Is every field the field of fractions of an integral domain which is not itself a field?
What about the field of real numbers?
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7Yes, every field is its own field of fractions! – Mariano Suárez-Álvarez Nov 23 '10 at 15:12
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1You' re write.I didn't mean the trivial case so i changed the question. – t.k Nov 23 '10 at 15:16
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5Then no: pick any field with a prime number of elements. – Mariano Suárez-Álvarez Nov 23 '10 at 15:17
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Thank you again. What about infinite fields?And the field of real numbers? – t.k Nov 23 '10 at 15:19
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Every field $F$ of characteristic zero or of prime characteristic but not algebraic over its prime field is the field of fractions of a proper subring of $F$. But no algebraic extension of $\mathbb F_p$ is, since its only subrings are fields.
If $F$ is not an algebraic extension of some $\mathbb F_p$ then $F$ contains a subring $A$ isomorphic to $\mathbb Z$ or $\mathbb F_p[X]$. Each of these rings $A$ has a nontrivial valuation $v$. The valuation $v$ can be prolonged to $F$. Its valuation ring is a proper subring of $F$ whose quotient field is $F$.
Robin Chapman
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11Variant of the argument: let $B$ be a transcendence basis of $F$. Let $A\subset F$ be the subring generated by $B$. This is a polynomial ring over $\mathbb{Z}$, or a nontrivial polynomial ring over $\mathbb{F}_p$. Now take the integral closure of $A$ in $F$. – Laurent Moret-Bailly Nov 23 '10 at 15:47
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2The existence of a prolongation of a valuation to a non algebraic extension field doesn't sem to be mentioned in the standard references (Atiyah, Bourbaki, Matsumura,...). A proof is provided here. – Georges Elencwajg Apr 27 '16 at 11:44