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Let $k$ be a field and $K$ be its extension field.

Suppose $y \in K$ is algebraic over $k(x)$ for some $x \in K$ and $y$ is transcendental over $k$.

Then $x$ is algebraic over $k(y)$. I think one way is the following,

Let $ f_0(x)/g_0(x) + f_1(x)/g_1(x)y + f_2(x)/g_2(x)y^2 + f_n(x)/g_n(x)y^n=0$.

so if we take the numerator after adding we get, $f_0(x)g_1(x)...g_n(x) + f_1(x)g_0(x)...g_n(x)y + f_2(x)g_0(x)...g_n(x)y^2+...+f_n(x)g_0(x)..g_{n-1}(x)y^n=0$

Is there any other cleaner way to prove the above statement?

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up vote 0 down vote accepted

Not really a cleaner way, essentially your argument in fancy language: If $x$ is not algebraic over $k(y)$, then $k(Y)(X)=k(X,Y)\rightarrow k(x,y)$ would be an isomorphism of fields. But the condition $y$ algebraic over $k(x)$ shows that the kernel is not $0$.

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Yes, there is a cleaner way, in some sense, using the transcendence degree.

Consider $k \subset k(y) \subset k(x,y)$ and $k \subset k(x) \subset k(x,y).$ We will have




Merging them, we will obtain, $tr.deg(k(x,y)/k(y))+1=1.$ Hence, $tr.deg(k(x,y)/k(y))=0,$ and $x$ is algebraic over $k(y).$

Note that, I have used the fact that an extension is algebraic $\iff$ its transcendence degree is zero.

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