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
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.
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