Question regarding fields.

[daon]

This was on my final, and I couldn't get it entriely.

Prove that \(\displaystyle \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 $\displaystyle d | n$ then $\displaystyle GF(p^d)$ is a subfield of $\displaystyle GF(p^n)$. And hence all fields $\displaystyle GF(p^k)$ with k<n would be subfields of $\displaystyle GF(p^{n!})$ 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]

Isn't this just the field of formal power series mod p?

 

[daon]

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

But after a bit more research, it seem the Frobenius Automorphism might be of use. I'll have to think about it a bit longer.