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As the title says I wonder what the galois group of $x+3$ is. Or even if that exists ? Since $x+3 = 0$ has only one zero/element I assume its the trivial group ? And what is the galois group of $(x+1)(x+2)$ ?

Assume we have (integer) polynomials $A(x)$ and $B(x)$ and we know their galois group as $A'$ and $B'$. Is the galois group of $A(x)B(x)$ the group $A' \times B'$ ?

Is the galois group of $A(B(x))$ the group $A' \times B'$ ?

Are there tricks for products or compositions ?

Sorry I am new to Galois theory. My questions are not random as they might appear. I know what a group is, but I have never seen this addressed specifically, and solving quintics is a bit much for a newby. I know I should read as much as possible, but answering these questions would probably help me get rid of my misunderstandings.

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Does that giant $X$ mean multiplication? –  rschwieb Sep 18 '12 at 20:37
    
Yes. So A' X B' = B' X A'. –  mick Sep 18 '12 at 21:16
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Use \times instead. –  rschwieb Sep 19 '12 at 0:00
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1 Answer

up vote 2 down vote accepted

Yes, the Galois group of x+3 is the trivial group because the splitting field of that polynomial is $\mathbb{Q}$ itself.

The idea that the Galois group of a product of polynomials is the direct product of the groups is false, however (assuming that is what you mean). Consider $A(x)=B(x)=x^2+3$. Then $A(x)B(x)=(x^2+3)^2$ and this polynomial also splits completely over $\mathbb{Q}(\sqrt{-3})$. So the Galois group of $A(x)B(x)$ is still $C_2 \not \cong C_2 \times C_2$.

For the composition consider $A(x)=x-3$, $B(x)=x^2+3$. Then $A(B(x))=x^2$ has trivial Galois group, but $B$ has Galois group $C_2$.

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Ah right, I actually had another counterexample in mind, I will edit. –  Oliver Braun Sep 18 '12 at 20:45
    
What about (x+1)(x+2) ? Trivial ? –  mick Sep 18 '12 at 22:12
    
@mick Exactly like your last example was explained, this polynomial splits in $\mathbb{Q}$, so you are asking what automorphisms of the field extension $\mathbb{Q}$ over $\mathbb{Q}$ which fix $\mathbb{Q}$, there is only one possible: the identity function on $\mathbb{Q}$. So, the Galois group has one element. It is really hard to tell if you've ever even read the definition of the Galois group... or even the previous solutions given to you... –  rschwieb Sep 19 '12 at 0:07
    
@mick And to preempt any more questions along the lines "does $(x-a)(x-b)$ have a trivial Galois group for rational $a$ and $b$?" The answer is still "yes", for the exactly the same reasons. The first time you will encounter a nontrivial extension will be with something that doesn't factor into linear factors over the rationals. Take $x^2+1$, for example. –  rschwieb Sep 19 '12 at 0:09
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