# Finite Field Extensions and the Sum of the Elements in Proper Subextensions

Let $F$ be a finite field, and let $u,v$ be algebraic over $F$. Consider the fields $F(u,v),F(u)$ and $F(v)$. Must it be the case that $F(u,v) = F(u)$ or $F(u,v) = F(u+v)$?

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No: if $\mathbb F_4=\mathbb F_2(a)$ and $\mathbb F_8=\mathbb F_2(b)$ then
$$\mathbb F_2(a, b-a)=\mathbb F_{64} \neq F_2(a)=\mathbb F_4 \quad \text {and}\quad \mathbb F_2(a, b-a)=\mathbb F_{64}\neq F_2(a+(b-a))=\mathbb F_8$$

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$F$ needs be a finite field. – Isaac Solomon May 13 '12 at 8:54
Sorry for missing that: I have modified my example in order to satisfy your requirement that $F$ be a finite field. – Georges Elencwajg May 13 '12 at 9:05
This works as a counterexample. But I think that a more interesting question is, whether we always have that $F(u,v)$ is equal to at least one of $F(u),F(v),F(u+v)$? For example, here obviously $F_2(b-a)=F_{64}$. – Jyrki Lahtonen May 13 '12 at 9:09
My goodness! I spent so long thinking about this that but I somehow missed this example. Thanks! :) – Isaac Solomon May 13 '12 at 9:11
@JyrkiLahtonen: I agree that is a more interesting question, but after playing around with possibility it appears to be more difficult. I was reticent to post a more difficult question when I was looking for a more immediate solution to a simpler one. – Isaac Solomon May 13 '12 at 9:13

No, neither statement is true. If the first equality were true, then we'd have to have $v \in F(u)$, so in particular, the degree of $v$ over $F$ would be $\leq$ the degree of $u$ over $F$. So if the degree of $v$ over $F$ is > degree of $u$ over $F$, then the first equality fails.

For the second one, choose $u$ to be any element algebraic over $F$, but $u \notin F$ and choose $v = -u$. Then $F(u,v) = F(u)$ while $F(u+v) = F(0) = F$ and $F(u) \neq F$ since $u \notin F$. Hope this helps.

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In the second case, $F(u) = F(u,v)$. If the assertion is false, one would need to produce an example where both criterion fail. – Isaac Solomon May 13 '12 at 8:53