Algebraic Extensions, vector spaces

joeblow

Suppose that F/K is an algebraic extension, S is a subset of F with S/K a vector space and [itex] s^n \in S [/itex] for all s in S. I want to show that if char(K) isn't 2, then S is a subfield of F.

Since F/K is algebraic, we know that [itex]\text{span} \lbrace 1,s,s^2,...\rbrace [/itex] is a field for any s in S. Thus, I want to define [itex] E= \bigcup _{s \in S} \text{span} \lbrace 1,s,s^2,...\rbrace [/itex]. Since S/K is a v.s., we have [itex] \text{span}\lbrace 1,s,s^2,...\rbrace \subseteq S[/itex] and since E is just a union of subsets of S, [itex] E \subset S [/itex]. Also, by the way we defined E, [itex] S \subseteq E [/itex] so E = S.

(E,+) is a group by the fact that S/K is a v.s. Each nonzero element has an inverse in S since [itex] \text{span} \lbrace 1,s,s^2,...\rbrace [/itex] is a field and each s belongs to one of those fileds (so that field would have the inverse of s).

The hard part seems to be proving closure of multiplication. Suppose that [itex] s_1 s_2 = s + v [/itex] for [itex] s_1,s_2,s\in S [/itex] and [itex] v\in F [/itex]. We have already shown that S contains all inverses, so [itex] s_2 = s_1^{-1} s + s_1^{-1} v \in S [/itex]. Whence, [itex] s_1^{-1} v \in S [/itex]. Obviously, I want to show that [itex] v\in S [/itex], but progress has stopped at this point.

Any suggestions would be appreciated.

morphism

Think about how you can get an expression involving [itex]s_1+s_2[/itex] and [itex]s_1s_2[/itex].

joeblow

Since S is a v.s., we get [itex]s_1+s_2[/itex] by vector addition.

For [itex] s_1 s_2[/itex] you have to multiply two linear combos in S (whose basis must be a subset of the basis for F/K). If the product of two basis elements of S is again in S, then the product of any two elements of S would be in S.

However, I do not see how to show this.

morphism

I'm not sure how to phrase my hint differently, so I apologize if this is giving away too much: Consider [itex](s_1+s_2)^2[/itex].

joeblow

Yay. That makes sense why char(K) can't be 2 then.

Thanks a lot... why couldn't I think of that.