Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 0 down vote accepted

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 $F/K$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.