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I took particular case when p=3 then I find that $\mathbb{Q}$ is subfield of index 2 in this particular case.

I also find that splitting field of given polynomial is $\mathbb{Q(a) }$ where a is pth root of unity such that a is not 1.

How can I find unique subfield of index 2 in general?

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  • $\begingroup$ Your special case $p=3$ though correct, it is over simplified. "A group order 2 has a subgroup of index 2, namely the trivial subgroup" is not a hard statement. Try $p=5$ at least. $\endgroup$ – P Vanchinathan Jul 3 at 0:43
  • $\begingroup$ It isn't hard to compute the Galois group very explicitly in this case. Hint: a permutation is determined by where you send your $a$, and you can send it to any root. What does this imply about the Galois group? $\endgroup$ – RghtHndSd Jul 3 at 2:02

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