[proofplan] The forward direction is immediate: isomorphic fields share the same characteristic and transcendence degree. For the converse, we fix two algebraically closed fields $\Omega_1$ and $\Omega_2$ of the same characteristic with equal transcendence degree over their common prime field $k_0$. We choose transcendence bases of equal cardinality, use them to construct an isomorphism of the purely transcendental subfields, and then extend this isomorphism to the algebraic closures by invoking the uniqueness (up to isomorphism) of algebraic closures. [/proofplan] [step:Verify that isomorphic algebraically closed fields have the same characteristic and transcendence degree] Suppose $\varphi: \Omega_1 \to \Omega_2$ is a field isomorphism. Since $\varphi$ is a ring isomorphism, $\operatorname{char}(\Omega_1) = \operatorname{char}(\Omega_2)$. Let $k_0$ denote the prime field (which is $\mathbb{Q}$ if $\operatorname{char} = 0$ and $\mathbb{F}_p$ if $\operatorname{char} = p > 0$). Since $\varphi$ fixes $k_0$ (every ring homomorphism fixes the prime subfield), $\varphi$ restricts to a bijection between transcendence bases: if $\mathcal{B}$ is a transcendence basis of $\Omega_1$ over $k_0$, then $\varphi(\mathcal{B})$ is a transcendence basis of $\Omega_2$ over $k_0$. Therefore $\operatorname{tr.deg}_{k_0}(\Omega_1) = |\mathcal{B}| = |\varphi(\mathcal{B})| = \operatorname{tr.deg}_{k_0}(\Omega_2)$. [/step] [step:Identify the common prime field and choose transcendence bases of equal cardinality] For the converse, let $\Omega_1$ and $\Omega_2$ be algebraically closed fields with $\operatorname{char}(\Omega_1) = \operatorname{char}(\Omega_2)$ and $\operatorname{tr.deg}_{k_0}(\Omega_1) = \operatorname{tr.deg}_{k_0}(\Omega_2)$, where $k_0$ denotes their common prime field. Let $\mathcal{B}_1 \subset \Omega_1$ and $\mathcal{B}_2 \subset \Omega_2$ be transcendence bases of $\Omega_1$ and $\Omega_2$ over $k_0$, respectively. By hypothesis, $|\mathcal{B}_1| = |\mathcal{B}_2|$. Fix a bijection $\psi: \mathcal{B}_1 \to \mathcal{B}_2$. [/step] [step:Extend the bijection to an isomorphism of purely transcendental subfields] The bijection $\psi: \mathcal{B}_1 \to \mathcal{B}_2$ extends uniquely to a field isomorphism \begin{align*} \Psi: k_0(\mathcal{B}_1) \to k_0(\mathcal{B}_2) \end{align*} between the purely transcendental extensions. To verify this: since $\mathcal{B}_1$ is algebraically independent over $k_0$, the field $k_0(\mathcal{B}_1)$ is the fraction field of the polynomial ring $k_0[\mathcal{B}_1]$. The map $\psi$ induces a ring isomorphism $k_0[\mathcal{B}_1] \to k_0[\mathcal{B}_2]$ (sending each generator $b \in \mathcal{B}_1$ to $\psi(b)$ and fixing $k_0$), which extends to a field isomorphism of their fraction fields. [guided] Why does a bijection between transcendence bases extend to an isomorphism of the corresponding purely transcendental extensions? The point is that a transcendence basis $\mathcal{B}$ over $k_0$ means the elements of $\mathcal{B}$ are algebraically independent over $k_0$, so the polynomial ring $k_0[\mathcal{B}]$ is a free commutative $k_0$-algebra generated by $\mathcal{B}$. Any set map $\psi: \mathcal{B}_1 \to \mathcal{B}_2$ extends uniquely to a $k_0$-algebra homomorphism $k_0[\mathcal{B}_1] \to k_0[\mathcal{B}_2]$ by the universal property of polynomial rings. When $\psi$ is a bijection between algebraically independent sets, this homomorphism is an isomorphism (it has an inverse induced by $\psi^{-1}$). Passing to fraction fields, we get a field isomorphism $\Psi: k_0(\mathcal{B}_1) \to k_0(\mathcal{B}_2)$. This isomorphism $\Psi$ fixes $k_0$ pointwise and sends each element $b \in \mathcal{B}_1$ to $\psi(b) \in \mathcal{B}_2$. [/guided] [/step] [step:Observe