Prove that two algebraically closed fields of the same characteristic are isomorphic if and only if they have the same transcendence degree over their prime fields.

The prime field is isomorphism to $\mathbb{Q}$ or $\mathbb{F}_p.$ Denote the prime field by $k$. Then there is a injective $\phi:F\rightarrow \bar k$, for some algebraic extension $F/k$. By the number of transcendental elements.

If $M$ isomorphic to $N$. So they have the same transcendental elements (under isomorphism), tr-deg$M/k=$ tr-deg$N/k$?

For the converse. tr-deg$M/k=$ tr-deg$N/k$. Then $M$ and $N$ have the same number of transcendental elements. This leads to a isomorphism.

Is this right? Or what's the right way?


1 Answer 1


Let $M, N$ be isomorphic algebraically closed fields over the prime field $k$. For simplicity we assume that $k \subseteq M, N$. Say $\sigma : M \mapsto N$ is our isomorphism. Notice that $\sigma$ preserves the prime field $k$. Moreover, the image of a transcendence basis of $M$ over $k$ is a transcendence basis of $N$ over $k$ (see proof bellow). Hence, $M$ isomorphic to $N$ implies that $M$ and $N$ have the same transcendence degree.

Let $t_1, \dots, t_n$ be a transcendence basis of $M$ over $k$.
First, we check that $\sigma(t_1), \dots, \sigma(t_n)$ is algebraically independent over $k$ :

If not, there is a non zero polynomial $P \in k[X_1, \dots, X_n]$ such that $P(\sigma(t_1), \dots, \sigma(t_n) ) = 0$. But since $P$ has coefficients in $k$, this implies that $P(t_1, \dots, t_n) = 0$, a contradiction!

Then, we show that any element in $N$ is algebraic over $\sigma(t_1), \dots, \sigma(t_n)$ :
Let $b \in N$. The element $a := \sigma^{-1}(b)$ is algebraic over $t_1, \dots, t_n$ i.e. there is a polynomial $P(X) \neq 0$ with coefficients in $k(t_1, \dots, t_n)$ such that $P(a) = 0$.
Write $P^\sigma$ for the polynomial with coefficients in $k(\sigma(t_1), \dots, \sigma(t_n))$ obtained by applying $\sigma$ to all coefficients of $P$. We have $P^\sigma \neq 0$ and $P^\sigma(b) = P^\sigma(\sigma(a)) = \sigma(P(a)) = 0$ : $b$ is algebraic over $\sigma(t_1), \dots, \sigma(t_n)$, whence the claim. $\square$

For the other implication, consider two algebraically closed fields over the prime field $k$ having the same transcendence degree $n$. Let $t_1, \dots, t_n$ (resp. $u_1,\dots, u_n$) be a transcendence basis of $M$ (resp. $N$). Consider the isomorphism $\tau : k(t_1, \dots, t_n) \mapsto k(u_1, \dots, u_n)$ sending $t_i$ to $u_i$ for all $i$. This isomorphism extends to an isomorphism $\sigma : M \mapsto N$ by successive adjunction of algebraic elements.

  • 2
    $\begingroup$ This answers assumes that the transcendence degrees are finite, but the general case works just the same. $\endgroup$ Apr 13, 2020 at 17:08

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