Thread: Extension fields

Hello,
I have been asked to prove that the cardinality of any algebraic extension of a finite field k is the same as that of Z(the integers)
And if k is not finite, the cardinality of any algebraic extension is the same as that of the field k.

I am not really sure what to do.

If k is a finite field say Fp, then an element z is algebraic over Fp, if it satisfies a polynomial with Fp coefficients say a(0) + a(1)x + a(2)x^2 +....a(n)x^n.

I dont know how to go about it. I was wondering if you could help.

Originally Posted by poorna

Hello,
I have been asked to prove that the cardinality of any algebraic extension of a finite field k is the same as that of Z(the integers)
And if k is not finite, the cardinality of any algebraic extension is the same as that of the field k.

I am not really sure what to do.

If k is a finite field say Fp, then an element z is algebraic over Fp, if it satisfies a polynomial with Fp coefficients say a(0) + a(1)x + a(2)x^2 +....a(n)x^n.

I dont know how to go about it. I was wondering if you could help.

So the idea is simple enough. In any extension field any polynomial has at most the number of roots as its degree, right? So, the largest it could possible be is . If is infinite then clearly this can be reduced in cardinality to using basic cardinal arithmetic. Thus, every algebraic extension field has at most elements, but since it clearly contains it obviously has exactly elements. If is finite it's clear that our bounding above set is countable, and since our algebraic extension field is infinite we may clearly conclude it's countable.



NOTE: I intentionally left out a lot of detail. You fill it in.

Thanks a lot! I now get it. We go about this, the same way like we do to prove, the set of algebraic numbers in R is countable (one of the initial proofs in Real analysis).

Thanks a lot for your help. I was quite stumped.