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It seems to me that this argument is only half right and Hartshorne is also half right and there is a way to correct Hartshorne's statement in a simple way to make it right, so at the end it qualifies indeed to be a typo.

So, everything is obviously fine for $t\neq 0$ and also for $h^0$, but I think that for $t=0$ you get a global nilpotent on $\mathscr X_0$ (the closed subscheme defined by $\mathscr I_0\subset \mathscr O_{\mathbb P^4\times \mathbb A^1}$) from $x_2$. Notice that $x_2$ is obviously not defined on the entire $\mathbb P^4_{a=0}$ and is zero on $\mathbb P^3=(x_2=0)\subset \mathbb P^4_{a=0}$, but $\mathscr X_0\not\subset \mathbb P^3=(x_2=0)$ and it seems to me that $x_2\in H^0(\mathscr X_0,\mathscr O_{\mathscr X_0})$. In any case there are certainly no other global sections, so this means that $h^0(\mathscr X_0,\mathscr O_{\mathscr X_0})=2$ and hence $H^0(\mathbb P^4,\mathscr O_{\mathbb P^4})\to H^0(\mathscr X_0,\mathscr O_{\mathscr X_0})$ is not surjective and has a $1$-dimensional cokernel. Now the fact that $H^1(\mathbb P^4,\mathscr O_{\mathbb P^4})=0$ implies that then $h^1(\mathbb P^4,\mathscr I_0)=1$ as claimed by Hartshorne.

Furthermore, by flatness the Euler characteristic of $\mathscr O_{\mathscr X_0}$ is $1$ (i.e., the arithmetic genus is $0$), it has dimension $1$, so it follows that also $h^1(\mathscr X_0,\mathscr O_{\mathscr X_0})=1$. Again, the fact that $H^2(\mathbb P^4,\mathscr O_{\mathbb P^4})=0$ implies that then $h^2(\mathbb P^4,\mathscr I_0)=1$.

The corresponding calculation for $t\neq 0$ gives $h^1(\mathscr X_t,\mathscr O_{\mathscr X_t})=0$ and so the fact that $H^2(\mathbb P^4,\mathscr O_{\mathbb P^4})=0$ in this case implies that then $h^2(\mathbb P^4,\mathscr I_t)=0$.

This gives us a simple way, in fact two simple ways, to correct the statement.

1. Change $h^0, h^1$ to $h^1, h^2$. The spirit of the problem remains the same.
2. Change $\mathscr I$ to $\mathscr O_{\mathscr X}$ (where $\mathscr X$ is the subscheme defined by $\mathscr I$). Via this correction we get the jump for $h^0, h^1$, but for the cokernel of the ideal sheaf and the numbers are not exactly right, but still the point of the example is there.

I personally like choice #1 just because its proof requires one extra step. I would even make a wild guess that this may have been where the typo has come from: Hartshorne may have made the computation for $\mathscr O_{\mathscr X}$ and then decided to add an additional twist by "moving" the jump to the ideal sheaf, but forgot to correct the cohomology.

Addendum Here is an argument to prove that $x_2$ is indeed a global regular function on $\mathscr X_0$:

Using choa's description of the ideal we have that $$ \mathscr I_0=\mathscr J + (x_0x_2,x_1x_2,x_2^2,x_3x_2,x_4x_2). $$ where $\mathscr J$ is the ideal sheaf of a rational normal quartic curve in $\mathbb P^3_{x_0,x_1,x_3,x_4}$.

