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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.