# Determine the galois group of $x^5+sx^3+t$

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

• It is not always $S_5$. For $s=10, t=\pm 5$ the group is $F_{20}$. – i. m. soloveichik Jul 27 '15 at 15:57
• If $s=0$ then the Galois group is not $S_5$. – i. m. soloveichik Jul 27 '15 at 16:05
• If $s=15, t=81$ then the group is $D_5$. – i. m. soloveichik Jul 27 '15 at 16:07

## 1 Answer

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 $S_5$.

• Also, can't you do a base change over $\Bbb C$ to reduce it to a geometry problem ? – mercio Jul 27 '15 at 19:00
• @mercio Sure...though I'm not entirely sure how that would help.. – oxeimon Jul 27 '15 at 19:05