Consider the affine charts $U_i=(x_i\neq 0)\subset \mathbb P^3$ for $i=0,1,3,4$. Observe that $U_i\simeq \mathbb A^3$ with coordinates $y_j=\dfrac{x_j}{x_i}$ ($j\neq i$). In these coordinates $\mathscr J$ becomes very simple. For simplifying the notation I will work on $U_0$, but the other charts work the exact same way. So, $\mathscr J|_{U_0}=(y_3-y_1^3, y_4-y_1^4)$ and hence the affine coordinate ring of $\mathscr X_0\cap U_0$ is isomorphic to $k[y_1,x_2]/(y_1x_2,x_2^2)$. In particular, $x_2\in \Gamma(U_0,\mathscr O_{\mathscr X_0})$. Similarly, the affine coordinate ring of $\mathscr X_0\cap U_1$ is isomorphic to $k[y_0, y_0^{-1},x_2]/(y_0x_2,x_2^2)$ and so $x_2\in \Gamma(U_1,\mathscr O_{\mathscr X_0})$, the affine coordinate ring of $\mathscr X_0\cap U_3$ is isomorphic to $k[y_4, y_4^{-1},x_2]/(y_4x_2,x_2^2)$ and so $x_2\in \Gamma(U_3,\mathscr O_{\mathscr X_0})$ and the affine coordinate ring of $\mathscr X_0\cap U_4$ is isomorphic to $k[y_3,x_2]/(y_3x_2,x_2^2)$ and so $x_2\in \Gamma(U_4,\mathscr O_{\mathscr X_0})$.
Therefore $x_2$ is regular on each affine chart of a covering and hence it is a global regular function.

Note that the arithmetic genus of $\mathscr X_0$ is still $0$ since $\chi(\mathscr O_{\mathscr X_0})=h^0-h^1=2-1=1$.

It seems to me that this argument is only half right and Hartshorne is also half right and there is a way to correct Hartshorne's statement in a simple way to make it right, so at the end it qualifies indeed to be a typo.

So, everything is obviously fine for $t\neq 0$ and also for $h^0$, but I think that for $t=0$ you get a global nilpotent on $\mathscr X_0$ (the closed subscheme defined by $\mathscr I_0\subset \mathscr O_{\mathbb P^4\times \mathbb A^1}$) from $x_2$. Notice that $x_2$ is obviously not defined on the entire $\mathbb P^4_{a=0}$ and is zero on $\mathbb P^3=(x_2=0)\subset \mathbb P^4_{a=0}$, but $\mathscr X_0\not\subset \mathbb P^3=(x_2=0)$ and it seems to me that $x_2\in H^0(\mathscr X_0,\mathscr O_{\mathscr X_0})$. In any case there are certainly no other global sections, so this means that $h^0(\mathscr X_0,\mathscr O_{\mathscr X_0})=2$ and hence $H^0(\mathbb P^4,\mathscr O_{\mathbb P^4})\to H^0(\mathscr X_0,\mathscr O_{\mathscr X_0})$ is not surjective and has a $1$-dimensional cokernel. Now the fact that $H^1(\mathbb P^4,\mathscr O_{\mathbb P^4})=0$ implies that then $h^1(\mathbb P^4,\mathscr I_0)=1$ as claimed by Hartshorne.

Furthermore, by flatness the Euler characteristic of $\mathscr O_{\mathscr X_0}$ is $1$ (i.e., the arithmetic genus is $0$), it has dimension $1$, so it follows that also $h^1(\mathscr X_0,\mathscr O_{\mathscr X_0})=1$. Again, the fact that $H^2(\mathbb P^4,\mathscr O_{\mathbb P^4})=0$ implies that then $h^2(\mathbb P^4,\mathscr I_0)=1$.

The corresponding calculation for $t\neq 0$ gives $h^1(\mathscr X_t,\mathscr O_{\mathscr X_t})=0$ and so the fact that $H^2(\mathbb P^4,\mathscr O_{\mathbb P^4})=0$ in this case implies that then $h^2(\mathbb P^4,\mathscr I_t)=0$.

This gives us a simple way, in fact two simple ways, to correct the statement.

1. Change $h^0, h^1$ to $h^1, h^2$. The spirit of the problem remains the same.
2. Change $\mathscr I$ to $\mathscr O_{\mathscr X}$ (where $\mathscr X$ is the subscheme defined by $\mathscr I$). Via this correction we get the jump for $h^0, h^1$, but for the cokernel of the ideal sheaf and the numbers are not exactly right, but still the point of the example is there.

