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Let $x$ be an element not algebraic over $F$, and $K \subset F(x)$ a subfield that strictly contains $F$. Why is $x$ algebraic over $K$?

Thanks a lot!

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closed as off-topic by user26857, C. Falcon, Daniel W. Farlow, M. Vinay, Jon Mark Perry Jun 26 at 5:12

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up vote 11 down vote accepted

The thing is you need to understand what are the subfields in between $F$ and $F(x)$. If you have a subfield $K$ of $F(x)$ that contains $F$, then to allow $K$ to contain "more" elements than $F$ (i.e. to be distinct from $F$, but lie inside $F(x)$), $K$ must contain some element of $F(x)$, say $\frac{p(x)}{q(x)} \in K \cap F(x)$. But then this means $x$ is a root of $p(t) - q(t)\frac{p(x)}{q(x)} \in K[t]$, hence $x$ is algebraic over $K$.

EDIT : @Aspirin : I assumed in my question that $F \subset K \subset F(x)$ because I thought it was clear from the context, but perhaps it is not. I think that in this case, if you assume that $K$ is not a subfield of $F$, you can replace $F$ by $F' = K \cap F$ so that $F' \subsetneq K \subsetneq F'(x)$. You clearly have $F' \subsetneq K$ by assumption on $K$ in this case (i.e. that $K$ not contained in $F$) and you get $K \subsetneq (K \cap F)(x) = K(x) \cap F(x)$ since $K \subseteq K(x)$ and $K \subsetneq F(x)$, hence is in the intersection, but if you had equality, it would mean $K = K(x) \cap F(x)$, and since $x \in K(x) \cap F(x)$, that would mean $x \in K$. In that case OP's question is trivial, so we can suppose we are not in that case. Also note that since $x$ is not algebraic over $F$ it is certainly not over $F'$. Therefore $F' \subsetneq K \subsetneq F'(x)$ and you can use the arguments above to show that $x$ is algebraic over $K$.

EDIT2 : I am leaving the preceding edit there because I still don't understand why it breaks down in the case $K \subseteq F$, which shouldn't work if we are still sane around here. I am more convinced that you need to assume $F \subsetneq K$ than the fact that my edit gives something good.

EDIT3 : After all those editings and reading the comments, I modified EDIT so that the case where $K \subseteq F$ is clearly not possible. The most general case where it would work is when $K \subseteq F(x)$ but $K$ is not a subfield of $F$. A slighty small detail dodged my eye and Dylan Moreland caught it, thanks to him. Note that this also proves that $x$ would be algebraic over $K$ if and only if $K$ is not a subfield of $F$ (assuming $K$ is described as in OP's question).

Hope that helps,

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2  
The OP can hopefully provide an argument for why this is a non-zero element of $K[t]$. – Dylan Moreland Jun 28 '12 at 7:06
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The question is technically easy but difficult psychologically because one has to think of introducing a new indeterminate $t$.My experience is that beginners don't see that spontaneously . In the same vein the proof of Lüroth's theorem (classifying such intermediate extensions) is not that easy to understand. @Dylan's comment is an excellent test for self-checking the understanding of the solution. – Georges Elencwajg Jun 28 '12 at 7:29
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@Patrick: Can you explain case $F\not\subset K$? i.e. $\exists\alpha$ such that $\alpha\in F,\alpha\not\in K$ – Aspirin Jun 28 '12 at 8:06
    
@Aspirin : I edited my question to answer your very pertinent question. – Patrick Da Silva Jun 29 '12 at 1:57
    
@Dylan : I don't think it is that hard, but I must admit I didn't think about it when I wrote my proof. I'll let OP answer that. =) – Patrick Da Silva Jun 29 '12 at 2:05

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