1

For some cases, we can use Sylows theorem to determine that theres a unique quadratic extension or how many there are.

But I have a homework, where I am supposed to show that there are exactly 3 quadratic extensions contained in $Q(\zeta_{143})$, since $\Phi(143)=120$ we know, using the Galois Correspondence that the quadratic extensions over Q appear when we have a group of order 60. How can I show that there are 3 such groups?

Also my Galois Group corresponds to $C_{10}$ $X$ $C_{12}$, can I use that?

How do you tackle these generally?

peterh
  • 2,756
Not_Bernhard_Riemann
  • 81

1 Answers1

1

Your Galois group is isomorphic, thanks to CRT, to $C_2\times C_4\times C_{15}$.

It is a standard fact (and a good exercise to prove it) that if $G_1,G_2$ are finite groups with coprime orders, then the subgroups of $G_1\times G_2$ are exactly the subgroups of the form $H_1\times H_2$, where $H_i$ is a subgroup of $G_i$.

Now, you want a subgroup of order $60$, so you have no choice but taking subgroups of the form $H\times C_{15}$, where $H$ is a subgroup of order $4$ of $C_2\times C_4$. But there are exactly 3 such subgroups.

GreginGre
  • 16,641