$\newcommand{\Q}{\mathbb{Q}}$ Problem: Let $A = \Q[x,y,w,z]/(xy-wz)$. Show that $\Q[x,y,w] \subseteq A$ is not an integral extension. Exhibit $A$ as an integral extension of $\mathbb Q[a,b,c]$ for some $a,b,c$ algebraically independent elements.
Noether's Normalization Lemma: Let $k$ be a field and $A$ a finitely generated $k$-algebra. There exist $y_1,\dots,y_r$ algebraically independent over $k$, such that $A$ is integral over $k[y_1,\dots,y_r]$.
Some notation first. If $t_1,\dots,t_n$ are variables and $I = (i_i,\dots,i_n)$ with $i_j \in \mathbb{N}$ a multi-index, we let $|I| = i_1+\dots+i_n$ and $t^I = t_1^{i_1}\dots t_n^{i_n}$. With this notation, if $f \in k[t_1,\dots,t_n]$ then $f = \sum_I a_It^I = \sum_{|I| = e} a_It^I+\sum_{|I|<e}a_It^I$ where $a_I\neq 0$ and $|I| = e$, and we then define $F = \sum_{|I|=e}a_It^I$.
Proof (of Normalization Lemma): Assume that $k$ is infinite. Write $A = k[x_1,\dots,x_n]$ as $A$ is a finitely generated $k$-algebra. Without loss of generality we can assume that $x_1,\dots,x_r$ are algebraically independent and $x_{r+1},\dots,x_n$ are algebraic over $k[x_1,\dots,x_r]$. We induct on $n$, noting that if $n = r$ we are finished. Suppose $n>r$. Because $x_n$ is algebraic over $k[x_1,\dots,x_{n-1}]$ there is some $f \in k[t_1,\dots,t_n]$ for which $f(x_1,\dots,x_n) = 0$ and $f \neq 0$. Since $k$ is infinite, there are some $\lambda_1,\dots,\lambda_{n-1}$ for which $F(\lambda_1,\dots,\lambda_{n-1},1) \neq 0$. Now let $$x_1' = x_1 -\lambda_1x_n,\dots, x_{n-1}'=x_{n-1}-\lambda_{n-1}x_n.$$ Then $0 = f(x_1'+\lambda_1x_n,\dots,x_{n-1}'+\lambda_{n-1}x_n,x_n) = x_n^eF(\lambda_1,\dots,\lambda_{n-1},1)+\Delta$ where $\Delta$ is lower order terms. Because $F(\lambda_1,\dots,\lambda_{n-1},1)$ is non-zero we can invert it, giving a relation $$ 0 = x_n^e+\alpha_{e-1}x_n^{e-1}+\dots+\alpha_0, $$ with $\alpha_i \in k[x_1',\dots,x_{n-1}']$. Therefore $x_n$, and hence $A$ is integral over the subring $k[x_1',\dots,x_{n-1}']$. By induction, there exist $y_1,\dots,y_r \in k[x_1',\dots,x_{n-1}']$ such that $y_i$ are algebraically independent over $k$ and $k[x_1',\dots,x_{n-1}']$ is integral over $k[y_1,\dots,y_r]$. Then $A$ is integral over $A[y_1,\dots,y_r]$.
Attempt: I can show that $A$ is not integral, note that $(w)$ doesn't go up since every ideal of $A$ containing $w$ also contains $x,y$. I know that the second half of this problem should utilize the normalization lemma, but I can't see how. It seems that I am trying to find coefficients analogous to the $x_i'$ defined in the proof, but I don't see where to start. Any hints would be greatly appreciated.
