I'm new to field theory and face the following exercise:

Give an example of a field extension $F/K$ such that $u,v \in F$ are transcendental over $K$, but $K(u,v) \not\simeq K(X_1, X_2).$ Hint: Consider $v$ over the field $K(u)$.

I don't have any idea of how to even start with this exercise and would appreciate any kind of help.

Thanks for any answers.

  • $\begingroup$ You could take $v=u$, or, say, $u^2$. $\endgroup$ – Orest Bucicovschi Oct 16 '17 at 23:26
  • $\begingroup$ If I understand you correctly you say $v = u^2$, then applies: $K(u,v) =K(u) \simeq K(X_1)$ since u is transcendent. Am I right so far ? But how can I show $K(X_1) \not \simeq K(X_1,X_2)$ ? $\endgroup$ – 3nondatur Oct 16 '17 at 23:44
  • $\begingroup$ Update: I guess you have to take a special field to show that, I think $\mathbb{R}$ and $\mathbb{C}$ are certainly wrong for that task, but what could be a good fit? $\endgroup$ – 3nondatur Oct 16 '17 at 23:51
  • $\begingroup$ $\pi$ and $2\pi$ are both transcendental over $\mathbb{Q}$, but $\mathbb{Q}(\pi, 2\pi)$ is obviously not isomorphic to $\mathbb{Q}(X_1,X_2)$. $\endgroup$ – D_S Oct 16 '17 at 23:58
  • $\begingroup$ Look at the elliptic curve $E: y^2= x^3+x$ whose function field is $\mathbb{Q}(E)\cong\mathbb{Q}(x,\sqrt{x^3+x})\cong \mathbb{Q}(x)[y]/(y^2-x^3-x)$ $\endgroup$ – reuns Oct 17 '17 at 3:06

Too long for a comment: why $K(X)$ is not $K$-isomorphic to $K(X_1, X_2)$ ?

Take $r=R(X) \in K(X) \backslash K$. It is not hard to see that $X$ satisfies an algebraic equation over $K(r)$. Therefore, the extension $K(r)\subset K(X)$ is finite ( it is interesting to find the exact degree, but we won't concern with it now). So, if $r_1$, $r_2$ in $k(X) \backslash K$, the extension $K(r_1)\subset K(r_1, r_2) \subset K(X)$ is finite, and so $r_2$ is algebraically over $K(r_1)$. This is equivalent to $r_1$, $r_2$ algebraically dependent over $K$. Now, this will be true also if one of the $r_i$ is in $K$ ( that is even simpler to see).

But $X_1$, $X_2$ from $K(X_1, X_2)$ are not algebraically dependent over $K$.

(We showed that the two extensions have a different transcendent degree over $K$).

As example to the question, you can take $K= \mathbb{Q}$, and $u=v= \pi$, or any other transcendental number.

  • $\begingroup$ One last question: Why do we know that $K(r_1, r_2) \subset K(X)$ holds, is it because of $r_1, r_2 \in K(x) \backslash K$ ? $\endgroup$ – 3nondatur Oct 17 '17 at 14:40

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