb{Q}(\alpha_1)$ is the splitting field of your polynomial (since you would have $p(X)=(X-\alpha_1)(X-\alpha_2)f(X)$ where $f(X)$ must have degree $1$, hence it should contain also the third root of your polynomial). Then, your $\text{Gal}(p(X))\cong H \leq \mathbb{S}_3$. Since in this case we would have $[\mathbb{Q}(\alpha):\mathbb{Q}]=3$, and it would be a Galois extension (it is the splitting field of $\alpha$), $|\text{Gal}(p(X))| = 3$. Therefore, $$\text{Gal}(p(X)) \cong A_3$$ which is the only subgroup of $\mathbb{S}_3$ with order equal to $3$.

• If this is not the case, then you would still have $\text{Gal}(p(X))\cong H \leq \mathbb{S}_3$ but you would not have $[\mathbb{Q}(\alpha):\mathbb{Q}]=3$. Let $E$ be the splitting field of $p(X)$. Then, $E=\mathbb{Q}(\alpha_1,\alpha_2,\alpha_3)$. However, by the tower of extensions formula, we know that $[E:\mathbb{Q}] > 3$. Thus, $|\text{Gal}(p(X))| > 3$ (still Galois since it is the splitting field of $p(X)$). Therefore, $$\text{Gal}(p(X)) \cong \mathbb{S}_3$$ which is the only subgroup of $\mathbb{S}_3$ with order bigger than $3$.

You may could apply the following theorem:

Proposition 1. Let $p(X)\in K[X]$ with all its roots different. Let $A?\{\alpha_1,...,\alpha_n\}$ be the set of its roots. Then, $\text{Gal}(p(x)) \cong \text{Alt}(A)$ if and only if $\delta \in K$ where $$\delta = \prod_{i<j}(\alpha_i - \alpha_j).$$