Is $\mathbb{R}(X+Y)\subseteq\mathbb{R}(X,Y)$ a purely transcendental extension? 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.
 A: 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 $\mathbb{R}(X^2 + Y^2).$
A: 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. 
