Galois group of $(x^2+1)(x^2+3)$ Both roots lines on imaginary axis which confuses me. Can I say that $(x^2+1)(x^2+3)$ splits in $\Bbb Q(i, \sqrt 3)$ in which $(x^2+1)(x^2-3)$ also splits so Galois group is $V_4$?
 A: It is true that $f(x) = (x^2+1)(x^2+3)$ splits in $\mathbb{Q}(i, \sqrt{3})$ and that this field has automorphism group $V_4$.  It is necessary, however, to check that $\mathbb{Q}(i, \sqrt{3})$ is the smallest field in which this polynomial splits.  If it is, then indeed $\text{Gal}(f) \cong V_4$.
If it is not the smallest field in which $f$ splits, then $f$ will split in one of its subfields: $\mathbb{Q}(i \sqrt{3})$, $\mathbb{Q}(i)$, or $\mathbb{Q}(\sqrt{3})$.  If this is the case, then the Galois group will be one of the subgroups of $V_4$.  One can manually confirm that none of these contain all the roots of $f$.  For example, if $f$ splits in $\mathbb{Q}(i)$, then we should be able to write $i\sqrt{3} = a + bi$ for some $a, b \in \mathbb{Q}$.  This means $-3 = (a + bi)^2 = a^2 + 2abi + b^2 \implies $ $a =0 \text{ or } b=0 \implies -3$ has a square root in $\mathbb{Q}$, a contradiction.

As a general rule$^\dagger$, if $p$ and $q$ are irreducible polynomials over a field $F$ with splitting fields $K/F$ and $L/F$ and Galois groups $G_1$ and $G_2$ respectively, then the polynomial $pq$ has Galois group $G_1 \times G_2$ whenever $K \cap L = F$.  We can apply this rule to this problem:  $\mathbb{Q}(i\sqrt{3}) \cap \mathbb{Q}(i) = \mathbb{Q}$.

$^\dagger$ Click here for further discussion.
