# computing the galois group of a polynomial

Compute the Galois group of the splitting field of the polynomial $t^4-3t^2+4$ over $\mathbb{Q}$. I don't know how can I do this problem, the roots are very "ugly" maybe if I consider another basis (a more workable basis). The roots of the polynomial are: $$\pm \sqrt {\frac{1} {2}3 - i\sqrt 7 } ,\qquad \pm \sqrt {\frac{1} {2}3 + i\sqrt 7 }$$

• – Watson Dec 26 '16 at 13:18

## 2 Answers

There is a standard way to find Galois groups of degree 4 irreducible separable polynomials. Say your polynomial is

$f(x) = x^4+bx^3+cx^2+dx+e$

Call $x_1,x_2,x_3,x_4$ its (pairwise distinct) roots. Let

$\alpha := x_1x_2 + x_3x_4$

$\beta := x_1x_3 + x_2x_4$

$\gamma := x_1x_4 + x_2x_3$

The polynomial $g(x) := (x-\alpha)(x-\beta)(x-\gamma)$ is called the resolvent cubic of $f(x)$. Computation shows that

$g(x) = x^3 - cx^2 + (bd-4e)x - b^2e + 4ce - d^2$.

Call $K$ the field $\mathbb{Q}(\alpha,\beta,\gamma)$, the splitting field of $g(x)$ over $\mathbb{Q}$. Call $G$ the Galois group of $f(x)$. The natural action of $G$ on the four roots of $f$ gives an embedding of $G$ inside the symmetric group $S_4$ (identify $G$ with its image in $S_4$). $G$ is transitive in $S_4$ being $f(x)$ irreducible.

Call $V$ the Klein subgroup of $S_4$ (i.e. the subgroup consisting of the elements which are product of two disjoint transpositions). Note that the intermediate field $K$ corresponds (through the Galois correspondence) to the subgroup $G \cap V$ of $G$, therefore $G/G \cap V$ is isomorphic to the Galois group of $g(x)$.

With this information it is possible to detect $G$ among the transitive subgroups of $S_4$ (provided you can compute Galois groups of cubics, but that's easy: the Galois group of an irreducible separable degree 3 polynomial is $A_3$ if the discriminant is a square in the base field, it is $S_3$ otherwise).

In your case the cubic resolvent splits completely as $(x+3)(x+4)(x-4)$, so $K=\mathbb{Q}$ and $G=G \cap V$, in other words $G \subseteq V$. Having $V$ no proper transitive subgroup, it must be that $G=V$.

The roots are $\pm\sqrt{\frac{3\pm\sqrt{-7}}{2}}$.

We can obtain the root field (splitting field) by first adjoining $\sqrt{-7}$ to the rationals $Q$, then take appropriate square roots to get the root field of the polynomial. So the degree of the extension for the splitting field is 4.

The root field has degree 4 so the Galois group has 4 elements. Hence the Galois group is either cyclic of order 4 or $Z_2^2$. For the first case there is a unique quadratic subfield and for the second there are 3 quadratic subfields.

Consider the conjugate roots $a=\sqrt{(3+\sqrt{-7})/2}$, $b=\sqrt{(3-\sqrt{-7})/2}$; then $a+b=\sqrt{7}$, $a-b=\sqrt{-1}$, $a^2-b^2=\sqrt{-7}$

Since there are 3 quadratic subfields, the galois group is $Z_2^2$.

• How are you getting $(9-7)/4=-1/2$? – The Substitute Jul 17 '15 at 22:38
• @TheSubstitute Thanks, I made revisions. – i. m. soloveichik Jul 18 '15 at 14:24
• @i.m.soloveichik could you please elaborate on this sentence "then take appropriate square roots to get the root field of the polynomial"? Do we have to make some sort of choice here? Is it clear that the splitting field for this polynomial is equal to $\mathbb{Q}(\sqrt{(3+\sqrt{-7})/2})$ (say)? Is it clear that $\sqrt{(3-\sqrt{-7})/2} \in \mathbb{Q}(\sqrt{(3+\sqrt{-7})/2})$? – abc Nov 3 '16 at 1:54
• @abc The field generated by $\sqrt{(3+\sqrt{-7})/2}$ is degree 4 over Q (any easy check shows it doesn't satisfy a quadratic polynomial over Q). – i. m. soloveichik Nov 5 '16 at 12:41