Examples of correspondence theorem
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- Опубликовано: 5 янв 2025
- Lecture 13
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should we not have bar over f(t) instead of f(bar over t) at 23:00
Sir how can it be possible that ker(phi) =x^2+1(because phi: phi( f(x)) |------>f(i) ) and if f(x) = 0 then this imply f(i) =0 hence 0€ker(phi)
Even later in proof u concluded that f(x) is x^ 2+1 so any element of ker(phi) is of form ( x^2 +1) g(x) {so if g(x) =0, (x^4+1+x) etc.) Then these all elements are also belonging to ker(phi) so how can it be preciously ( x^2 +1)
Please answer
I think it should be f bar because the zero element on the quotation is not really zero but zero bar I.e f(t)+(t^2+1) where f(t) is generated by t^2+1 then the coset is the zero element.
I take this back after looking because t bar^2+1 is in fact zero because t bar is coset and t will be squared after multiplying the t bar and t^2 +1 will be obsorbed by the kerner coset
1. Ideal I in F[x] is generated by a monic polynomial corresponding to that particular ideal.
As C (t) is a field it should contain only two ideals O and C. Then why it has 4 ideals?
good question but observe that C is a field but C(t) is not a field.
@@chinmaypadhan5936 Thank you, I was confused too so Ct is polynomial ring and polynomials have no inverses. So the earlier theorem he used it was K that was field and not Kx