I personally like choice #1 just because its proof requires one extra step. I would even make a wild guess that this may have been where the typo has come from: Hartshorne may have made the computation for $\mathscr O_{\mathscr X}$ and then decided to add an additional twist by "moving" the jump to the ideal sheaf, but forgot to correct the cohomology.

Addendum Here is an argument to prove that $x_2$ is indeed a global regular function on $\mathscr X_0$:

Using choa's description of the ideal we have that $$ \mathscr I_0=\mathscr J + (x_0x_2,x_1x_2,x_2^2,x_3x_2,x_4x_2). $$ where $\mathscr J$ is the ideal sheaf of a rational normal quartic curve in $\mathbb P^3_{x_0,x_1,x_3,x_4}$.

Consider the affine charts $U_i=(x_i\neq 0)\subset \mathbb P^3$ for $i=0,1,3,4$. Observe that $U_i\simeq \mathbb A^3$ with coordinates $y_j=\dfrac{x_j}{x_i}$ ($j\neq i$). In these coordinates $\mathscr J$ becomes very simple. For simplifying the notation I will work on $U_0$, but the other charts work the exact same way. So, $\mathscr J|_{U_0}=(y_3-y_1^3, y_4-y_1^4)$ and hence the affine coordinate ring of $\mathscr X_0\cap U_0$ is isomorphic to $k[y_1,x_2]/(y_1x_2,x_2^2)$. In particular, $x_2\in \Gamma(U_0,\mathscr O_{\mathscr X_0})$. Similarly, the affine coordinate ring of $\mathscr X_0\cap U_1$ is isomorphic to $k[y_0, y_0^{-1},x_2]/(y_0x_2,x_2^2)$ and so $x_2\in \Gamma(U_1,\mathscr O_{\mathscr X_0})$, the affine coordinate ring of $\mathscr X_0\cap U_3$ is isomorphic to $k[y_4, y_4^{-1},x_2]/(y_4x_2,x_2^2)$ and so $x_2\in \Gamma(U_3,\mathscr O_{\mathscr X_0})$ and the affine coordinate ring of $\mathscr X_0\cap U_4$ is isomorphic to $k[y_3,x_2]/(y_3x_2,x_2^2)$ and so $x_2\in \Gamma(U_4,\mathscr O_{\mathscr X_0})$.
Therefore $x_2$ is regular on each affine chart of a covering and hence it is a global section.

Note that the arithmetic genus of $\mathscr X_0$ is still $0$ since $\chi(\mathscr O_{\mathscr X_0})=h^0-h^1=2-1=1$.

It seems to me that this argument is only half right and Hartshorne is also half right and there is a way to correct Hartshorne's statement in a simple way to make it right, so at the end it qualifies indeed to be a typo.

So, everything is obviously fine for $t\neq 0$ and also for $h^0$, but I think that for $t=0$ you get a global nilpotent on $\mathscr X_0$ (the closed subscheme defined by $\mathscr I_0\subset \mathscr O_{\mathbb P^4\times \mathbb A^1}$) from $x_2$. Notice that $x_2$ is obviously not defined on the entire $\mathbb P^4_{a=0}$ and is zero on $\mathbb P^3=(x_2=0)\subset \mathbb P^4_{a=0}$, but $\mathscr X_0\not\subset \mathbb P^3=(x_2=0)$ and it seems to me that $x_2\in H^0(\mathscr X_0,\mathscr O_{\mathscr X_0})$. In any case there are certainly no other global sections, so this means that $h^0(\mathscr X_0,\mathscr O_{\mathscr X_0})=2$ and hence $H^0(\mathbb P^4,\mathscr O_{\mathbb P^4})\to H^0(\mathscr X_0,\mathscr O_{\mathscr X_0})$ is not surjective and has a $1$-dimensional cokernel. Now the fact that $H^1(\mathbb P^4,\mathscr O_{\mathbb P^4})=0$ implies that then $h^1(\mathbb P^4,\mathscr I_0)=1$ as claimed by Hartshorne.

