The scheme $y^n = x^{2n}$ for $n$ a rational number [closed]

Question

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.

Answer 1

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 $(x, y)\mapsto(x^2, y^2)$.