Compute Krull dimension

For $A$ a principal ideal domain, $a\in A$ non-zero, $R:=A[X,Y]/(Y^2-aX)$. Show that $R$ is an integral domain of dimension $\operatorname{dim}(A)+1$.

$R\cong A[X,\sqrt{aX}]$ as $A$-algebras, and $R\cong A[X]\oplus A[X]\sqrt{aX}$ as $A$-modules. For $f(X),f'(X),g(X),g'(X)\in A[X]$, suppose: $$(f(X)+g(X)\sqrt{aX})(f'(X)+g'(X)\sqrt{aX})=0.$$ It follows: $$f(X)f'(X)+aXg(X)g'(X)=0$$ $$f(X)g'(X)+f'(X)g(X)=0.$$ Being $A$ an integral domain, we find $\operatorname{deg}f=\operatorname{deg}g+\frac 12$, $\operatorname{deg}f'=\operatorname{deg}g'+\frac 12$, that is impossible because the degree is a non-negative integer.

I've been thinking to the dimension for a while, but I really don't see what to do. Using directly the definition with a prime ideal of $A[X,Y]$ containing $Y^2-aX$ (to prove that is maximal) doesn't seem useful; I can't even apply the transcendence degree theorem. I'd like to have just a hint, thank you in advance

Sometimes seems easier to handle the dimension of some particular rings by ad-hoc methods, but could not be.

In this case one can use the following general result:

If $S$ is a noetherian ring and $s\in S$ is a non-zerodivisor, then $\dim S\ge\dim S/(s)+1$.

This result is an immediat consequence of Krull Principal Theorem.

In your case we set $S=A[X,Y]$ and $s=Y^2-aX$ and get $\dim R\le(\dim A +2)-1$, that is, $\dim R\le\dim A+1$.

For the converse, one considers the case when $a$ is not invertible. (When $a$ is invertible everything is clear.) Then $R_a\simeq A_a[Y]$ and therefore $\dim R_a=\dim A_a+1$. But $\dim R\ge\dim R_a$, so $\dim R\ge\dim A_a+1$. Since $A$ is PID we have $\dim A_a=\dim A=1$.