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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.

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I would be interested to see your proof "using automorphisms from the Galois group" –  jspecter Jul 2 '11 at 16:08
    
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? –  Damian Sobota Jul 2 '11 at 18:14
    
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 ? –  user10676 Jul 2 '11 at 18:43
    
@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. –  Pete L. Clark Jul 2 '11 at 18:58
    
@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? –  Damian Sobota Jul 2 '11 at 19:01

2 Answers 2

up vote 6 down vote accepted

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 $\mathbb{R}(X^2 + Y^2).$

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Perhaps you could expand a little on the "It follows that..."? It seems enough to rewrite $\mathbb{R}(X,Y)$ as $\mathbb{R}(X+Y,Y)$. –  Pete L. Clark Jul 2 '11 at 16:57
    
@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. –  jspecter Jul 2 '11 at 18:42
    
$@$jspecter: ah, I see it now. Thanks. –  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.

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