- #1
Ad123q
- 19
- 0
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.
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.