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daon
05-16-2007, 02:42 PM
This was on my final, and I couldn't get it entriely.

Prove that \L \,\, \,\, \bigcup_{k \ge 0} GF$$p^k$$ \,\, \,\, is a field.

So many thoughts ran though my head. 1) Induct on k? 2) Try a proof by contradiction? 3) Show directly that it is a unitary, commutative ring with all non-zero elements being units?

This idea also came to me, but I felt like I was digging myself into a hole:

If d | n then GF(p^d) is a subfield of GF(p^n). And hence all fields GF(p^k) with k<n would be subfields of GF(p^{n!}) and hence there is always some "larger finite field" which contains all other fields in the union chain. I'm not sure where to go from here with this.

After spending a few minutes on each possibility, I gave up. Any ideas?

Count Iblis
05-16-2007, 05:30 PM
Isn't this just the field of formal power series mod p?

daon
05-16-2007, 05:52 PM
Could you elaborate a little? We didn't cover the topic you mentioned, and I can't seem to fiund information on the internet regarding the formal power series mod P you mentioned.

The only information I have access to for this proof is basic Galois theory, Field Extensions (including algebraic, root) and the various alegbraic structures one would encounter in an undegraduate course in abstract/modern algebra.

Another thought I had was that since GF(p^k) is exactly the roots of x^{p^k}-x, we get that this union is the roots of the polynomial (x^p^1-x)(x^p^2-x)(x^p^3-x) \cdot \cdot \cdot. Not sure if this helps me or not.

But after a bit more research, it seem the Frobenius Automorphism (http://en.wikipedia.org/wiki/Frobenius_automorphism) might be of use. I'll have to think about it a bit longer.