0

I tried to find galois group of polynomial $p(x) = x^5 + 2x + 2$ and I managed after a while using the fact that it has only $1$ real root and the others are complex so they correspond to transposition and real root to cycle of length $5$. Combining everything gives Galois group is going to be $S_5$.

Now, my question is what if I shift my polynomial with $1$ i.e $p(x+1) = (x+1)^5 + 2(x+1) + 2$. Is still the galois group $S_5$?. I checked in programme, YES it is.

Actually, I checked the Galois group $p(x+a)=(x+a)^5 + 2(x+a) + 2$ where $1\le a\le 35$ is still $S_5$. But I didn't see the reason.

So, my question is pick arbitrary polynomial $f(x)$ with certain galois group then does galois group of iterated polynomial $f(x+a),a\in \mathbb Z$ be same?

Elise9
  • 105
  • 7
  • 3
    The question in the title is different from that in the body. – Alex B. Mar 19 '24 at 16:25
  • The Galois group is almost always $S_n$, see this post. No need of other "reasons" (which are probably not true). So the Galois group of $x^5+rx+s$ is almost always $S_5$, see here, or also here. – Dietrich Burde Mar 19 '24 at 17:20
  • Actually, I am not interested that polynomial $p(x)$. That is why I am asking the question more general. Pick any polynomial $f(x)$ whose galois group is $S_n$ then can I say Galois group of $f(x+a)$ where $a \in \mathbb Z$ is also $S_n$? – Elise9 Mar 19 '24 at 18:09
  • Yes. The splitting fields of both polynomiasl are equal (easy exercise), so they have same galois group – GreginGre Mar 19 '24 at 19:08
  • Could give me a hint? – Elise9 Mar 21 '24 at 19:21

1 Answers1

0

Let the Galois group of polynomial $p(x)$ is $S_n$. Any iteration of $f$ by $\pm c$ does not change anything. Reason follows from isomorphism of iteration.

nozalp10
  • 108