Cohomology is insensitive to field extensions. Let $X$ be a quasicompact separated scheme over a field $k$ so that we have a well-defined Čech cohomology for any coherent sheaf $\mathscr{F}$ on $X.$ For any field extension $k \hookrightarrow K,$ we have $$H^{i}(X, \mathscr{F}) \otimes_{k} K \simeq H^{i}(X_{K}, \mathscr{F}_{K}),$$ where $X_{K} := X \times_{\mathrm{Spec}(k)} \mathrm{Spec}(K)$ and $\mathscr{F}_{K}$ means the pullback of $\mathscr{F}$ under the map $X_{K} \rightarrow X.$
Moral. Taking $K = \bar{k}$ in the above fact, we see that the cohomology (or more precisely, the set of Betti numbers) does not change when we replace $X$ with $X_{\bar{k}}.$ To see why this fact is useful in practice, we consider the following exercise.
Exercise. Let $C$ be an one-dimensional closed subscheme of $\mathbb{P}^{3}_{k}$ (for a given field $k$). If $\deg(C) = 1,$ then $C \simeq \mathbb{P}^{1}_{k}.$
How to reduce to algebraically closed case. Suppose that we know that the statement in the above exercise is true when $k$ is algebraically closed. We will see below that $\deg(C) = \deg(C_{\bar{k}}).$ Since dimension does not change when we base change to $\bar{k}$ (as we can see it affine-locally as transcendence degree), we will then apply the exercise for the algebraically closed field case to prove $C_{\bar{k}} \simeq \mathbb{P}^{1}_{\bar{k}}.$ This does not prove that $C$ is isomorphic to $\mathbb{P}^{1}_{k}$ just yet, but this is great amount of information to deduce something like it.
Base changing the inclusion $i : C \hookrightarrow \mathbb{P}^{3}_{k}$ over $\bar{k}$ gives the inclusion $\bar{i} : C_{\bar{k}} \hookrightarrow \mathbb{P}^{3}_{\bar{k}}$. (Probably, the most painless way to see this is to check this affine locally.) In particular, denoting $\pi_{1} : C_{\bar{k}} \rightarrow C$ and $\pi_{2} : \mathbb{P}^{1}_{\bar{k}} \rightarrow \mathbb{P}^{1}_{k}$ for projection maps, we have $$\pi_{2} \circ \bar{i} = i \circ \pi_{1}.$$ From explicit descriptions of $\mathscr{O}(m)$ of $\mathbb{P}^{3}_{k}$ and $\mathbb{P}^{3}_{\bar{k}},$ we note that $$\mathscr{O}_{\mathbb{P}^{3}_{\bar{k}}}(m) = \pi_{2}^{*}\mathscr{O}_{\mathbb{P}^{3}_{k}}(m).$$ This implies that $$\bar{i}^{*}\mathscr{O}_{\mathbb{P}^{3}_{\bar{k}}}(m) = \pi_{1}^{*}i^{*}\mathscr{O}_{\mathbb{P}^{3}_{k}}(m),$$ which are line bundles over $C_{\bar{k}}.$ Thus, we have (for all $i \geq 0$) $$H^{i}(C_{\bar{k}}, \bar{i}^{*}\mathscr{O}_{\mathbb{P}^{3}_{\bar{k}}}(m)) = H^{i}(C_{\bar{k}}, \pi_{1}^{*}i^{*}\mathscr{O}_{\mathbb{P}^{3}_{k}}(m)) \simeq H^{i}(C, i^{*}\mathscr{O}_{\mathbb{P}^{3}_{k}}(m)) \otimes_{k} \bar{k}.$$ In particular, we have $$h^{i}(C_{\bar{k}}, \bar{i}^{*}\mathscr{O}_{\mathbb{P}^{3}_{\bar{k}}}(m)) = h^{i}(C, i^{*}\mathscr{O}_{\mathbb{P}^{3}_{k}}(m)).$$ Thus, the Hilbert polynomial of $C$ in $\mathbb{P}^{3}_{k}$ is equal to that of $C_{\bar{k}}$ in $\mathbb{P}^{3}_{\bar{k}}.$ In particular, we have $\deg(C) = \deg(C_{\bar{k}}),$ so applying the statement for the base field $\bar{k},$ we have $C_{\bar{k}} \simeq \mathbb{P}^{1}_{\bar{k}}.