Galois theory: Norm and trace
Вставка
- Опубліковано 16 жов 2024
- This lecture is part of an online graduate course on Galois theory.
We define the norm and trace of a finite extension of fields.
We give some examples of calculating the image of the norm map, and show how to use the norm and trace to find rings of algebraic integers.
Dear Sir,
I want to thank you for preparing and providing such beautiful lectures for global audience. I am from a remote part of India and due to these lectures I not only get the opportunity to learn higher mathematics but also get inspired to learn more.
Thank you
I second you Harsh... These lectures are gems.
I can't see how the degree of an extension [M:K] can be zero.
Take M of order p^p and K of order p
The degree isn't zero, considered as a natural number. However, if you wanted to perform arithmetic in, say, K, where K has characteristic q, then suddenly the nonzero integer [M:K] risks being equal to 0 in K (if q divides [M:K]) so you'd have to be careful
3:34.
First see that deg(chi_a)=deg(m_a) (since deg(chi_a)=[K(a):K] and deg(m_a)=[K(a):K]) where chi_a is the characteristic polynomial and m_a is the minimal polynomial. Now let T_a denote the linear map. The claim is that f(T_a)=T_f(a) for any polynomial \sum_i a_iX^i=f\in F[X]. Now, f(T_a)a^k=(\sum_i a_iT_a^i)a^k= (\sum_i a_i(a^k+i))=(\sum_i a_i(a^i))a^k=f(a)a^k=T_f(a)(a^k). Since {1,a,a^2\ldots a^n-1} is a basis and f(T_a) and T_f(a) are linear transformations, f(T_a)=T_f(a).
Now if f(a)=0, then T_f(a)=0, which means that f(T_a)=0. On the other hand, if f(T_a)=0, then Tf(a)=0, which means that f(a)x=0 for all x, so choosing nonzero x, this implies f(a)=0. Hence, f(a)=0 iff f(T_a)=0.
Now chi_a(a)=0 since the cayley hamilton theorem implies chi_a(T_a)=0. Since their degrees are the same, this implies that they are associate, but since they are monic, then they are equal.
yee