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The answer is the following and can be found in Hartshorne's book Deformation Theory, see in particular Exercise 10.5 page 83.

We work over an algebraically closed field $k$. Then there is an exact sequence of sheaves $$0 \to \beta_*T_X \to T_S \to k_s \oplus k_s \to 0,$$ inducing an exact sequence in cohomology $$0 \to H^0(X, \, T_X) \to H^0(S, \, T_S) \to k \oplus k \to H^1(X, \, T_X) \to H^1(S, \, T_S) \to 0$$ and an isomorphism $H^2(X, \, T_X) \cong H^2(S, \, T_S)$.

We can interpret this as follows.

First of all, the obstructions to first-order deformations of $X$ are the same as the obstructions to first-order deformations of $S$.

Next, the term $k \oplus k$ corresponds to the deformations of the point $s$ inside $S$. Then if the group of infinitesimal automorphisms of $S$ (that is, the term $H^0(S, \, T_S))$ maps surjectively onto the deformations of $s$ into $S$, then $H^1(X, \, T_X) \cong H^1(S, \, T_S)$, i.e. the first-order deformations of $X$ are just given by first-order deformations of $S$. Otherwise, moving $s$ gives nontrivial deformations of $S$.

The first case happens for instance when $S$ is an abelian surface. Then $\textrm{Aut}(S)$ is transitive, hence the map $H^0(S, \, T_S) \to k \oplus k$ is an isomorphism and this implies $$h^0(X, \, T_X)=0, \quad h^1(X, \, T_X) = h^1(S, \, T_S)=4.$$

The second case happens for instance when $S$ is a surface of general type. Then it is possible to prove that $S$ has at most finitely many automorphisms, hence $h^0(S, \, T_S)=0$. Then $$h^0(X, \, T_X)=0, \quad h^1(X, \, T_X) = h^1(S, \,T_S)+2.$$

The answer is the following and can be found in Hartshorne's book Deformation Theory, see in particular Exercise 10.5 page 83.

We work over an algebraically closed field $k$. Then there is an exact sequence of sheaves $$0 \to \beta_*T_X \to T_S \to k_s \oplus k_s \to 0,$$ inducing an exact sequence in cohomology $$0 \to H^0(X, \, T_X) \to H^0(S, \, T_S) \to k \oplus k \to H^1(X, \, T_X) \to H^1(S, \, T_S) \to 0$$ and an isomorphism $H^2(X, \, T_X) \cong H^2(S, \, T_S)$.

We can interpret this as follows.

First of all, the obstructions to first-order deformations of $X$ are the same as the obstructions to first-order deformations of $S$.

Next, the term $k \oplus k$ corresponds to the deformations of the point $s$ inside $S$. Then if the group of infinitesimal automorphisms of $S$ (that is, the term $H^0(S, \, T_S))$ maps surjectively onto the deformations of $s$ into $S$, then $H^1(X, \, T_X) \cong H^1(S, \, T_S)$, i.e. the first-order deformations of $X$ are just given by first-order deformations of $S$. Otherwise, moving $s$ gives nontrivial deformations of $S$.

The first case happens for instance when $S$ is an abelian surface. Then $\textrm{Aut}(S)$ is transitive, hence the map $H^0(S, \, T_S) \to k \oplus k$ is an isomorphism and this implies $$h^0(X, \, T_X)=0, \quad h^1(X, \, T_X) = h^1(S, \, T_S)=4.$$

The second case happens for instance when $S$ is a surface of general type. Then it is possible to prove that $S$ has at most finitely many automorphisms, hence $h^0(S, \, T_S)=0$. This yields $$h^0(X, \, T_X)=0, \quad h^1(X, \, T_X) = h^1(S, \,T_S)+2.$$

The answer is the following and can be found in Hartshorne's book Deformation Theory, see in particular Exercise 10.5 page 83.

We work over an algebraically closed field $k$. Then there is an exact sequence of sheaves $$0 \to \beta_*T_X \to T_S \to k_s \oplus k_s \to 0,$$ inducing an exact sequence in cohomology $$0 \to H^0(X, \, T_X) \to H^0(S, \, T_S) \to k_s \oplus k_s \to H^1(X, \, T_X) \to H^1(S, \, T_S) \to 0$$ and an isomorphism $H^2(X, \, T_X) \cong H^2(S, \, T_S)$.

We can interpret this as follows.

First of all, the obstructions to first-order deformations of $X$ are the same as the obstructions to first-order deformations of $S$.

Next, the term $k_s \oplus k_s$ corresponds to the deformations of the point $s$ inside $S$. Then if the group of infinitesimal automorphisms of $S$ (that is, the term $H^0(S, \, T_S))$ maps surjectively onto the deformations of $s$ into $S$, then $H^1(X, \, T_X) \cong H^1(S, \, T_S)$, i.e. the first-order deformations of $X$ are just given by first-order deformations of $S$. Otherwise, moving $s$ gives nontrivial deformations of $S$.

The first case happens for instance when $S$ is an abelian surface. Then $\textrm{Aut}(S)$ is transitive, hence the map $H^0(S, \, T_S) \to k_s \oplus k_s$ is an isomorphism and this implies $$h^0(X, \, T_X)=0, \quad h^1(X, \, T_X) = h^1(S, \, T_S)=4.$$

The second case happens for instance when $S$ is a surface of general type. Then it is possible to prove that $S$ has at most finitely many automorphisms, hence $h^0(S, \, T_S)=0$. Then $$h^0(X, \, T_X)=0, \quad h^1(X, \, T_X) = h^1(S, \,T_S)+2.$$

The answer is the following and can be found in Hartshorne's book Deformation Theory, see in particular Exercise 10.5 page 83.

