Transcendental extensions of R\mathbb{R} that are not purely transcendental

Purely transcendental extensions of R\mathbb{R} are those of the form R((Xi)i∈I)\mathbb{R}((X_i)_{i \in I}) where II is a set and the XiX_i’s are (distinct) indeterminates.

Now, I wonder if there is a transcendental extension of R\mathbb{R} that is not purely transcendental.

If such extensions exist, I wonder if there is a classical way to classify them in different categories.

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What about C(X)\Bbb C(X)?
– Stahl
Oct 20 at 23:22

  

 

@Stahl : Thank you. Can you answer my second question ?
– Vandrin
Oct 20 at 23:27

  

 

Note that algebraic closure or R\mathbb{R} is C\mathbb{C}.
– i707107
Oct 20 at 23:48

  

 

Instead of reals, if you work with complex numbers, this is birational classification of ALL complex algebraic varieties, in the case of finite transcendence degrees. Highly complicated problem and unresolved in any satisfactory manner.
– Mohan
Oct 21 at 1:34

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1 Answer
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If K/RK/\Bbb R is a field extension, you can always find a transcendence basis SS; i.e., a set S⊆KS\subseteq K such that KK is algebraic over R(S)\Bbb R(S). The extension is purely algebraic if and only if K=R(S)K = \Bbb R(S). So you’re asking for a classification of algebraic extensions of R((Xi)i∈I)\Bbb R((X_i)_{i\in I}). As I mentioned in the comments, C(X)\Bbb C(X) is such an extension, but so are R(X,√X)\Bbb R(X,\sqrt{X}) and R(X1/p∞):=R(X,X1/p,X1/p2,X1/p3,…)\Bbb R(X^{1/p^\infty}) := \Bbb R(X,X^{1/p},X^{1/p^2},X^{1/p^3},\dots). As you can see, even in one variable the extensions can be quite complicated. I don’t know of any classification of such fields beyond the description I gave here. The answer here discusses the classification of fields in general, and what subjects are involved in trying to obtain such a classification.