$ Thus, we have $$\begin{align*}1 &= \deg(C) \\ &= \deg(C_{\bar{k}}) \\ &= \deg(\bar{i}^{*}\mathscr{O}_{\mathbb{P}^{3}_{\bar{k}}}(1)) \\ &= \chi(C_{\bar{k}}, \bar{i}^{*}\mathscr{O}_{\mathbb{P}^{3}_{\bar{k}}}(1)) - \chi(C_{\bar{k}}, \mathscr{O}_{C_{\bar{k}}}) \\ &= \chi(C_{\bar{k}}, i^{*}\mathscr{O}_{\mathbb{P}^{3}_{\bar{k}}}(1)) - \chi(\mathbb{P}^{1}_{\bar{k}}, \mathscr{O}_{\mathbb{P}^{1}_{\bar{k}}}) \\ &= \chi(C_{\bar{k}}, i^{*}\mathscr{O}_{\mathbb{P}^{3}_{\bar{k}}}(1)) - 1. \end{align*}$$ That is, we have $$h^{0}(C_{\bar{k}}, i^{*}\mathscr{O}_{\mathbb{P}^{3}_{\bar{k}}}(1)) - h^{1}(C_{\bar{k}}, i^{*}\mathscr{O}_{\mathbb{P}^{3}_{\bar{k}}}(1)) = \chi(C_{\bar{k}}, i^{*}\mathscr{O}_{\mathbb{P}^{3}_{\bar{k}}}(1)) = 2.$$ Since $C_{\bar{k}} \simeq \mathbb{P}^{1}_{\bar{k}},$ so the genus of $C$ is $0.$ By Serre duality, we have a line bundle $\omega$ on $C_{\bar{k}}$ with degree $-2$ (18.5.A (b) in Vakil) such that $$h^{1}(C_{\bar{k}}, i^{*}\mathscr{O}_{\mathbb{P}^{3}_{\bar{k}}}(1)) = h^{0}(C_{\bar{k}}, \omega \otimes i^{*}\mathscr{O}_{\mathbb{P}^{3}_{\bar{k}}}(-1)) = 0$$ because we have the negative degree (19.2.3 in Vakil) $$\begin{align*}\deg(\omega \otimes i^{*}\mathscr{O}_{\mathbb{P}^{3}_{\bar{k}}}(-1)) &= \deg(\omega) + \deg(i^{*}\mathscr{O}_{\mathbb{P}^{3}_{\bar{k}}}(-1)) \\ &= -2 + \deg(i^{*}\mathscr{O}_{\mathbb{P}^{3}_{\bar{k}}}(-1)) \\ &= -3,\end{align*}$$ where we know the last equality holds because $$\begin{align*}\deg(i^{*}\mathscr{O}_{\mathbb{P}^{3}_{\bar{k}}}(-1)) - 1 &= \deg(i^{*}\mathscr{O}_{\mathbb{P}^{3}_{\bar{k}}}(-1)) + \deg(i^{*}\mathscr{O}_{\mathbb{P}^{3}_{\bar{k}}}(1)) \\ &= \deg(i^{*}\mathscr{O}_{\mathbb{P}^{3}_{\bar{k}}}) \\ &= \deg(\mathscr{O}_{C_{\bar{k}}}) \\ &= 0.\end{align*}.$$ Thus, we have $$h^{0}(C, i^{*}\mathscr{O}_{\mathbb{P}^{3}_{k}}(1)) = h^{0}(C_{\bar{k}}, i^{*}\mathscr{O}_{\mathbb{P}^{3}_{\bar{k}}}(1)) = \chi(C_{\bar{k}}, i^{*}\mathscr{O}_{\mathbb{P}^{3}_{\bar{k}}}(1)) = 2.$$ This may not look like much, but it involves quite a bit. Recall that the line bundle $i^{*}\mathscr{O}_{\mathbb{P}^{3}_{\bar{k}}}(1)$ has four global sections $s_{0}, s_{1}, s_{2}, s_{3} \in \Gamma(C, \mathscr{O}_{C})$ that do not have common have common zeros, which corresponds to the closed embedding: $$i : C \hookrightarrow \mathbb{P}^{3}_{k} = \mathrm{Proj}(k[x_{0}, x_{1}, x_{2}, x_{3}]).$$ Explicitly, for any nonempty affine open $\mathrm{Spec}(A) \subset C,$ this map is associated to the ring surjections (for $i = 0, 1, 2, 3$) $$k[x_{0}/x_{i}, x_{1}/x_{i}, x_{2}/x_{i}, x_{3}/x_{i}] \twoheadrightarrow A$$ given by $x_{j}/x_{i} \mapsto s_{j}/s_{i}.$ Since $h^{0}(C, i^{*}\mathscr{O}_{\mathbb{P}^{3}_{\bar{k}}}(1)) = 2,$ it must be the case that
$$s_{0}, s_{1}, s_{2}, s_{3} \in \Gamma(C, i^{*}\mathscr{O}_{\mathbb{P}^{3}_{\bar{k}}}(1))$$ are $k$-linearly dependent. Thus, we may write $$s_{3} = b_{0}s_{0} + b_{1}s_{1} + b_{2}s_{2}$$ for some $b_{0}, b_{1}, b_{2} \in k.$ Similarly, we see that $s_{0}, s_{1}, s_{2}$ must be $k$-linearly dependent, so we may write $$s_{2} = a_{0}s_{0} + a_{1}s_{1}$$ for some $a_{0}, a_{1} \in k$ as well. This implies that the ring maps $$k[x_{0}/x_{i}, x_{1}/x_{i}] \rightarrow A$$ given by $x_{j}/x_{i} \mapsto s_{j}/s_{i}$ are surjective as well, which induces the closed embedding $$C \hookrightarrow \mathbb{P}^{1}_{k} \subset \mathbb{P}^{3}_{k}.$$ Since $\mathbb{P}^{1}_{k}$ is one-dimensional and irreducible, this closed embedding must be an isomorphism, so we get $C \simeq \mathbb{P}^{1}_{k}.$
Thus, we only need to do the exercise for the case where $k$ is algebraically closed.
