4 deleted 6 characters in body

I have tried to generalize the Exercise referenced by Karl, even though he told me that it shouldn't be possible this way. I think, however, it works:

Edit: I made a mistake concerning $J_i$ - it cannot be equal to $I_i\oplus\bigoplus_{d\ge 1} I_i^dT^d$ because that is not necessarily an ideal - it might not be closed under multiplication by elements from the ring $S$. The version below looks better.

Proposition. Let $Z_0,\ldots,Z_r$ be closed subschemes of a Noetherian scheme $X$ such that $Z_i\not\subset Z_j$ for $i\ne j$. Let $I_i:=I(Z_i)$ and denote by $\tilde{Z}_i$ the respective strict transform of $Z_i$ under the blow-up $\beta:\tilde{X}\to X$ of $X$ along $I:=\sum_{i=0}^rI_i$. Then, $\bigcap_{i=0}^r\tilde{Z}_i=\emptyset$.

Proof. The statement can be checked locally, so we may assume that $X=\mathrm{Spec}(A)$ is affine. Let $f_i:Z_i\hookrightarrow X$ be the respective closed immersion, so $Z_i=\mathrm{Spec}(A/I_i)$ and $f_i^\sharp:A\twoheadrightarrow A/I_i$. Then, the inverse image ideal sheaf of $I$ under $f_i$ is $I\cdot A/I_i$ and hence,

$\displaystyle\tilde{Z}_i=\mathrm{Proj}\left(\bigoplus_{d\ge 0} \left(I\cdot A/I_i\right)^d\cdot T^d\right)$

With $S=\bigoplus_{d\ge 0} I^d\cdot T^d$, the homogeneous ideal defining $\tilde{Z}_i$ inside $\tilde{X}=\mathrm{Proj}(S)$ is equal to

$\displaystyle J_i = \bigoplus_{d\ge 0} (I^d\cap I\cap I_i)$

In particular, $J_0+\cdots+J_r\supseteq S_+$, so any point $P\in\tilde{Z}_0\cap\cdots\cap\tilde{Z}_r$ would correspond to a homogeneous prime ideal containing each of the $J_i$ and hence, the irrelevant ideal. There is no such point.

Did I miss something? Or is this correct?

3 Yet another correction.

I have tried to generalize the Exercise referenced by Karl, even though he told me that it shouldn't be possible this way. I think, however, it works:

Edit: I made a mistake concerning $J_i$ - it cannot be equal to $I_i\oplus\bigoplus_{d\ge 1} I_i^dT^d$ because that is not necessarily an ideal - it might not be closed under multiplication by elements from the ring $S$. The version below looks better.

Proposition. Let $Z_0,\ldots,Z_r$ be closed subschemes of a Noetherian scheme $X$ such that $Z_i\not\subset Z_j$ for $i\ne j$. Let $I_i:=I(Z_i)$ and denote by $\tilde{Z}_i$ the respective strict transform of $Z_i$ under the blow-up $\beta:\tilde{X}\to X$ of $X$ along $I:=\sum_{i=0}^rI_i$. Then, $\bigcap_{i=0}^r\tilde{Z}_i=\emptyset$.

Proof. The statement can be checked locally, so we may assume that $X=\mathrm{Spec}(A)$ is affine. Let $f_i:Z_i\hookrightarrow X$ be the respective closed immersion, so $Z_i=\mathrm{Spec}(A/I_i)$ and $f_i^\sharp:A\twoheadrightarrow A/I_i$. Then, the inverse image ideal sheaf of $I$ under $f_i$ is $I\cdot A/I_i$ and hence,

$\displaystyle\tilde{Z}_i=\mathrm{Proj}\left(\bigoplus_{d\ge 0} \left(I\cdot A/I_i\right)^d\cdot T^d\right)$

With $S=\bigoplus_{d\ge 0} I^d\cdot T^d$, the homogeneous ideal $J_i$ defining $\tilde{Z}_i$ inside $\tilde{X}=\mathrm{Proj}(S)$ is the one generated by equal to

$I_i\oplus I_iT$.\displaystyle J_i = \bigoplus_{d\ge 0} (I^d\cap I\cap I_i)$ In particular,$J_0+\cdots+J_r\supseteq S_+$because it is generated by$(I_0+\cdots+I_r)T=S_1$, , so any point$P\in\tilde{Z}_0\cap\cdots\cap\tilde{Z}_r$would correspond to a homogeneous prime ideal containing each of the$J_i$and hence, the irrelevant ideal. There is no such point. Did I miss something? Or is this correct? 2 Corrected a mistake in the proof. I have tried to generalize the Exercise referenced by Karl, even though he told me that it shouldn't be possible this way. I think, however, it works: Edit: I made a mistake concerning$J_i$- it cannot be equal to$I_i\oplus\bigoplus_{d\ge 1} I_i^dT^d$because that is not necessarily an ideal - it might not be closed under multiplication by elements from the ring$S$. The version below looks better. Proposition. Let$Z_0,\ldots,Z_r$be closed subschemes of a Noetherian scheme$X$such that$Z_i\not\subset Z_j$for$i\ne j$. Let$I_i:=I(Z_i)$and denote by$\tilde{Z}_i$the respective strict transform of$Z_i$under the blow-up$\beta:\tilde{X}\to X$of$X$along$I:=\sum_{i=0}^rI_i$. Then,$\bigcap_{i=0}^r\tilde{Z}_i=\emptyset$. Proof. The statement can be checked locally, so we may assume that$X=\mathrm{Spec}(A)$is affine. Let$f_i:Z_i\hookrightarrow X$be the respective closed immersion, so$Z_i=\mathrm{Spec}(A/I_i)$and$f_i^\sharp:A\twoheadrightarrow A/I_i$. Then, the inverse image ideal sheaf of$I$under$f_i$is$I\cdot A/I_i$and hence, $\displaystyle\tilde{Z}_i=\mathrm{Proj}\left(\bigoplus_{d\ge 0} \left(I\cdot A/I_i\right)^d\right)$A/I_i\right)^d\cdot T^d\right)$

With $S=\bigoplus_{d\ge 0} I^d$I^d\cdot T^d$, the homogeneous ideal$J_i$defining$\tilde{Z}_i$inside$\tilde{X}=\mathrm{Proj}(S)$is equal to the one generated by$\displaystyle J_i := I_i\oplus\bigoplus_{d\ge 1} I_i^d.$Observe I_i\oplus I_iT$.

In particular, $J_0+\cdots+J_r\supseteq S_+$ , because it is generated by $(I_0+\cdots+I_r)T=S_1$, so any point $P\in\tilde{Z}_0\cap\cdots\cap\tilde{Z}_r$ would correspond to a homogeneous prime ideal containing each of the $J_i$ and hence, the irrelevant ideal. There is no such point.

Did I miss something? Or is this correct?

1