Degree of fields query


by Ad123q
Tags: degree, fields, query
Ad123q
Ad123q is offline
#1
Apr10-12, 10:35 AM
P: 19
I'm not sure if my reasoning below is correct or not.

If a=e[itex]\stackrel{\underline{2πi}}{5}[/itex], then Q(a) = {r + sa + ta2 + ua3 +va4 : r,s,t,u,v [itex]\in[/itex] 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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DonAntonio
DonAntonio is offline
#2
Apr10-12, 11:35 AM
P: 606
Quote Quote by Ad123q View Post
I'm not sure if my reasoning below is correct or not.

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


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

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

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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