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?
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?