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im trying to show that the galois group of $x^5+sx^3+t$ over $\mathbb{Q(s,t)}$ is $S_5$. By just looking at the discriminant, it has to be $S_5$ or $F_{20}$.

I know i could distinguish between those 2 using the weber sextic resolvent, but this is a degree 6 polynomial with 2 parameters and i have no idea how to show that there is no root in $\mathbb{Q(s,t)}$. Is there another way to see this more easily?

Something else i would like to know is:

I think it s true under some conditions that if u specialize s,t in a polynomial as above the galois group can only become smaller. Does anyone know if this is a basic fact or needs more work?

Thx in advance

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I can only answer your second question.

What you’re looking at geometrically is the finite cover of degree 5:

$$f : \text{Spec }\mathbb{Q}[s,t][x]/(x^5+sx^3+t)\rightarrow \text{Spec }\mathbb{Q}[s,t]$$

Away from the ramification locus (this is basically the values of $(s,t)$ which make the polynomial $x^5+sx^3+t$ have multiple roots) in $\text{Spec }\mathbb{Q}[s,t]$, the map is unramified, though not galois. The galois group you’re referring to is also called the “monodromy group” (this is geometric language). Let $U\subset\text{Spec }\mathbb{Q}[s,t]$, and let $X := f^{-1}(U)$. By the theory of the fundamental group, the cover $X\rightarrow U$ corresponds to a homomorphism

$$\pi_1(U)\longrightarrow S_5$$

The image of this map is called the monodromy group (for you, the galois group).

Specializing to a particular value of $s,t$ corresponds to pulling back the cover $X\rightarrow U$ via a map $p : \text{Spec }\mathbb{Q}\rightarrow U$. The pulled-back cover corresponds to the homomorphism

$$\pi_1(\text{Spec }\mathbb{Q})\stackrel{p_*}{\longrightarrow}\pi_1(U)\longrightarrow S_5$$

The galois group (monodromy group) of the specialization is then just the image of the above homomorphism, which can obviously cannot be larger than the image of the original homomorphism $\pi_1(U)\longrightarrow S_5$.

Therefore, if you can find a single value of $s,t$ for which the galois group is all of $S_5$, then you can conclude that the galois group of the unspecified polynomial is also $S_5$.

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