Galois group of a product of irreducible polynomials

If you mean: does knowing the G_i tell you the Galois group of P, then no.

Examples:

$P = (X^2+1)(X^2-2)$ has Galois group $C_2 times C_2$, and both factors have Galois group $C_2$; this works because the splitting fields of the two factors intersect only in $mathbb{Q}$.

But $P = (X^2 + X + 1)(X^2+3)$ has Galois group $C_2$, although both factors again have Galois group $C_2$. Here both factors, though they're coprime, define the same extension $mathbb{Q}(sqrt{-3})$.

I've just seen Robin's answer, so to relate to that: in the first example, the Galois group of P is the whole of $G_1 times G_2$. In the second example, it is the diagonal subgroup of $G_1 times G_2$, which is smaller although still projects surjectively onto each factor.

• Next soft question - Too old for advanced mathematics?
• Previous linear algebra - Hermitian matrices with prescribed number of positive and negative eigenvalues