Algebraically closed case. If $k$ is algebraically closed, then the above exercise can be done as follows. Pick any two (distinct) closed points $p, q$ on $C,$ which exist because $\dim(C) = 1 > 0.$ Since $k$ is algebriacally closed, by Nullstellensatz, we know $p, q$ are $k$-points. Thus, we may write $$p = [p_{0} : p_{1} : p_{2} : p_{3}] \in \mathbb{P}^{3}(k)$$ and $$q = [q_{0} : q_{1} : q_{2} : q_{3}] \in \mathbb{P}^{3}(k).$$ Then there is a hyperplane $H$ of $\mathbb{P}^{3}_{k}$ that contains $p$ and $q.$
Why does such a hyperplane exist? We must find a $k$-point $a = [a_{0} : a_{1} : a_{2} : a_{3}] \in \mathbb{P}^{3}(k)$ such that $$a_{0}p_{0} + a_{1}p_{1} + a_{2}p_{2} + a_{3}p_{3} = a_{0}q_{0} + a_{1}q_{1} + a_{2}q_{2} + a_{3}q_{3} = 0.$$ These are precisely $k$-points of the intersection of the hyperplanes defined by the two linear equations $$p_{0}x_{0} + p_{1}x_{1} + p_{2}x_{2} + p_{3}x_{3} = 0$$ and $$q_{0}x_{0} + q_{1}x_{1} + q_{2}x_{2} + q_{3}x_{3} = 0.$$ Since $p \neq q,$ the intersection of these two hyperplanes is isomorphic to $\mathbb{P}^{1}_{k},$ which has many $k$-points (i.e., as many as $k \sqcup \{\infty\}$). In particular, there is such an $a = [a_{0} : a_{1} : a_{2} : a_{3}].$ This establishes that there is a hyperplane $H$ in $\mathbb{P}^{3}_{k}$ containing $p, q.$
Remark. The above argument shows that (for an algebraically closed field $k$) there are infinitely many hyperplanes in $\mathbb{P}^{3}_{k}$ passing through $p$ and $q.$ We will use this remark soon.
If $C$ is reduced and irreducible, by Bézout's theorem, we must have $C \subset H$ because otherwise $$\deg(C \cap H) = \deg(C)\deg(H) = 1$$ so that there is one point in the intersection $C \cap H,$ contradicting the fact that there are at least two points in the intersection, namely $p, q \in C \cap H.$
If $C$ is irreducible but not necessarily reduced, then its associated reduced scheme $C'$ is in $H,$ but the underlying topological spaces for $C$ and $C'$ are the same, so $C \subset H$ as well. Since there are infinitely many such $H$ (and they all contain $C$), we must have $C \hookrightarrow \mathbb{P}^{1}_{k}$ as a closed embedding. Thus, we must have $C \simeq \mathbb{P}^{1}_{k}.$
If $C$ is not irreducible, write $$C = C_{1} \cup \cdots \cup C_{r},$$ where each $C_{i}$ is an irreducible component of $C.$ Then (by 18.6.F in Vakil) $$\deg(C_{i}) \leq \deg(C) = 1,$$ and since the Hilbert polynomial of $C_{i}$ has degree equal to $\dim(C_{i}) = 1$ (Theorem 18.6.2 in Vakil), we cannot have $\deg(C_{i}) = 0.$ Since any degree is an integer (18.6.G in Vakil), this implies that $$\deg(C_{i}) = 1.$$ Similarly, any finite union of them has degree $1.$