Furthermore, by flatness the Euler characteristic of $\mathscr O_{\mathscr X_0}$ is $1$, it has dimension $1$, so it follows that also $h^1(\mathscr X_0,\mathscr O_{\mathscr X_0})=1$. Again, the fact that $H^2(\mathbb P^4,\mathscr O_{\mathbb P^4})=0$ implies that then $h^2(\mathbb P^4,\mathscr I_0)=1$.

The corresponding calculation for $t\neq 0$ gives $h^1(\mathscr X_t,\mathscr O_{\mathscr X_t})=0$ and so the fact that $H^2(\mathbb P^4,\mathscr O_{\mathbb P^4})=0$ in this case implies that then $h^2(\mathbb P^4,\mathscr I_t)=0$.

This gives us a simple way, in fact two simple ways, to correct the statement.

1. Change $h^0, h^1$ to $h^1, h^2$. The spirit of the problem remains the same.
2. Change $\mathscr I$ to $\mathscr O_{\mathscr X}$ (where $\mathscr X$ is the subscheme defined by $\mathscr I$). Via this correction we get the jump for $h^0, h^1$, but for the cokernel of the ideal sheaf and the numbers are not exactly right, but still the point of the example is there.

I personally like choice #1 just because its proof requires one extra step. I would even make a wild guess that this may have been where the typo has come from: Hartshorne may have made the computation for $\mathscr O_{\mathscr X}$ and then decided to add an additional twist by "moving" the jump to the ideal sheaf, but forgot to correct the cohomology.

It seems to me that this argument is only half right and Hartshorne is also half right and there is a way to correct Hartshorne's statement in a simple way to make it right, so at the end it qualifies indeed to be a typo.

So, everything is obviously fine for $t\neq 0$ and also for $h^0$, but I think that for $t=0$ you get a global nilpotent on $\mathscr X_0$ (the closed subscheme defined by $\mathscr I_0\subset \mathscr O_{\mathbb P^4\times \mathbb A^1}$) from $x_2$. Notice that $x_2$ is obviously not defined on the entire $\mathbb P^4_{a=0}$ and is zero on $\mathbb P^3=(x_2=0)\subset \mathbb P^4_{a=0}$, but $\mathscr X_0\not\subset \mathbb P^3=(x_2=0)$ and it seems to me that $x_2\in H^0(\mathscr X_0,\mathscr O_{\mathscr X_0})$. In any case there are certainly no other global sections, so this means that $h^0(\mathscr X_0,\mathscr O_{\mathscr X_0})=2$ and hence $H^0(\mathbb P^4,\mathscr O_{\mathbb P^4})\to H^0(\mathscr X_0,\mathscr O_{\mathscr X_0})$ is not surjective and has a $1$-dimensional cokernel. Now the fact that $H^1(\mathbb P^4,\mathscr O_{\mathbb P^4})=0$ implies that then $h^1(\mathbb P^4,\mathscr I_0)=1$ as claimed by Hartshorne.

Furthermore, by flatness the Euler characteristic of $\mathscr O_{\mathscr X_0}$ is $1$ (i.e., the arithmetic genus is $0$), it has dimension $1$, so it follows that also $h^1(\mathscr X_0,\mathscr O_{\mathscr X_0})=1$. Again, the fact that $H^2(\mathbb P^4,\mathscr O_{\mathbb P^4})=0$ implies that then $h^2(\mathbb P^4,\mathscr I_0)=1$.

The corresponding calculation for $t\neq 0$ gives $h^1(\mathscr X_t,\mathscr O_{\mathscr X_t})=0$ and so the fact that $H^2(\mathbb P^4,\mathscr O_{\mathbb P^4})=0$ in this case implies that then $h^2(\mathbb P^4,\mathscr I_t)=0$.

This gives us a simple way, in fact two simple ways, to correct the statement.

1. Change $h^0, h^1$ to $h^1, h^2$. The spirit of the problem remains the same.
2. Change $\mathscr I$ to $\mathscr O_{\mathscr X}$ (where $\mathscr X$ is the subscheme defined by $\mathscr I$). Via this correction we get the jump for $h^0, h^1$, but for the cokernel of the ideal sheaf and the numbers are not exactly right, but still the point of the example is there.

