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Taking an algebraically closed field $K$ with $char(K)=0$ and $|K|\geq |\mathbb{C}|$, we have that $\bar{\mathbb{Q}}\subset K$. Let $T$ be a transcendence basis of $K\setminus \bar{\mathbb{Q}}$. Writing $|T|=\alpha$ and $[K:\bar{\mathbb{Q}}(T)]=n$, I wonder what is $|K|$ in terms of $\alpha$ and $n$. Does $|K|$ determine $\alpha$ and $n$ uniquely?

Actually, I want to prove that any two algebraically closed fields with $0$ characteristic and same cardinality greater or equal then $|\mathbb{C}|$ are isomorphic and I already know that $\alpha$ and $n$ determine $K$ up to isomorphisms. The question is the missing step.

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  • $\begingroup$ I'm not sure what's going on with your $n$. Given that $K$ is algebraically closed, $n$ will actually always be just $|T|+\aleph_0$ (or just $|T|$ given that $K$ is uncountable). I don't know how $n$ would ever come up in the uniqueness result you are referring to. Also, presumably in your second paragraph you mean for your fields to be algebraically closed? $\endgroup$ – Eric Wofsey Dec 8 '17 at 4:46
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Algebraic extensions never change the cardinality of an infinite field. Indeed, if $E$ is an infinite field and $F$ is an algebraic extension, every $x\in F$ is a root of some polynomial over $E$. There are only $|E|$ polynomials over $E$ and each polynomial has only finitely many roots, so $F$ cannot have more than $|E|$ elements.

So, in your setup, we have $|K|=|\bar{\mathbb{Q}}(T)|$. Moreover, $|\bar{\mathbb{Q}}(T)|=\aleph_0+|T|$, since every element of $\bar{\mathbb{Q}}(T)$ is a rational function in elements of $T$ with coefficients in $\bar{\mathbb{Q}}$ and $\bar{\mathbb{Q}}$ is countable. In your case, $T$ must be uncountable since $|K|\geq|\mathbb{C}|$, so we get $|K|=|\bar{\mathbb{Q}}(T)|=|T|$.

More generally, the same argument shows that if $K$ is any uncountable field, then its cardinality is equal to its transcendence degree over the prime field.

What you call $n$ must also always be equal to $|T|$. For instance, for any $t\in T$, there is a square root $\sqrt{t}$ in $K$, and these elements are linearly independent over $\bar{\mathbb{Q}}(T)$. So $n\geq |T|$, but also $n\leq |K|=|T|$ so $n=|T|$

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  • $\begingroup$ Actually, I want to prove that any two fields with $0$ characteristic and same cardinality greater or equal then $|\mathbb{C}|$ are isomorphic and I already know that $\alpha$ and $n$ determine $K$ uniquely up to isomorphisms. I was hoping that showing that $|K|$ determines $\alpha$ and $n$ uniquely would end the prove. However, what you are saying allows the construction of two non-isomorphic fields with same cardinality by taking different values for $n$. What am I getting wrong? $\endgroup$ – Roland Dec 8 '17 at 4:50
  • $\begingroup$ Assuming $K$ is algebraically closed, there is only one possible value of $n$: it must always be equal to $\alpha$. As I commented, though, I have no idea how $n$ is coming up in your work--it does not show up in any natural argument I know of. The standard classification of algebraically closed field says that any two algebraically closed fields of the same characteristic and transcendence degree are isomorphic. $\endgroup$ – Eric Wofsey Dec 8 '17 at 4:52
  • $\begingroup$ If you're not assuming your fields are algebraically closed then what you are trying to prove is horribly false (as is the statement that $\alpha$ and $n$ determined the field up to isomorphism). $\endgroup$ – Eric Wofsey Dec 8 '17 at 4:53
  • $\begingroup$ I added a brief expanation for that to my answer. As for your example, $\mathbb{R}$ is not a field of the form $\bar{\mathbb{Q}}(T)$. $\endgroup$ – Eric Wofsey Dec 8 '17 at 5:20
  • $\begingroup$ I would reiterate that I don't know why you're interested in $n$ in the first place. I strongly encourage you to revisit whatever argument you think makes $n$ relevant to this discussion. I suspect that argument is incorrect or you are misunderstanding it. $\endgroup$ – Eric Wofsey Dec 8 '17 at 5:23

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