Find a group isomorphic to the Galois group of the polynomial $x^3+4$?

Question

Find a group isomorphic to the Galois group of the polynomial $x^2+4$? Am I correct that it will be isomorphic to $S_3$?

Answer 1

Consider $f(x) = x^{3} + 4$. Let $E/\mathbb{Q}$ be its splitting field, and $G$ its Galois group, which regarded as a group of permutation on the roots, will be a subgroup of $S_{3}$.

Substitute $x = y -1$ to get $$ f(y - 1) = y^{3} - 3 y^{2} + 3 y + 3, $$ which is irreducible by Eisenstein's criterion. Therefore the degree $3$ of $f(x)$ divides $\lvert E : \mathbb{Q} \rvert = \lvert G \rvert$.

Now study the graph of $f(x)$ to see that it has one real root, as $f(x)$ is non-decreasing, and thus $f(x)$ has two complex ones, which must be conjugates.

It follows that complex conjugation induces an element of order $2$ in $G$, so that $G$ has order at least $6$, and thus $G = S_{3}$.

Answer 2

If $f(x) = x^2 + 4$, then $G\cong \mathbb{Z}_2$ (why?)

If $f(x) = x^3+4$, then you have the following useful fact : Let $f(x) \in k[x]\setminus k$ have (not necessarily distinct) roots $\alpha_1, \alpha_2,\ldots, \alpha_n$ is some splitting field $K$. Write $$ D := \Delta^2 \text{ where } \Delta := \prod_{i<j} (\alpha_i-\alpha_j) $$ Then $D$ is called the discriminant of $f$.

Now one has the following fact

If $\text{char}(k) \neq 2$, and $f(x) \in k[x]\setminus k$ is irreducible and separable of degree 3 with Galois group $G$, then $G\cong \mathbb{Z}/3\mathbb{Z}$ iff $D$ is a perfect square in $k$. Otherwise, $G\cong S_3$.

Furthermore, if $f(x) = x^3+4$, one checks that $$ D = -27(4)^2 $$ which is not a perfect square in $k = \mathbb{Q}$. Hence the Galois group of $f$ is $S_3$.

Answer 3

Without using the discriminant:

The Galois group of $f=x^3+4$ is (isomorphic to) a transitive subgroup of $S_3$ hence (isomorphic to) $S_3$ itself or $C_3$ depending on whether the splitting field of $f$ over $\mathbb{Q}$ has degree $6$ or $3$. Now $f$ has $3$ roots in $\mathbb{C}$ exactly one of which is real. Hence the field obtained by adjoining this (real) root to $\mathbb{Q}$ can not contain the other two roots. But this field already has dimension $3$($=\mathrm{deg}(f)$) and the splitting field therefore has degree greater than $3$.

We conclude: The Galois group of $f$ is isomorphic to $S_3$.