Calculating the integral closure (normalizatrion) of a quotient over a polynomial ring in three variables. Let $A=\mathbb{Q}[x,y,z]/((x+a)^2-z(y+z^b)^2)$ for some $a,b\in\mathbb{N}$. Assume that $A$ is an integral domain (this is easy to show, you can just use Generalised Eisenstein over $\mathbb{Q}[y,z][x+a]$ with prime ideal $(z)$ I think). The question is to compute the integral closure of $A$.
My initial guess was to find an integral element say $t$ not in $A$ and adjoin to find $A[t]$ and hopefully this would be my integral closure. Letting $t=\frac{x+a}{y+z^b}$ seems to be the obvious choice, and belongs to $Frac(A)$.
My attempt: Shift the variables of $A$ to be over $u,v,z$ where $u=x+a$, $v=y+z^b$ (First question, why can I so this? I haven't found a reason I trust enough, though I'm sure its just isomorphisms that allow me to do this? Is there a way to do this without shifting?).
Then I have the string of isomorphisms as below:
\begin{align*}
    A[t] \cong A\left[\frac{u}{v}\right] &\cong \left(\mathbb{Q}[u, v, z] /\left(u^{2}-z v^{2}\right)\right)\left[\frac{u}{v}\right]\\
    &\cong \mathbb{Q}\left[u, v, \frac{u^2}{v^2}\right]\left[\frac{u}{v}\right]\\
    &\cong \mathbb{Q}\left[v,\frac{u}{v} \right]
\end{align*}
Now clearly $A[t]$ is integrally closed if you note that $v, u/v$ is independent over $\mathbb{Q}$ and thus we have it is isomorphic to the UFD $\mathbb{Q}\left[v,\frac{u}{v} \right]$. My second question, is why exactly is this $A[t]$ the integral closure? Is it becasue $A[t]$ is contained in the integral closure of $A$ or is it other way around, i.e. it contains it. And why? It doesn't seem obvious to me.
Lastly, is there an easier/simpler way to tackle this? My course is working through Atiyah and MacDonald's Commutative Algebra.
(I haven't had the most solid basis for ring theory, so even simple explanations are helpful.)
 A: You can shift the variables in such a way by what is called the universal property of polynomial rings: a ring homomorphism from a polynomial ring $R[X_i:i\in I]$ is completely determined on $R$ and $X_i$.
We have two morphisms $\Psi, \Phi$ between polynomial rings $\mathbb{Q}[x,y,z] \rightleftarrows \mathbb{Q}[u,v,w]$ such that
$$\Psi(x)=u-a, \Psi(y)=v-w^b, \Psi(z)=w,$$
$$\Phi(u)=x-a, \Phi(v)=y+z^b, \Phi(w)=z.$$
Since $\Psi\circ\Phi$ and $\Phi\circ\Psi$ map indeterminates to themselves, they must be identity, i.e. $\Psi$ is an isomorphism. It follows that $\Psi$ maps the ideal $((x+a)^2-z(y+z^b)^2)$ to $(u^2-v^2w)$ and by quotienting we get $A\simeq \mathbb{Q}[u,v,w]/(u^2-v^2w)=\mathbb{Q}[u,v,u^2/v^2]$. We can see that $B=\mathbb{Q}[u,u/v]$ is integral over $A$ since it is obtained by adjoining an integral element $\in \operatorname{Frac}A$ so that $B 
$ is contained in the integral closure, and it is integrally closed since it is UFD. As in the case of fields, if we have a tower of ring extensions $A\subset B\subset \operatorname{Frac}A=\operatorname{Frac} B$, where $A\subset B$ is integral extension, and an element $a\in \operatorname{Frac}A$ integral over $A$, then $a$ is also integral over $B$ so that $a\in B$. This shows $B$ contains eny element $\in \operatorname{Frac}A$ integral over $A$, i.e. contains the integral closure.
Lastly, this is the most usual and common procedure to find the integral clousure of an integral domain as far as I know.
