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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7$\begingroup$ Yes, every field is its own field of fractions! $\endgroup$– Mariano Suárez-ÁlvarezCommented Nov 23, 2010 at 15:12
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1$\begingroup$ You' re write.I didn't mean the trivial case so i changed the question. $\endgroup$– t.kCommented Nov 23, 2010 at 15:16
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5$\begingroup$ Then no: pick any field with a prime number of elements. $\endgroup$– Mariano Suárez-ÁlvarezCommented Nov 23, 2010 at 15:17
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$\begingroup$ Thank you again. What about infinite fields?And the field of real numbers? $\endgroup$– t.kCommented Nov 23, 2010 at 15:19
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1 Answer
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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 $F$.
answered Nov 23, 2010 at 15:23
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11$\begingroup$ Variant 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 $F$. $\endgroup$ Commented Nov 23, 2010 at 15:47
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2$\begingroup$ The 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. $\endgroup$ Commented Apr 27, 2016 at 11:44