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Let $K$ be a field and $L = K(x, y)$, where $x$ is transcendental over $K$ and $y$ is such that $f(x, y) = 0$, for $f \in K[X, Y]$ irreducible. I have to prove that if $f$ is also irreducible over $\overline{K}[X, Y]$, where $\overline{K}$ is the algebraic closure of $K$, then $K$ is algebraically closed in $L$, i.e., if $g(x, y)/h(x, y) \in L$ is algebraic over $K$, then $g(x, y)/h(x, y) \in K$.

Does anyone have any hint? Thanks.

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Let us denote $\alpha$ by your $y$ (Since the symbols $x,y$ always let me feel that they are indeterminate). Suppose $\beta\in L$ is algebraic over $K$, by assumption the minimal polynomial of $\alpha$ in $K(x)[Y]$ is still irreducible over $K(\beta)(x)$, then $[K(\beta)(x,\alpha):K(\beta)(x)]=[K(x,\alpha):K(x)]=deg(\alpha)$, but $K(x,\alpha)=K(\beta)(x,\alpha)$, and $[K(\beta)(x):K(x)]=[K(\beta):K]=deg(\beta)$. Hence $[K(\beta)(x,\alpha):K(x)]=[K(\beta)(x,\alpha):K(\beta)(x)][K(\beta)(x):K(x)]=deg(\alpha)deg(\beta)$. This implies $deg(\beta)=1$, so $\beta\in K$.

Maybe you should show that $[F(x):K(x)]=[F:K]$ for a finite algebraic field extension $F/K$.

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your proof seems work well, but you did not use the hypothesis of $f$ been irreducible over $\overline{K}[X, Y]$, did you? – rla Apr 28 '12 at 18:30
I did. Because the hypothesis we can deduce $[K(\beta)(x,\alpha):K(\beta)(x)]=deg_y(f(x,y))$ which I denote $deg(\alpha)$, if not the hypothesis, this may be false. – wxu Apr 29 '12 at 3:49

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