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Let $\mathbb{C}(x)$ be the field of rational functions over the complex numbers and $F$ a finite extension of $\mathbb{C}(x)$. Suppose $B$ is the integral closure of $\mathbb{C}[x]$ (the ring of polynomials over $\mathbb{C}$). Is $B$ always a Unique Factorization Domain (UFD)?

  • I meant "B is the integral closure of C[x] in F" –  Dec 21 '18 at 16:15
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    Please edit your post rather than amend it in the comments. Please use latex code for math symbols. – YCor Dec 21 '18 at 16:19
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    Every integrally closed one dimensional domain, finitely generated over $\mathbb{C}$ is the integral closure of $\mathbb{C}[x]$ for a suitable $x$ (known as Noether Normalization). Most of these are not UFDs. – Mohan Dec 21 '18 at 19:33
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    In fact, $F$ is the field of rational functions of a smooth projective cuve $C$. As soon as $g(C)\geq 1$, $B$ is not a UFD. –  Dec 21 '18 at 20:11
  • I see we can construct rather trivial counterexamples! Even for genus 0:$x^2 - 1 = (x-1)(x+1) = \sqrt{x^2 -1} \sqrt{x^2 -1}$ (unless I miss something). –  Dec 24 '18 at 11:17
  • Hint: What happens when you look at the function field of an elliptic curve, say, $$E:y^2=x^3-x?$$ does that equation not suggest two different factorizations for an element of the integral closure $\Bbb{C}[x,y]$? – Jyrki Lahtonen Dec 28 '18 at 11:00

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