RNT1.2.2. Order of a Finite Field

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Abstract Algebra: Let F be a finite field. Prove that F has p^m elements, where p is prime and m gt 0. We note two approaches: one uses the Fundamental Theorem of Finite Abelian Groups, while the other uses linear algebra.
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I you have any specific problems that you want to see, let me know. - Bob

MathDoctorBob
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Good morning children. Today you will learn mathematics. Or else...

Thank you for making the video series!

ViatoremDiEfa
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@Nazlijelveh That's step 1: Every finite field has characteristic of prime order q. Since F is a vector space over this prime field, F has order q^m. Since p and q are prime, p=q ad m=n. - Bob

MathDoctorBob
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@Nazlijelveh The only assumption we have is that F is finite. The first step is to show that there is a subfield of prime order (which we call p). Sorry for not being clear. - Bob

MathDoctorBob
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Bob, it is very nice to have this lectures, but are the RNT lectures pack is available as it is available in group theory??

notunkabita
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Two operations: 1) under multiplication, the nonzero elements form a cyclic group, and 2) under addition, F is a direct product of Z/p's, where p is the characteristic. To decompose, 1 generates a Z/p. Then F is a vector space over Z/p, and choosing a basis over Z/p gives the product. - Bob

MathDoctorBob
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Since F is a finite field, then it is a cyclic group (and thus generated by one element). the issue is that i don't get how to make that decomposition: F is product of Z/ni where ni|n(i+1)...

Ramzythemaster
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could you add more lectures on Galois theory ??!!, please

mathWisdom
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yes I want to understand better how to find element of Galois group ( be exactly automorphism)

mathWisdom
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Everything is in the Abstract Algebra playlist in order. I haven't done problems for Ring Theory yet. Maybe in the future.

MathDoctorBob
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@Thatdarnmonkeylove You're welcome! - Bob

MathDoctorBob
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@porkinmycorkhole Admittedly this video is mostly abstract nonsense. To give examples with order p^2 or higher, we need to develop some theory for irreducible polynomials over Z/p. - Bob

MathDoctorBob
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Sounds a bit like the ultimate warrior.

Freddyphelp