This comes from my attempt to solve Exercise 1.1c in Hartshorne's Deformation Theory book, which says that a family of conics in $\mathbb{P}^2$ parameterised by some finitely generated $k$-algebra $A$ ($k$ is algebraically closed of course, and let's say characteristic 0 for simplicity), say $X \subseteq \mathbb{P}^2_A$, flat over $A$, is defined by a single degree 2 polynomial. More precisely, the ideal in $A[x,y,z]$ corresponding to $X$ is principal, of degree 2.

There's a somewhat complicated story behind this question; in particular this exercise has been asked on overflow, and the conclusion was that the question is wrong, and is only true locally. To me, that seems like it should be true for any finitely generated local $k$-algebra. However, I came up with the following example over $A = k[t]_{\langle t-1 \rangle}$; consider $$ \operatorname{Proj}A[x,y,z]/\langle tyz - x^2, yz - tx^2 \rangle.$$ Obviously the only closed fibre is $\operatorname{Proj} k[x,y,z]/yz = x^2$, but I am in the awkward situation of not being able to show that the ideal of the $X$ is principally generated, yet also not being able to show it is not flat. Indeed, one can show that $$(t+1)(yz-x^2) = 0$$ and since $t+1$ is invertible in $A$, it follows $yz-x^2 = 0$. However, I am not able to show that $tyz - x^2$ is in the ideal $\langle yz - x^2 \rangle$, and not able to find any torsion element either: obviously we have $$(t-1)(yz - x^2) = 0,$$ but as mentioned $yz-x^2$ is already zero, so this doesn't help!

And on that note, could I please have some hints on how to approach (the local version of) the exercise? Unfortunately the quoted overflow page only has the answer as to why it's not true.

Thanks in advance!

This comes from my attempt to solve Exercise 1.1c in Hartshorne's Deformation Theory book, which says that a family of conics in $\mathbb{P}^2$ parameterised by some finitely generated $k$-algebra $A$ ($k$ is algebraically closed of course, and let's say characteristic 0 for simplicity), say $X \subseteq \mathbb{P}^2_A$, flat over $A$, is defined by a single degree 2 polynomial. More precisely, the ideal in $A[x,y,z]$ corresponding to $X$ is principal, of degree 2.

There's a somewhat complicated story behind this question; in particular this exercise has been asked on overflow, and the conclusion was that the question is wrong, and is only true locally. It seems to me that means it should be true for any finitely generated local $k$-algebra. However, I came up with the following example over $A = k[t]_{\langle t-1 \rangle}$; consider $$ \operatorname{Proj}A[x,y,z]/\langle tyz - x^2, yz - tx^2 \rangle.$$ Obviously the only closed fibre is $\operatorname{Proj} k[x,y,z]/yz = x^2$, but I am in the awkward situation of not being able to show that the ideal of the $X$ is principally generated, yet also not being able to show it is not flat. Indeed, one can show that $$(t+1)(yz-x^2) = 0$$ and since $t+1$ is invertible in $A$, it follows $yz-x^2 = 0$. However, I am not able to show that $tyz - x^2$ is in the ideal $\langle yz - x^2 \rangle$, and not able to find any torsion element either: obviously we have $$(t-1)(yz - x^2) = 0,$$ but as mentioned $yz-x^2$ is already zero, so this doesn't help!

And on that note, could I please have some hints on how to approach (the local version of) the exercise? Unfortunately the quoted overflow page only has the answer as to why it's not true.

Thanks in advance!

In my attempt to solve Exercise 1.1c in Hartshorne's Deformation Theory book, which says that a family of conics in $\mathbb{P}^2$ parameterised by some finitely generated $k$-algebra $A$ ($k$ is algebraically closed of course, and let's say characteristic 0 for simplicity), say $X \subseteq \mathbb{P}^2_A$, flat over $A$, is defined by a single degree 2 polynomial. More precisely, the ideal in $A[x,y,z]$ corresponding to $X$ is principal, of degree 2.

There's a somewhat complicated story behind this question; in particular this exercise has been asked on overflow, and the conclusion was that the question is wrong, and is only true locally. To me, that seems like it should be true for any finitely generated local $k$-algebra. However, I came up with the following example over $A = k[t]_{\langle t-1 \rangle}$; consider $$ \operatorname{Proj}A[x,y,z]/\langle tyz - x^2, yz - tx^2 \rangle.$$ Obviously the only closed fibre is $\operatorname{Proj} k[x,y,z]/yz = x^2$, but I am in the awkward situation of not being able to show that the ideal of the $X$ is principally generated, yet also not being able to show it is not flat. Indeed, one can show that $$(t+1)(yz-x^2) = 0$$ and since $t+1$ is invertible in $A$, it follows $yz-x^2 = 0$. However, I am not able to show that $tyz - x^2$ is in the ideal $\langle yz - x^2 \rangle$, and not able to find any torsion element either: obviously we have $$(t-1)(yz - x^2) = 0,$$ but as mentioned $yz-x^2$ is already zero, so this doesn't help!

And on that note, could I please have some hints on how to approach (the local version of) the exercise? Unfortunately the quoted overflow page only has the answer as to why it's not true.

Thanks in advance!

This comes from my attempt to solve Exercise 1.1c in Hartshorne's Deformation Theory book, which says that a family of conics in $\mathbb{P}^2$ parameterised by some finitely generated $k$-algebra $A$ ($k$ is algebraically closed of course, and let's say characteristic 0 for simplicity), say $X \subseteq \mathbb{P}^2_A$, flat over $A$, is defined by a single degree 2 polynomial. More precisely, the ideal in $A[x,y,z]$ corresponding to $X$ is principal, of degree 2.

There's a somewhat complicated story behind this question; in particular this exercise has been asked on overflow, and the conclusion was that the question is wrong, and is only true locally. To me, that seems like it should be true for any finitely generated local $k$-algebra. However, I came up with the following example over $A = k[t]_{\langle t-1 \rangle}$; consider $$ \operatorname{Proj}A[x,y,z]/\langle tyz - x^2, yz - tx^2 \rangle.$$ Obviously the only closed fibre is $\operatorname{Proj} k[x,y,z]/yz = x^2$, but I am in the awkward situation of not being able to show that the ideal of the $X$ is principally generated, yet also not being able to show it is not flat. Indeed, one can show that $$(t+1)(yz-x^2) = 0$$ and since $t+1$ is invertible in $A$, it follows $yz-x^2 = 0$. However, I am not able to show that $tyz - x^2$ is in the ideal $\langle yz - x^2 \rangle$, and not able to find any torsion element either: obviously we have $$(t-1)(yz - x^2) = 0,$$ but as mentioned $yz-x^2$ is already zero, so this doesn't help!

And on that note, could I please have some hints on how to approach (the local version of) the exercise? Unfortunately the quoted overflow page only has the answer as to why it's not true.

Thanks in advance!