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.

  • $\begingroup$ 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

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 $n=|T|$

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.