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 $s^n \in S$ 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 $\text{span} \lbrace 1,s,s^2,...\rbrace$ is a field for any s in S. Thus, I want to define $E= \bigcup _{s \in S} \text{span} \lbrace 1,s,s^2,...\rbrace$. Since S/K is a v.s., we have $\text{span}\lbrace 1,s,s^2,...\rbrace \subseteq S$ and since E is just a union of subsets of S, $E \subset S$. Also, by the way we defined E, $S \subseteq E$ 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 $\text{span} \lbrace 1,s,s^2,...\rbrace$ 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 $s_1 s_2 = s + v$ for $s_1,s_2,s\in S$ and $v\in F$. We have already shown that S contains all inverses, so $s_2 = s_1^{-1} s + s_1^{-1} v \in S$. Whence, $s_1^{-1} v \in S$. Obviously, I want to show that $v\in S$, but progress has stopped at this point.

Any suggestions would be appreciated.

 

[morphism]

Think about how you can get an expression involving $s_1+s_2$ and $s_1s_2$.

 

[joeblow]

Since S is a v.s., we get $s_1+s_2$ by vector addition.

For $s_1 s_2$ 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 $(s_1+s_2)^2$.

 

[joeblow]

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

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