Is the affine cone of a flat projective scheme again flat? I'm trying to solve Hartshorne Chap.III Ex.9.5.(c):





The biggest problem is to show the existence of a closed subscheme $\tilde{X}\subset \mathbb{P}_{T}^{n+1}$ such that $\tilde{X}_{t} = \operatorname{C}(X_{t})$.

I guess the projective cone $\operatorname{C}(X) \subset \mathbb{P}_{T}^{n+1}$ of $X$ works.

Note that an inclusion $\operatorname{C}(X_{t}) \subset \operatorname{C}(X)_{t}$ always holds but an equality doesn't necessarily hold in a general case.

However, one can easily see that $\operatorname{C}(X_{\eta}) = \operatorname{C}(X)_{\eta}$, where $\eta$ is the generic point of $T$.

Moreover, it is clear that the Hilbert polynomial of $\operatorname{C}(X_{t})$ is independent of $t \in T$.

Hence it suffices to show that the Hilbert polynomial of $\operatorname{C}(X)_{t}$ is independent of $t$, i.e., $\operatorname{C}(X)$ is flat over $T$.

(Then from the inclusion $\operatorname{C}(X_{t}) \subset \operatorname{C}(X)_{t}$ and the agreement of their Hilbert polynomials we conclude that $\operatorname{C}(X_{t}) = \operatorname{C}(X)_{t}$.)

Note that $\operatorname{C}(X)$ is almost everywhere an $\mathbb{A}^{1}-$bundle over $X$, so almost everywhere flat.

Therefore we only have to check the flatness of the affine cone $\operatorname{C}_{\operatorname{aff}}(X)$ of $X$.

So we have reduced the original problem to the following:


  Let $A$ be a noetherian domain and write $T:= \operatorname{Spec}A$.
  
  Let $X \subset \mathbb{P}_{T}^{n}$ be a closed subscheme which is flat
  over $T$.
  
  Then, is the affine cone $\operatorname{Spec} A[x_0,\cdots x_n] /
\operatorname{I}(X)$ flat over $T$?


Thank you!
WARNING: The below is not completely correct and I am in the process of attempting to fix it.

$\renewcommand{\Proj}{\operatorname{Proj}} \renewcommand{\Spec}{\operatorname{Spec}}$ Actually, $C(X_t)\cong C(X)_t$, always. It suffices to work affine-locally on $T$, so suppose $T= \Spec R$ and write $X=\Proj(S_X)$ where $S_X=R[x_0,\cdots,x_n]/\Gamma_*(\mathcal{I}_X)$.
By definition, $C(X) = \Proj(S_X[x_{n+1}])$, $X_t=\Proj(S_X)\times_T \Spec k(t)$, and $C(X)_t = \Proj(S_X[x_{n+1}])\times_T \Spec k(t)$. Since Proj is compatible with base extensions along the degree-zero part of our algebra, we have $X_t\cong \Proj(S_X\otimes_R k(t))$, which is Proj of $$k(t)[x_0,\cdots,x_n]/Im(I_X\otimes_R k(t)\to k(t)[x_0,\cdots,x_n]).$$
Thus $C(X_t)$ is Proj of $$k(t)[x_0,\cdots,x_{n+1}]/(Im(I_X\otimes_R k(t)\to k(t)[x_0,\cdots,x_n])),$$ which can be written as $$\left(k(t)[x_0,\cdots,x_n]/Im(I_X\otimes_R k(t)\to k(t)[x_0,\cdots,x_n])\right)[x_{n+1}],$$ or $(S_X\otimes_R k(t))[x_{n+1}]$, so $C(X_t)\cong \Proj((S_X\otimes_R k(t))[x_{n+1}])$.
On the other hand, $C(X)_t\cong \Proj((S_X[x_{n+1}])\otimes_R k(t))$. As $(S_X\otimes_R k(t))[x_{n+1}]\cong (S_X[x_{n+1}])\otimes_R k(t)$, we see that $C(X_t)\cong C(X)_t$, resolving your difficulty.
WARNING: The below is not completely correct and I am in the process of attempting to fix it.

$\renewcommand{\Proj}{\operatorname{Proj}} \renewcommand{\Spec}{\operatorname{Spec}}$ Actually, $C(X_t)\cong C(X)_t$, always. It suffices to work affine-locally on $T$, so suppose $T= \Spec R$ and write $X=\Proj(S_X)$ where $S_X=R[x_0,\cdots,x_n]/\Gamma_*(\mathcal{I}_X)$.
By definition, $C(X) = \Proj(S_X[x_{n+1}])$, $X_t=\Proj(S_X)\times_T \Spec k(t)$, and $C(X)_t = \Proj(S_X[x_{n+1}])\times_T \Spec k(t)$. Since Proj is compatible with base extensions along the degree-zero part of our algebra, we have $X_t\cong \Proj(S_X\otimes_R k(t))$, which is Proj of $$k(t)[x_0,\cdots,x_n]/Im(I_X\otimes_R k(t)\to k(t)[x_0,\cdots,x_n]).$$
Thus $C(X_t)$ is Proj of $$k(t)[x_0,\cdots,x_{n+1}]/(Im(I_X\otimes_R k(t)\to k(t)[x_0,\cdots,x_n])),$$ which can be written as $$\left(k(t)[x_0,\cdots,x_n]/Im(I_X\otimes_R k(t)\to k(t)[x_0,\cdots,x_n])\right)[x_{n+1}],$$ or $(S_X\otimes_R k(t))[x_{n+1}]$, so $C(X_t)\cong \Proj((S_X\otimes_R k(t))[x_{n+1}])$.
On the other hand, $C(X)_t\cong \Proj((S_X[x_{n+1}])\otimes_R k(t))$. As $(S_X\otimes_R k(t))[x_{n+1}]\cong (S_X[x_{n+1}])\otimes_R k(t)$, we see that $C(X_t)\cong C(X)_t$, resolving your difficulty.