that $\Omega_1$ and $\Omega_2$ are algebraic closures of the purely transcendental subfields] Since $\mathcal{B}_1$ is a transcendence basis of $\Omega_1$ over $k_0$, every element of $\Omega_1$ is algebraic over $k_0(\mathcal{B}_1)$. Therefore $\Omega_1 / k_0(\mathcal{B}_1)$ is an algebraic extension. Moreover, $\Omega_1$ is algebraically closed by hypothesis. Hence $\Omega_1$ is an algebraic closure of $k_0(\mathcal{B}_1)$. By the same reasoning, $\Omega_2$ is an algebraic closure of $k_0(\mathcal{B}_2)$. [/step] [step:Extend $\Psi$ to an isomorphism $\Omega_1 \cong \Omega_2$ using uniqueness of algebraic closures] We have a field isomorphism $\Psi: k_0(\mathcal{B}_1) \to k_0(\mathcal{B}_2)$, and $\Omega_i$ is an algebraic closure of $k_0(\mathcal{B}_i)$ for $i = 1, 2$. By the [Extension of Embeddings to Algebraic Closures](/theorems/3322) applied to the embedding $\Psi: k_0(\mathcal{B}_1) \hookrightarrow \Omega_2$ (viewing $\Omega_2$ as the algebraically closed target) with $K = \Omega_1$ (an algebraic extension of $k_0(\mathcal{B}_1)$), there exists a field embedding $\tau: \Omega_1 \hookrightarrow \Omega_2$ extending $\Psi$. It remains to show $\tau$ is surjective. The image $\tau(\Omega_1)$ is an algebraically closed subfield of $\Omega_2$ (since $\Omega_1$ is algebraically closed and $\tau$ is an embedding). Moreover, $\tau(\Omega_1) \supset \tau(k_0(\mathcal{B}_1)) = \Psi(k_0(\mathcal{B}_1)) = k_0(\mathcal{B}_2)$, so $\Omega_2 / \tau(\Omega_1)$ is an algebraic extension (since $\Omega_2$ is algebraic over $k_0(\mathcal{B}_2) \subset \tau(\Omega_1)$). By the [No Proper Algebraic Extensions of an Algebraically Closed Field](/theorems/3320), since $\tau(\Omega_1)$ is algebraically closed and $\Omega_2 / \tau(\Omega_1)$ is algebraic, we have $\Omega_2 = \tau(\Omega_1)$. Therefore $\tau: \Omega_1 \to \Omega_2$ is a field isomorphism. [guided] We have constructed an isomorphism $\Psi: k_0(\mathcal{B}_1) \to k_0(\mathcal{B}_2)$ between the two purely transcendental subfields, and we know that $\Omega_i$ is an algebraic closure of $k_0(\mathcal{B}_i)$. The final step uses the fact that algebraic closures are unique up to isomorphism, but we need to be precise about how the extension works. By the [Extension of Embeddings to Algebraic Closures](/theorems/3322): given a field embedding $\sigma: k \to \Omega$ where $\Omega$ is algebraically closed, and an algebraic extension $K/k$, there exists a field embedding $\tau: K \to \Omega$ extending $\sigma$. We apply this with $k = k_0(\mathcal{B}_1)$, $\sigma = \Psi: k_0(\mathcal{B}_1) \hookrightarrow \Omega_2$ (this is an embedding since $\Psi$ is an isomorphism onto $k_0(\mathcal{B}_2) \subset \Omega_2$), $K = \Omega_1$ (which is algebraic over $k_0(\mathcal{B}_1)$), and $\Omega = \Omega_2$ (which is algebraically closed). The theorem yields an embedding $\tau: \Omega_1 \hookrightarrow \Omega_2$ with $\tau|_{k_0(\mathcal{B}_1)} = \Psi$. But is $\tau$ surjective? Since $\tau$ is a field embedding and $\Omega_1$ is algebraically closed, the image $\tau(\Omega_1)$ is algebraically closed (being isomorphic to $\Omega_1$). Furthermore, $\tau(\Omega_1) \supset k_0(\mathcal{B}_2)$ because $\tau$ extends $\Psi$ and $\Psi$ maps onto $k_0(\mathcal{B}_2)$. Since $\Omega_2$ is algebraic over $k_0(\mathcal{B}_2)$ and $k_0(\mathcal{B}_2) \subset \tau(\Omega_1)$, the extension $\Omega_2 / \tau(\Omega_1)$ is algebraic. By the [No Proper Algebraic Extensions of an Algebraically Closed Field](/theorems/3320), an algebraically closed field admits no proper algebraic extensions. Since $\tau(\Omega_1)$ is algebraically closed, $\Omega_2 = \tau(\Omega_1)$. Therefore $\tau: \Omega_1 \to \Omega_2$ is a field isomorphism, completing the proof. [/guided] [/step]