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This is a very general question. I am currently working on algorithms to compute Galois groups using the resolvent method, and I don't understand the connection between the resolvent and the Galois group.

Let $f(x)$ be a separable polynomial over a field $K$ with Galois group $G$, and let $H$ be a normal subgroup of $G$. Let $y=y(x_1,….,x_n)$ be a rational expression in $x_1,…,x_n$ that remains invariant under all permutations of the roots $x_1,…,x_n$ belonging to $H$ , and let $y$ not be in $K$. Then $y$ is a root of some equation $g(y)=0$ with coefficients from $K$, the Galois group of which is a proper quotient group of $G$. Thus, solving the equation $f(x)=0$ reduces to solving the equation $g(y) = 0$ and solving the equation $f(x) =0$ over the field $K(y_1,…,y_n)$.

[http://www.encyclopediaofmath.org/index.php/Resolvent]

So my question is:

Assuming I am implementing Cohen's degree 4 algorithm, $H$ would be my stabilzer of $y$ and $g(y)$ would be my resolvent, and by looking at the field $K(y_1,…,y_n)$ (where I assume the $y_i$ are the roots of $g$) this will give me the Galois group of $g$. But how does this tell me about the Galois group of $f$? I think that the $Gal(g) \leq Gal(f)$ is this correct?

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From your own post: "Then $y$ is a root of some equation $g(y)=0$ with coefficients from $K$, the Galois group of which is a proper quotient group of $G$." –  Alex B. Dec 10 '12 at 19:20
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