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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: $$ \pm \sqrt {\frac{1} {2}3 - i\sqrt 7 } ,\qquad \pm \sqrt {\frac{1} {2}3 + i\sqrt 7 } $$

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up vote 3 down vote accepted

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 $G=V$.

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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 $Z_2^2$.

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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

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