Let $n\geq 1$ be an integer. If $A$ is a ring, then the spectrum of $A[x,y]/(y^n - x^{2n})$ is a well-defined (affine) scheme, say $X\_n$. This scheme describes the "variety" given by the equation $y^n = x^{2n}$. In particular, the parabola $y=x^2$ can be given the structure of a scheme. Note that there is a natural morphism of schemes $X\_2\to X\_1$ given by squaring the coordinates.


Now, what if $n=1/2$? Then can we still make sense of $\sqrt{y} = x$ in some algebraic geometric way? Do we get a "generalized scheme" $X\_{1/2}$ and a natural morphism $X\_{1}\to X\_{1/2}$ by squaring the coordinates?


I have a feeling there should be a way to incorporate these type of equations. I'm merely asking how.


Sorry for the vague question.



The scheme $y^n = x^{2n}$ for $n$ a rational number
BEGIN\_EDIT


in view of the comment, let us assume from the very beginning that $A$ is a field whose characteristic is not equal to $2$.


END\_EDIT


The scheme $X\_1$ is nothing but the affine line: $X\_1 = Spec(A[x]) = 
\mathbb{A}^1$.


The scheme $X\_2$ is nothing but two copies of the affine line: $X\_2 = Spec(A[y]\oplus A[z]) = \mathbb{A}^1\sqcup\mathbb{A}^1$.


Your morphism $X\_2\rightarrow X\_1$ corresponds to the morphism $A[y]\oplus A[z]\rightarrow A[x]$ given by sending both $y$ and $z$ to $x^2$. In geometric terms, on both pieces of $\mathbb{A}^1$ in $X\_2$, it's given by $x\mapsto x^2$.


There is no similar morphism $X\_1\rightarrow X\_{1/2}$, because the scheme $X\_1$, unlike $X\_2$, is irreducible.


The only thing you can do is to give another morphism $\mathbb{A}^1\mapsto\mathbb{A}^1$ sending $x$ to $x^2$.


In your language, this means the following: let $X\_{1/2}$ be the same as $X\_1$, i.e. $Spec(A[x, y]/(y-x^2))$. Then there is a morphism $X\_1\rightarrow X\_{1/2}$ given (in terms of points) by $(x, y)\mapsto(x^2, y^2)$.


