Show the following properties of $A$ as an $R$-algebra 
Let $R = k[X,Y]$, where $k$ is a field. Let $A = R[Z]/\langle XZ-Y\rangle$.



*

*I wish to show that $A$ is a finitely generated $R$-algebra.
So my thoughts are as follows; We could try to consider a "simpler" problem $A' = R[Z]$ and ask the same question. Then by definition I need to produce a surjective homomorphism $R[t_1] \rightarrow A'$, but this is easy, just send $X \mapsto X, Y\mapsto Y, t_1\mapsto Z$. Since this "bigger" ring is finitely generated as an $R$-algebra, then can I conclude the same for the quotient ?

*I wish to show $A$ is not finitely generated as an $R$-module.
I am having some trouble seeing this; are there any things I can always consider for such problems ?

*I wish to show $A$ is not a flat $R$-module.
I need to come up with an injective map that after tensoring with $A$, fails injectivity; once again, I'm stuck in this regard.
Any help  or hints is deeply appreciated.
Cheers
 A: $1.$
Yes, from $A'$ being finitely generated as an $R$-algebra it follows that $A$ is finitely generated as an $R$-algebra: in fact, can you find a surjective homomorphism from $R[t]$ to $R[Z]/\left<XZ-Y\right>$?
$2.$
Assume $A$ is a finitely generated $R$-module. let $x_1,\ldots,x_n$ be generators. Then notice that the $R$-linear combinations of $x_i$ cannot have degree higher than $\max\deg_Z(x_i)$.
$3.$
Let's see what happens when $A$ is assumed to be flat.
Let $I\subseteq R$ be the ideal generated by $X, Y$.
Consider the map $\varphi: R^{\oplus 2}\rightarrow I$ sending $(a, b)\in R^{\oplus
  2}$ to $X\cdot a-Y\cdot b\in I$. Denote $\operatorname{Ker}(\varphi)$ as
$K$.
Then there is an exact sequence
$$0\rightarrow K \rightarrow R^{\oplus 2} \rightarrow I\rightarrow 0.$$
Since $A$ is assumed to be flat, we obtain the exact sequence
$$
  0 \rightarrow K\otimes_{R}A\rightarrow A^{\oplus 2} \rightarrow I\otimes
  A\rightarrow 0.
$$
Now consider the element $x:=X\otimes_{R}\overline Z-Y\otimes_{R}1\in
I\otimes_{R}A$, where $\overline Z$ is the image of $Z$ in $A$.
Clearly $x$ is sent to $0$ in $R\otimes_{R}A\cong A$. Since $A$ is
assumed to be flat, the map $I\otimes_{R}A\rightarrow R\otimes_{R}A$ is an
injection. Thus $x=0$ in $I\otimes_{R}A$.
We also see that $(\varphi\otimes A)((\overline Z, 1)) = x = 0$. By the
above-mentioned exact sequence, this means $(\overline Z, 1)\in K\otimes_{R}A$.
As a consequence, there exist finitely many elements $(a_{i}, b_{i})\in K$ and
$\alpha_{i}\in A$ such that $\sum (a_{i}, b_{i})\otimes_{R}\alpha_{i}$ is mapped
to $(\overline Z, 1)\in A^{\oplus 2}$.
Since $(a_{i}, b_{i})\in K$, we have $X\cdot a_{i} - Y\cdot b_{i}=0,\,\forall
i$. Moreover, the assumption that $\sum (a_{i}, b_{i})\otimes_{R}\alpha_{i}$ is
mapped to $(\overline Z, 1)$ means that
$$
  \sum_{i} a_{i}\alpha_{i}=\overline Z,\quad
  \sum_{i} b_{i}\alpha_{i}=1.
$$
However, $X\cdot a_{i} - Y\cdot b_{i}=0$ implies that $X$ divides $b_{i}$.
Thus $X$ generates $A$ as an $A$-ideal. But this is impossible, as
$$
  A/XA \cong R[Z]/\left<X, XZ-Y\right>\cong k[Z]\not\cong 0.
$$
Therefore $A$ is not flat.
Remark:
We can produce non-flat modules easily by finding non-trivial relations, and then taking the quotients by them.
This proof is derived from Stacks Project Lemma 00HK. But I think maybe it is too much as a reference, so I merged the necessary parts in the lemma into the proof.

If any error is found, please inform me, thanks.
