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I have been struggling with this specific confusion throughout my galois course, lets say we have like $x^{11}-42$. Then this will have degree 110, by adjoining $\zeta_{11}$ and $\sqrt[11](42)$. We know that $Gal(Q(\zeta_{11})/Q)$ is isomorphic to $Z/10Z$, but the Galois correspondence , says that a group of order should be 11 for this extension?

So basicly does the field extension of order 10 correspond to a group of order 11 or 10? And if it is 11, when why not 10? Also $(Q(\zeta_{11})/Q)$ is a galois extension over the 11th cyclotomic polynomial.

Please helpp

Not_Bernhard_Riemann
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    That cyclotomic extension has degree 10, e.g. $x^{11}-1=(x-1)(x^{10}+...+1)$, so there is no problem. – okabe rintarou Dec 15 '24 at 11:04
  • Ok and does it correspond to a group of order 11? – Not_Bernhard_Riemann Dec 15 '24 at 11:18
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    Given a finite Galois extension $L/K$, the Galois correspondence realizes a bijection between the subextensions $M$ and the subgroups of $Gal(L/K)$. If a subgroup $H$ corresponds to a subextension $M$, then $H=Gal(L/M)$, so the degree of $M$ over $K$ is $\frac{[L:K]}{|H|}$! – Aphelli Dec 15 '24 at 11:23
  • Oh ok thank you , it makes sense now, but so if L/K has degree 110, M is a field extension of degree 10 over K, then it corresponds to Gal(L/M) with degree 11? – Not_Bernhard_Riemann Dec 15 '24 at 11:31
  • Captain Lama's answer and Aphelli's comment explain this specific problem very well. This old thread discusses a natural follow up -question of figuring out what that Galois group of order $110$ looks like. My answer is mostly about book keeping, but Tom Oldfield's answer ties the question to a very concrete group. Caveat, you may need two more weeks of lectures to be able to follow the arguments. Have fun! – Jyrki Lahtonen Dec 16 '24 at 03:53

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I'm not sure that is what you mean, but $\mathbb{Q}(\zeta_{11})/\mathbb{Q}$ is indeed a Galois extension of degree 10, so there is no contradiction that its Galois group has order 10.

The minimal polynomial of $\zeta_{11}$ is $X^{10}+X^9+\dots+X+1$, which is of degree 10.

Captain Lama
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  • Yes but then the galois theorem sas if we have : L ${1}$ K H Q G , and L/K is galois , then like the extension K/Q is [G:H] , and in my case its 10, and G has order 110, so H has order 11, not 10. – Not_Bernhard_Riemann Dec 15 '24 at 11:12
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    I think what makes you confused is that you think that $Gal(\mathbb{Q}(\zeta_{11})/\mathbb{Q}$ is the subgroup of $Gal(K/\mathbb{Q})$ that appears in the Galois correspondance, but this is not the case: the subgroup in question is $H = Gal(K/\mathbb{Q}(\zeta_{11})$, which has order 11, and $Gal(\mathbb{Q}(\zeta_{11})/\mathbb{Q}$ is the quotient of $Gal(K/\mathbb{Q})$ by $H$. – Captain Lama Dec 15 '24 at 14:17