Galois group of a product of irreducible polynomials - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T19:55:18Z http://mathoverflow.net/feeds/question/19435 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/19435/galois-group-of-a-product-of-irreducible-polynomials Galois group of a product of irreducible polynomials Justin 2010-03-26T15:35:22Z 2010-03-26T16:25:34Z <p>Hello</p> <p>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 $Q_i$.</p> <p>Is there an easy correlation between the Galois group of $P$, and the $G_i$?</p> http://mathoverflow.net/questions/19435/galois-group-of-a-product-of-irreducible-polynomials/19437#19437 Answer by Robin Chapman for Galois group of a product of irreducible polynomials Robin Chapman 2010-03-26T15:44:08Z 2010-03-26T15:44:08Z <p>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 $G_i$.</p> http://mathoverflow.net/questions/19435/galois-group-of-a-product-of-irreducible-polynomials/19438#19438 Answer by Martin Bright for Galois group of a product of irreducible polynomials Martin Bright 2010-03-26T15:47:26Z 2010-03-26T15:47:26Z <p>If you mean: does knowing the <em>G<sub>i</sub></em> tell you the Galois group of <em>P</em>, then no.</p> <p>Examples:</p> <p>$P = (X^2+1)(X^2-2)$ 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 $\mathbb{Q}$.</p> <p>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 $\mathbb{Q}(\sqrt{-3})$.</p> <p>I've just seen Robin's answer, so to relate to that: in the first example, the Galois group of <em>P</em> 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.</p>