Is there a nice, short and elementary argument that the field extension $\mathbb{R}(X+Y)\subseteq\mathbb{R}(X,Y)$ is purely transcendental?

Obviously, $\mbox{tr deg}_{\mathbb{R}(X+Y)}\mathbb{R}(X,Y)\le1$, because $\mathbb{R}(X+Y)\subseteq\mathbb{R}(X+Y,Y)=\mathbb{R}(X,Y)$, so it is left to show that $Y$ is not algebraic over $\mathbb{R}(X+Y)$.

I don't see any nice proofs of this fact, only some brute force methods of summing degrees of powers of $Y$ in polynomials from $\mathbb{R}(X+Y)[\mathbb{X}]$.

Similar question concerns the transcendence degree of the extension $\mathbb{R}(X^2+Y^2)\subseteq\mathbb{R}(X,Y)$. This extension is not purely transcendal (an easy proof using automorphisms from Galois group). $X$ is algebraic over $\mathbb{R}(X^2)$, so again $\mbox{tr deg}_{\mathbb{R}(X^2+Y^2)}\mathbb{R}(X,Y)\le1$, because $\mathbb{R}(X^2+Y^2)\subseteq\mathbb{R}(X^2+Y^2,Y)=\mathbb{R}(X^2,Y)\subseteq\mathbb{R}(X,Y)$. But how to show that $Y$ is not algebraic over $\mathbb{R}(X^2+Y^2)$?

I don't know algebraic geometry, thus please don't use it in your answer.

  • 2
    $\begingroup$ I would be interested to see your proof "using automorphisms from the Galois group" $\endgroup$ – jspecter Jul 2 '11 at 16:08
  • $\begingroup$ By the "Galois group" I meant the group $Aut(\mathbb{R}(X,Y)/\mathbb{R})$ (the group of automorphisms fixing $\mathbb{R}$). Suppose it is purely transcendental: $\mathbb{R}(X^2+Y^2,F)=\mathbb{R}(X,Y)$ for some $F\in\mathbb{R}(X,Y)$. Take $\varphi\in Aut(\mathbb{R}(X,Y)/\mathbb{R})$ st. $\varphi(X)=X^2+Y^2,\varphi(Y)=F$. Then $X=\varphi^{-1}(X^2+Y^2)=\varphi^{-1}(X)^2+\varphi^{-1}(Y)^2$, so $X$ is a sum of squares, which is impossible. A contradiction. Exactly in the same way one can show eg. that $\mathbb{R}(X^2,Y+Z)\subseteq\mathbb{R}(X,Y,Z)$ is not purely transcendental. Or am I wrong? $\endgroup$ – Damian Sobota Jul 2 '11 at 18:14
  • $\begingroup$ Why is $X$ not sum of 2 squares in $R(X,Y)$ ? I want to say that the function $X : R \times R \rightarrow R$ takes negative values, but not all elements of $R(X,Y)$ are functions on $R^2$. Can you deal with other fields ? $\endgroup$ – user10676 Jul 2 '11 at 18:43
  • $\begingroup$ @user10676: Suppose $x = p_1^2/q_1^2 + p_2^2/q_2^2$. Then $ (q_3)^2 x := (q_1^2 q_2^2) x = (q_2 p_1)^2 + (q_1 p_2)^2$. For this to hold, $q_3(x,y)$ must be zero at every point at which $x$ is negative, but this is a Zariski-dense subset of $\mathbb{R}^2$, so $q_3 = 0$, contradiction. $\endgroup$ – Pete L. Clark Jul 2 '11 at 18:58
  • $\begingroup$ @user10676: Yes, if $X=\varphi^{-1}(X)^2+\varphi^{-1}(Y)^2$, then $X$ as a function into $\mathbb{R}$ would take only nonnegative values. What do you mean by dealing with other fields? $\endgroup$ – Damian Sobota Jul 2 '11 at 19:01

The key to observe is that transcendence degree is additive in towers. Therefore as the transcendence degree of $\mathbb{R}(X,Y)/\mathbb{R}$ is $2$ and $\mathbb{R}(X+Y)/\mathbb{R}$ is $1,$ the transcendence degree of $\mathbb{R}(X,Y)/\mathbb{R}(X+Y)$ is $1$ and $Y$ can satisfy no algebraic relation over $\mathbb{R}(X+Y).$ It follows $\mathbb{R}(X,Y)/\mathbb{R}(X+Y)$ is purely transcendental.

The same method can be used to show $Y$ is not algebraic over $\mathbb{R}(X^2 + Y^2).$

  • $\begingroup$ Perhaps you could expand a little on the "It follows that..."? It seems enough to rewrite $\mathbb{R}(X,Y)$ as $\mathbb{R}(X+Y,Y)$. $\endgroup$ – Pete L. Clark Jul 2 '11 at 16:57
  • $\begingroup$ @Pete L. Clark. Yes, I skipped that step because it was covered in the original posters question. But of course your way is how one would do it. $\endgroup$ – jspecter Jul 2 '11 at 18:42
  • $\begingroup$ $@$jspecter: ah, I see it now. Thanks. $\endgroup$ – Pete L. Clark Jul 2 '11 at 18:45

Let $A=R[Y]$ and $a=Y \in A$. Then $R[X,Y] = A[X]$ and $R[X+Y,Y]=A[X+a]$. The map $$A[X] \rightarrow A[X+a], P(X) \mapsto P(X+a)$$ is an isomorphism, i.e $$R[U,V] \rightarrow R[X+Y,Y], P(U,V) \mapsto P(X+Y,Y)$$ is an isomorphism. So $(X+Y,Y)$ is a transcendental basis of $R(X+Y,Y) =R(X,Y)$.

Now let $A=R[Y]$ and $a=Y^2$. Then $R[X^2+Y^2] = A[X^2+a]$. It is easy to see that the map $A[X] \rightarrow A[X^2+a], P(X) \mapsto P(X^2+a)$ is injective. Which means that $X^2+Y^2$ and $Y$ are algebraicaly indepandant over $R$. The proof that $A[X] \rightarrow A[X^2+a]$ is injective is quite similar to your "brute force method of summing degrees", but maybe it is clearer in this point of view.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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