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Is every field the field of fractions of an integral domain which is not itself a field?

What about the field of real numbers?

edited Nov 23 2010 at 17:08

Cam McLeman
5,780●1●22●42

asked Nov 23 2010 at 15:10

t.k
85●1●6

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Yes, every field is its own field of fractions! – Mariano Suárez-Alvarez Nov 23 2010 at 15:12

You' re write.I didn't mean the trivial case so i changed the question. – t.k Nov 23 2010 at 15:16

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Then no: pick any field with a prime number of elements. – Mariano Suárez-Alvarez Nov 23 2010 at 15:17

Thank you again. What about infinite fields?And the field of real numbers? – t.k Nov 23 2010 at 15:19

Robin Chapman · Accepted Answer · 2010-11-23 15:23:39Z UTC

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 $F$.

Is every field the field of fractions of an integral domain?

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