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Hello 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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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. 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. |
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