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Let $x$ be trancendental over $\mathbb{C}$. Let $K$ be the algebraic closure of $\mathbb{C}(x)$.

How to show that $K$ is isomorphic to $\mathbb{C}$?

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2 Answers

up vote 11 down vote accepted

If $B$ is a trascendence basis of $\mathbb C$ over $\mathbb Q$, then $B$ is uncountable, and $\mathbb C$ is an algebraic closure of $\mathbb Q(B)$.

Now $B\cup\{x\}$ is a transcendence basis of $K$ over $\mathbb Q$. Since $B$ and $B\cup\{x\}$ have the same cardinal, $K$ and $\mathbb C$ are isomorphic: they are both algebraic closures of purely trascendental extensions of $\mathbb Q$ of the same trascendence degree.

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I do not understand why B union {x} is a basis .I understand why B union {x} is linearly independent over Q. Secondly I do not understand why "Since B and B∪{x} have the same cardinal,K and C are isomorphic. I have done an only introductory course in galois theory –  Jishnu Ray Apr 28 '12 at 8:25
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Notice that I am not talking of bases of vector spaces: throughout, I am talking about trascendence bases. –  Mariano Suárez-Alvarez Apr 28 '12 at 8:29
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Suppose K1, K2 are two algebraically closed fields of the same characteristic and #K1 = #K2 is uncountable. Then K1 is isomorphic to K2.If we apply this theorem for K and C then we get the ans. Is it so? –  Jishnu Ray Apr 28 '12 at 16:30
    
That is the point of Bill's answer, not mine. In any case, do you know how to prove that, which is Steinitz's theorem? The argument I sketched in my answer is considerably more elementary and depends on a little of basic field theory. –  Mariano Suárez-Alvarez Apr 28 '12 at 18:01
    
I have one doubt in the proof.If S1 and S2 are uncountable sets,Does there exists a set bijection between x(i) {i runs over S1} and x(i) {i runs over S2}. x(i)`s are variables –  Jishnu Ray Apr 28 '12 at 20:08
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This is an immediate consequence of a famous theorem of Steinitz that uncountable algebraically closed fields are isomorphic iff they have equal characteristic and equal cardinality (a prototypical example of Morley's categoricity theorem).

This fails in the countable case, e.g. if $\mathbb A = $ algebraic numbers then, though countable, $\overline{\mathbb A(x)}\not\cong \mathbb A$ since they have unequal transcendence degree over $\mathbb Q,\:$ viz. $1$ vs. $0$, resp.

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