Q: Show that no finite field is algebraically closed.
A: This is how I went about doing it.
We know from the previous chapter that if a field is algebraically closed then by definition it has no extension field.
However given GF(p^n) we can always find GF(p^m) for which k is a multiple of n. Then it follows that since k is a multiple of n then n is a divisor of k.
By theorem 22.3 then that GF(p^n) contains GF(p^n) as a subfield or something similar to GF(p^n) in order.
Note: This does not contradict Ernst Steinitz theorem, namely every field has a unique* algebraic extension that is algebraically closed, since now we are begging the question whether or not the extension of a field is closed rather than the initial field itself. Or at least that's what I believe to be true.
--Bakey 14:06, 30 November 2012 (UTC)