Again, by Bézout's theorem, we either have $|C_{i} \cap H| = 1$ or $C_{i} \subset H.$ There cannot be more than one $C_{i}$ with $|C_{i} \cap H| = 1,$ because otherwise their union will violate Bézout's theorem. Thus, we have at most one such $C_{i},$ so let us assume that $C_{1}$ is such irreducible component so that $C_{2}, \dots, C_{r} \subset H.$ Writing $C_{1} \cap H = \{p_{1}\},$ we may assume that $p_{1} \neq q.$ Take any hyperplane $H_{1}$ in $\mathbb{P}^{3}_{k}$ that contains $p_{1}$ and $q.$ We may assume $q \in C_{2}.$ Applying Bézout's theorem in $H \simeq \mathbb{P}^{2}_{k},$ we see that $H \cap H_{1}$ meets $C_{1}$ at one point, namely $p_{1},$ and $C_{2}$ at one point, namely $q.$ Since $C_{2} \subset H,$ this means that $H_{1}$ meets $C_{2}$ only at $q,$ which implies that $C_{1} \subset H_{1}.$ We may take another hyperplane $H'_{1}$ that passes $p_{1}$ and $q$ so that $C_{1} \subset H_{1} \cap H'_{1} = H_{1} \cap H \subset H.$ Thus, we have shown that $$C = C_{1} \cup C_{2} \cup \cdots \cup C_{r} \subset H,$$ for any hyperplane $H$ containing $p$ and $q.$ Now, taking another such $H,$ we get $C \hookrightarrow \mathbb{P}^{1}_{k},$ which implies $C \simeq \mathbb{P}^{1}_{k},$ as desired. $\Box$
Why is cohomology insensitive to field extension? Denote by $$\pi : X_{K} = X \times_{\mathrm{Spec}(k)} \mathrm{Spec}(K) \rightarrow X$$ the projection map. For any affine open subset $\mathrm{Spec}(A) \subset X,$ we have $$\begin{align*}\Gamma(\mathrm{Spec}(A) \times_{\mathrm{Spec}(k)} \mathrm{Spec}(K), \pi^{*}\mathscr{F}) &= \Gamma(\mathrm{Spec}(A \otimes_{k} K), \pi^{*}\mathscr{F}) \\ &= \Gamma(\mathrm{Spec}(A) , \mathscr{F}) \otimes_{k} K. \end{align*}$$ Thus, we may construct a Čech cochain complex of $X_{K}$ by applying $(-) \otimes_{k} K$ to a Čech cochain complex of $X.$
Fix any cochain complex $$C : 0 \rightarrow C^{0} \rightarrow C^{1} \rightarrow C^{2} \rightarrow \cdots,$$ where each term is a $k$-vector space. Given a field extension $k \hookrightarrow K,$ we get $$C \otimes_{k} K: 0 \rightarrow C^{0} \otimes_{k} K \rightarrow C^{1} \otimes_{k} K \rightarrow C^{2} \otimes_{k} K \rightarrow \cdots,$$ which is a cochain complex of $K$-vector spaces. Our goal is to show that $$H^{i}(C) \otimes_{k} K \simeq H^{i}(C \otimes_{k} K).$$ The exact sequence $$C^{i-1} \overset{d^{i-1}}{\longrightarrow} \ker(d^{i}) \rightarrow H^{i}(C) \rightarrow 0$$ will stay exact if we apply $(-) \otimes_{k} K$ as it is a right exact functor, so we get the following exact sequence: $$C^{i-1} \otimes_{k} K \overset{d^{i-1} \otimes_{k} K}{\longrightarrow} \ker(d^{i}) \otimes_{k} K \rightarrow H^{i}(C) \otimes_{k} K \rightarrow 0.$$ This implies that $$H^{i}(C) \otimes_{k} K = \frac{\ker(d^{i}) \otimes_{k} K}{\mathrm{im}(d^{i-1} \otimes_{k} K)}.$$ Since $(-) \otimes_{k} K$ is left exact (because $k$ is a field), we can identify $\ker(d^{i}) \otimes_{k} K$ with $\ker(d^{i} \otimes_{k} K).$ Thus, we have $$H^{i}(C) \otimes_{k} K = \frac{\ker(d^{i}) \otimes_{k} K}{\mathrm{im}(d^{i-1} \otimes_{k} K)} \simeq \frac{\ker(d^{i} \otimes_{k} K)}{\mathrm{im}(d^{i-1} \otimes_{k} K)} = H^{i}(C \otimes_{k} K),$$ as desired. $\Box$
Remark. In general, it seems to be true that "cohomology commutes with any exact functor" (1.6.H (c) in Vakil). However, I am not in the mood for chasing diagrams tonight.
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