We work over an algebraically closed field $k$. Then there is an exact sequence of sheaves $$0 \to \beta_*T_X \to T_S \to k_s \oplus k_s \to 0,$$ inducing an exact sequence in cohomology $$0 \to H^0(X, \, T_X) \to H^0(S, \, T_S) \to k \oplus k \to H^1(X, \, T_X) \to H^1(S, \, T_S) \to 0$$ and an isomorphism $H^2(X, \, T_X) \cong H^2(S, \, T_S)$.

We can interpret this as follows.

First of all, the obstructions to first-order deformations of $X$ are the same as the obstructions to first-order deformations of $S$.

Next, the term $k \oplus k$ corresponds to the deformations of the point $s$ inside $S$. Then if the group of infinitesimal automorphisms of $S$ (that is, the term $H^0(S, \, T_S))$ maps surjectively onto the deformations of $s$ into $S$, then $H^1(X, \, T_X) \cong H^1(S, \, T_S)$, i.e. the first-order deformations of $X$ are just given by first-order deformations of $S$. Otherwise, moving $s$ gives nontrivial deformations of $S$.

The first case happens for instance when $S$ is an abelian surface. Then $\textrm{Aut}(S)$ is transitive, hence the map $H^0(S, \, T_S) \to k \oplus k$ is an isomorphism and this implies $$h^0(X, \, T_X)=0, \quad h^1(X, \, T_X) = h^1(S, \, T_S)=4.$$

The second case happens for instance when $S$ is a surface of general type. Then it is possible to prove that $S$ has at most finitely many automorphisms, hence $h^0(S, \, T_S)=0$. Then $$h^0(X, \, T_X)=0, \quad h^1(X, \, T_X) = h^1(S, \,T_S)+2.$$

The answer is the following and can be found in Hartshorne's book Deformation Theory, see in particular Exercise 10.5 page 83.

We work over an algebraically closed field $k$. Then there is an exact sequence of sheaves $$0 \to \beta_*T_X \to T_S \to k \oplus k \to 0,$$ inducing an exact sequence in cohomology $$0 \to H^0(X, \, T_X) \to H^0(S, \, T_S) \to k \oplus k \to H^1(X, \, T_X) \to H^1(S, \, T_S) \to 0$$ and an isomorphism $H^2(X, \, T_X) \cong H^2(S, \, T_S)$.

We can interpret this as follows.

First of all, the obstructions to first-order deformations of $X$ are the same as the obstructions to first-order deformations of $S$.

Next, the term $k \oplus k$ corresponds to the deformations of the point $s$ inside $S$. Then if the group of infinitesimal automorphisms of $S$ (that is, the term $H^0(S, \, T_S))$ maps surjectively onto the deformations of $s$ into $S$, then $H^1(X, \, T_X) \cong H^1(S, \, T_S)$, i.e. the first-order deformations of $X$ are just given by first-order deformations of $S$. Otherwise, moving $s$ gives nontrivial deformations of $S$.

The first case happens for instance when $S$ is an abelian surface. Then $\textrm{Aut}(S)$ is transitive, hence the map $H^0(S, \, T_S) \to k \oplus k$ is an isomorphism and this implies $$h^0(X, \, T_X)=0, \quad h^1(X, \, T_X) = h^1(S, \, T_S)=4.$$

The second case happens for instance when $S$ is a surface of general type. Then it is possible to prove that $S$ has at most finitely many automorphisms, hence $h^0(S, \, T_S)=0$. Then $$h^0(X, \, T_X)=0, \quad h^1(X, \, T_X) = h^1(S, \,T_S)+2.$$

The answer is the following and can be found in Hartshorne's book Deformation Theory, see in particular Exercise 10.5 page 83.

We work over an algebraically closed field $k$. Then there is an exact sequence of sheaves $$0 \to \beta_*T_X \to T_S \to k_s \oplus k_s \to 0,$$ inducing an exact sequence in cohomology $$0 \to H^0(X, \, T_X) \to H^0(S, \, T_S) \to k_s \oplus k_s \to H^1(X, \, T_X) \to H^1(S, \, T_S) \to 0$$ and an isomorphism $H^2(X, \, T_X) \cong H^2(S, \, T_S)$.

We can interpret this as follows.

First of all, the obstructions to first-order deformations of $X$ are the same as the obstructions to first-order deformations of $S$.

Next, the term $k_s \oplus k_s$ corresponds to the deformations of the point $s$ inside $S$. Then if the group of infinitesimal automorphisms of $S$ (that is, the term $H^0(S, \, T_S))$ maps surjectively onto the deformations of $s$ into $S$, then $H^1(X, \, T_X) \cong H^1(S, \, T_S)$, i.e. the first-order deformations of $X$ are just given by first-order deformations of $S$. Otherwise, moving $s$ gives nontrivial deformations of $S$.

The first case happens for instance when $S$ is an abelian surface. Then $\textrm{Aut}(S)$ is transitive, hence the map $H^0(S, \, T_S) \to k_s \oplus k_s$ is an isomorphism and this implies $$h^0(X, \, T_X)=0, \quad h^1(X, \, T_X) = h^1(S, \, T_S)=4.$$

The second case happens for instance when $S$ is a surface of general type. Then it is possible to prove that $S$ has at most finitely many automorphisms, hence $h^0(S, \, T_S)=0$. Then $$h^0(X, \, T_X)=0, \quad h^1(X, \, T_X) = h^1(S, \,T_S)+2.$$