Show the following properties of $A$ as an $R$-algebra

Question

Let $R = k[X,Y]$, where $k$ is a field. Let $A = R[Z]/\langle XZ-Y\rangle$.

1. 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 ?

2. 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 ?

3. 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

Answer 1

$1.$

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.

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