Is the Degree of Q(a) Over Q Correctly Understood?

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The degree of the field extension Q(a) over Q, where a = e^(2πi/5), is correctly determined to be 4, not 5. The basis for Q(a) as a vector space over Q consists of {1, a, a^2, a^3}, confirming that the minimal polynomial x^4 + x^3 + x^2 + x + 1 has a degree of 4. The Euler's totient function φ(5) also supports this conclusion, indicating that the dimension of the extension is indeed 4.

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Ad123q
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I'm not sure if my reasoning below is correct or not.

If a=e\stackrel{\underline{2πi}}{5}, then Q(a) = {r + sa + ta2 + ua3 +va4 : r,s,t,u,v \in Q} . [Is this correct?]

Then [Q(a):Q] = 5 as {1, a, a2, a3, a4} form a basis for Q(a) as a vector space over Q.

However I am not sure if my reasoning above is correct as I have just seen a proof that [Q(a):Q] = 4 for the same a above.

Thanks for your help.
 
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Ad123q said:
I'm not sure if my reasoning below is correct or not.

If a=e\stackrel{\underline{2πi}}{5}, then Q(a) = {r + sa + ta2 + ua3 +va4 : r,s,t,u,v \in Q} . [Is this correct?]


*** No, because \deg_{\mathbb Q}\mathbb Q(a)=\phi(5)=4 , so any basis has only 4 elements and not 5, as you wrote.

The minimal pol. of a over the rationals is x^4+x^3+x^2+x+1 .

DonAntonio

Then [Q(a):Q] = 5 as {1, a, a2, a3, a4} form a basis for Q(a) as a vector space over Q.

However I am not sure if my reasoning above is correct as I have just seen a proof that [Q(a):Q] = 4 for the same a above.

Thanks for your help.

...
 

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