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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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    @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?2011-07-02

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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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    $@$jspecter: ah, I see it now. Thanks.2011-07-02
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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.