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Suppose given a polynomial $P=Q_1\cdots Q_k$ of degree $n$, where each $Q_i$ is irreducible. Suppose also that I know the Galois group $G_i$ (over the rationals) of each irreducible factor $Q_i$.

Is there an easy correlation between the Galois group of $P$, and the $G_i$?

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No (assuming by the "Galois group of P" you mean the Galois group of its splitting field, and assuming by "correlation" you mean "can I work out the Galois group of P given only the G_i"), because you need to know how the splitting fields of the Q_i interact. Consider for example the cases P=(x^2-2)(x^2-8) and P=(x^2-2)(x^2-3). – Kevin Buzzard Mar 26 '10 at 15:45
This seems to be essentially the same as… – Franz Lemmermeyer Mar 26 '10 at 17:06
It's not the same question. The case $k=2$ is rather easy. – Martin Brandenburg Mar 27 '10 at 15:47
But in the other question, there was no requirement that $f$ and $g$ be irreducible, so I think the questions are the same. – Reid Barton Mar 27 '10 at 16:06
up vote 14 down vote accepted

The Galois group of $P$ will be a subdirect product of the $G_i$, that is a subgroup of $G_1\times\cdots\times G_k$ projecting surjectively onto each of the $G_i$.

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


$P = (X^2+1)(X^2-2)$ 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.

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