# Galois group of a reducible polynomial

I know if I have a polynomial $f(x) = g(x) \cdot h(x) \cdot k(x)$ and $g(x),h(x),k(x)$ are polynomials of degree 2, then the Galois group will be $Z/2Z \times Z/2Z \times Z/2Z$ if the roots of $g,h,k$ are distinct.

But what happens in the case when you have mixed factors... i.e. for example the polynomial

$(x^2 + 1)(x^2 -4x + 7)(x^2 - 2)$

the splitting field would be $Q(\sqrt{2},\sqrt{3},i)$ but what would be the Galois group?

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Did there use to be a question here? –  Arturo Magidin Nov 3 '11 at 2:03
This is kind of confusing, someone put the question mark somewhere back in there please? –  Patrick Da Silva Nov 3 '11 at 4:50
@Walter Vandalising your own questions is grounds for suspension on this site. Please refrain from doing that. Also, if the question was a homework problem, then you should tag it accordingly. –  Alex B. Nov 3 '11 at 8:55

It is easy to see that $x^2-4x+7$ has a root in a field $K$ if and only if $-3$ is a square in $K$ (because $x^2-4x+7=(x-2)^2+3$). Therefor the splitting field of $x^2-4x+7$ is the same as the splitting field of $x^2+3$. Hence we still have Galois group $\mathbb{Z}/2\mathbb{Z}^3$.
The result you quoted can be stated in a more general way: The Galois group of your polynomial will be $\mathbb{Z}/2\mathbb{Z}^3$ not if the roots are distinct (that is not a satisfying condition because eg. $\sqrt{2}$ and $\sqrt{2}+1$ yield the same extension of $\mathbb{Q}$) but if we have a trivial intersection between the splitting field of g and the compositum of the splitting fields of h and k (and analogous for h and k).
I might add that the result you stated is not even true in the current form. The Galois group of $f=(X^2+1)(x^2-2)(x^2+2)$ will only be $\mathbb{Z}/2\mathbb{Z}^2$ because the roots of the third polynomial are already elements of the compositum of the splitting fields of the first two polynomials.