I personally like choice #1 just because its proof requires one extra step. I would even make a wild guess that this may have been where the typo has come from: Hartshorne may have made the computation for $\mathscr O_{\mathscr X}$ and then decided to add an additional twist by "moving" the jump to the ideal sheaf, but forgot to correct the cohomology.

Addendum Here is an argument to prove that $x_2$ is indeed a global regular function on $\mathscr X_0$:

Using choa's description of the ideal we have that $$ \mathscr I_0=\mathscr J + (x_0x_2,x_1x_2,x_2^2,x_3x_2,x_4x_2). $$ where $\mathscr J$ is the ideal sheaf of a rational normal quartic curve in $\mathbb P^3_{x_0,x_1,x_3,x_4}$.

Consider the affine charts $U_i=(x_i\neq 0)\subset \mathbb P^3$ for $i=0,1,3,4$. Observe that $U_i\simeq \mathbb A^3$ with coordinates $y_j=\dfrac{x_j}{x_i}$ ($j\neq i$). In these coordinates $\mathscr J$ becomes very simple. For simplifying the notation I will work on $U_0$, but the other charts work the exact same way. So, $\mathscr J|_{U_0}=(y_3-y_1^3, y_4-y_1^4)$ and hence the affine coordinate ring of $\mathscr X_0\cap U_0$ is isomorphic to $k[y_1,x_2]/(y_1x_2,x_2^2)$. In particular, $x_2\in \Gamma(U_0,\mathscr O_{\mathscr X_0})$. Similarly, the affine coordinate ring of $\mathscr X_0\cap U_1$ is isomorphic to $k[y_0, y_0^{-1},x_2]/(y_0x_2,x_2^2)$ and so $x_2\in \Gamma(U_1,\mathscr O_{\mathscr X_0})$, the affine coordinate ring of $\mathscr X_0\cap U_3$ is isomorphic to $k[y_4, y_4^{-1},x_2]/(y_4x_2,x_2^2)$ and so $x_2\in \Gamma(U_3,\mathscr O_{\mathscr X_0})$ and the affine coordinate ring of $\mathscr X_0\cap U_4$ is isomorphic to $k[y_3,x_2]/(y_3x_2,x_2^2)$ and so $x_2\in \Gamma(U_4,\mathscr O_{\mathscr X_0})$.
Therefore $x_2$ is regular on each affine chart of a covering and hence it is a global regular function.

Note that the arithmetic genus of $\mathscr X_0$ is still $0$ since $\chi(\mathscr O_{\mathscr X_0})=h^0-h^1=2-1=1$.

It seems to me that this argument is only half right and Hartshorne is also half right and there is a way to correct Hartshorne's statement in a simple way to make it right, so at the end it qualifies indeed to be a typo.

So, everything is obviously fine for $t\neq 0$ and also for $h^0$, but I think that for $t=0$ you get a global nilpotent on $\mathscr X_0$ (the closed subscheme defined by $\mathscr I_0\subset \mathscr O_{\mathbb P^4\times \mathbb A^1}$) from $x_2$. Notice that $x_2$ is obviously not defined on the entire $\mathbb P^4_{a=0}$ and is zero on $\mathbb P^3=(x_2=0)\subset \mathbb P^4_{a=0}$, but $\mathscr X_0\not\subset \mathbb P^3=(x_2=0)$ and it seems to me that $x_2\in H^0(\mathscr X_0,\mathscr O_{\mathscr X_0})$. In any case there are certainly no other global sections, so this means that $h^0(\mathscr X_0,\mathscr O_{\mathscr X_0})=2$ and hence $H^0(\mathbb P^4,\mathscr O_{\mathbb P^4})\to H^0(\mathscr X_0,\mathscr O_{\mathscr X_0})$ is not surjective and has a $1$-dimensional cokernel. Now the fact that $H^1(\mathbb P^4,\mathscr O_{\mathbb P^4})=0$ implies that then $h^1(\mathbb P^4,\mathscr I_0)=1$ as claimed by Hartshorne.

