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Let $\mathbb{C}(x)$ be the field of rational functions over $\mathbb{C}$. Of course $\mathbb{C}(x)$ is a field extension of $\mathbb{C}$. My question now is: are there any intermediate fields between $\mathbb{C}$ and $\mathbb{C}(x)$? If so, what can we say about their dimension? Is it always infinite?

cip
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  • See egreg's answer at this duplicate. – Dietrich Burde Dec 26 '20 at 10:03
  • Thank you. So, if I understood it correctly, that means that all intermediate fields are infinite dimensional, right? – cip Dec 26 '20 at 10:11
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    @cip Yes: $\Bbb{C}$ is algebraically closed so has no finite or even algebraic field extensions. – Olivier Bégassat Dec 26 '20 at 10:14
  • Ok , thanks. What about vector subspaces of $\mathbb{C}(x)$? Are they all infinite dimensional as well? – cip Dec 26 '20 at 10:22
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    Vector spaces always have finite dimensional subspaces. Just take any nonzero vector not in the ground field and it will be the basis. – Gerry Myerson Dec 26 '20 at 11:25
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    If $K$ is an intermediate field then we always have $[K:\Bbb{C}]=\infty$, but $[\Bbb{C}(x):K]<\infty$. Furthermore, by Lüroth's theorem $K\simeq \Bbb{C}(x)$. Do you want me to flesh this out? – Jyrki Lahtonen Dec 26 '20 at 11:37
  • @Jyrki Lahtonen That would be great, thank you. – cip Dec 26 '20 at 12:21
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    @JyrkiLahtonen By the way I have a proof of Luroth's theorem looking at the normal closure $L$ of $K(x)/F$, $d=[L:K(x)],n=[K(x):F]$, let $Div(x) = \sum_{l=1}^d P_l-Q_l$ as a function in $L$,

    the $F$-minimal polynomial of $x$ is $\prod_{j=1}^n (T-\sigma_j(x))=\sum_{k=0}^n a_k(x) T^k$, we see that $Div(a_k(x))\ge -\sum_{j=1}^n \sum_{l=1}^d \sigma_j(Q_l)$, thus $a_k(x)$ has at most $nd$ poles on $L$ which means it has at most $n$ poles on $K(x)$, whence if $a_k(x)$ is non-constant then $[K(x):K(a_k(x))]\le n$ which implies that $F=K(a_k(x))$.

     – reuns Dec 26 '20 at 13:30

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A summary of the comments (excluding reuns result that they should post separately!) Below $K$ stands for an arbitrary intermediae field strictly in-between, $\Bbb{C}\subset K\subset\Bbb{C}(x)$.

  1. Because $\Bbb{C}$ is algebraically closed, it has no algberaic extensions. Hence no finite extensions. Therefore $[K:\Bbb{C}]=\infty$.
  2. On the other hand, if $u=f(x)/g(x)$ is an arbitrary element of $K\setminus\Bbb{C}$, $f,g\in\Bbb{C}[x]$, then $x$ is a zero of the polynomial $$ P(T):=f(T)-g(T)u\in K[T]. $$ Therefore $x$ is algebraic over $K$. Hence $[K(x):K]<\infty$. But, $K(x)=\Bbb{C}(x)$, so we can conclude that $[\Bbb{C}(x):K]<\infty$. Nothing more can be said, as we easily see that $[\Bbb{C}(x):\Bbb{C}(x^n)]=n$ for every positive integer $n$, so the extension degree can be arbitrarily high.
  3. By Lüroth's theorem every intermediate field $K$ is actually a simple transcendental extension of $\Bbb{C}$. In other words, $K$ is $\Bbb{C}$-isomorphic to $\Bbb{C}(x)$.
Jyrki Lahtonen
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