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Is there an integral domain $A$, with field of fractions $K$, such that $A$ is isomorphic to $K[X]$ ?

Such a ring $A$ has to be a PID, which is not a field (i.e. PID of Krull dimension $1$), and is infinite. Basic examples as $A = \Bbb Z, k[t], ...$ don't work. I can notice that $A^{\times} \cong K^{\times}$, which is a strange property, but this isomorphism is not necessarily coming from the inclusion $A \subset K$. Maybe I'm missing an easy obstruction for such a ring to exist, and anyway I don't find any counterexample.

Thank you!

Watson
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  • In the example below, we have a domain $A$, with field of fractions $B$, such that $A \not \cong A[X]$ but $B \cong B(X)$. – Watson Jun 29 '18 at 16:20

1 Answers1

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Set $B = k(x_1, x_2, x_3, \dots)$ and $A = B[x_0]$. Then $\mathrm{Frac}(A) \cong B$.

Billy
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