Furthermore, by flatness the Euler characteristic of $\mathscr O_{\mathscr X_0}$ is $1$, it has dimension $1$, so it follows that also $h^1(\mathscr X_0,\mathscr O_{\mathscr X_0})=1$. Again, the fact that $H^2(\mathbb P^4,\mathscr O_{\mathbb P^4})=0$ implies that then $h^2(\mathbb P^4,\mathscr I_0)=1$.

This gives us a simple way, in fact two simple ways, to correct the statement.

1. Change $h^0, h^1$ to $h^1, h^2$. The spirit of the problem remains the same.
2. Change $\mathscr I$ to $\mathscr O_{\mathscr X}$ (where $\mathscr X$ is the subscheme defined by $\mathscr I$). Via this correction we get the jump for $h^0, h^1$, but for the cokernel of the ideal sheaf and the numbers are not exactly right, but still the point of the example is there.

I personally like choice #1 just because its proof requires one extra step. I would even make a wild guess that this may have been where the typo has come from: Hartshorne may have made the computation for $\mathscr O_{\mathscr X}$ and then decided to add an additional twist by "moving" the jump to the ideal sheaf, but forgot to correct the cohomology.

It seems to me that this argument is only half right and Hartshorne is also half right and there is a way to correct Hartshorne's statement in a simple way to make it right, so at the end it qualifies indeed to be a typo.

So, everything is obviously fine for $t\neq 0$ and also for $h^0$, but I think that for $t=0$ you get a global nilpotent on $\mathscr X_0$ (the closed subscheme defined by $\mathscr I_0\subset \mathscr O_{\mathbb P^4\times \mathbb A^1}$) from $x_2$. Notice that $x_2$ is obviously not defined on the entire $\mathbb P^4_{a=0}$ and is zero on $\mathbb P^3=(x_2=0)\subset \mathbb P^4_{a=0}$, but $\mathscr X_0\not\subset \mathbb P^3=(x_2=0)$ and it seems to me that $x_2\in H^0(\mathscr X_0,\mathscr O_{\mathscr X_0})$. In any case there are certainly no other global sections, so this means that $h^0(\mathscr X_0,\mathscr O_{\mathscr X_0})=2$ and hence $H^0(\mathbb P^4,\mathscr O_{\mathbb P^4})\to H^0(\mathscr X_0,\mathscr O_{\mathscr X_0})$ is not surjective and has a $1$-dimensional cokernel. Now the fact that $H^1(\mathbb P^4,\mathscr O_{\mathbb P^4})=0$ implies that then $h^1(\mathbb P^4,\mathscr I_0)=1$ as claimed by Hartshorne.

Furthermore, by flatness the Euler characteristic of $\mathscr O_{\mathscr X_0}$ is $1$, it has dimension $1$, so it follows that also $h^1(\mathscr X_0,\mathscr O_{\mathscr X_0})=1$. Again, the fact that $H^2(\mathbb P^4,\mathscr O_{\mathbb P^4})=0$ implies that then $h^2(\mathbb P^4,\mathscr I_0)=1$.

The corresponding calculation for $t\neq 0$ gives $h^1(\mathscr X_t,\mathscr O_{\mathscr X_t})=0$ and so the fact that $H^2(\mathbb P^4,\mathscr O_{\mathbb P^4})=0$ in this case implies that then $h^2(\mathbb P^4,\mathscr I_t)=0$.

This gives us a simple way, in fact two simple ways, to correct the statement.

1. Change $h^0, h^1$ to $h^1, h^2$. The spirit of the problem remains the same.
2. Change $\mathscr I$ to $\mathscr O_{\mathscr X}$ (where $\mathscr X$ is the subscheme defined by $\mathscr I$). Via this correction we get the jump for $h^0, h^1$, but for the cokernel of the ideal sheaf and the numbers are not exactly right, but still the point of the example is there.

I personally like choice #1 just because its proof requires one extra step. I would even make a wild guess that this may have been where the typo has come from: Hartshorne may have made the computation for $\mathscr O_{\mathscr X}$ and then decided to add an additional twist by "moving" the jump to the ideal sheaf, but forgot to correct the cohomology.