Intermediate ring of an arbitrary field extension is also a field?

Question

Suppose $F$ is a subfield of a field $K$, and let $R$ be a ring contained in $K$ and containing $F$. Can we conclude that $R$ is a subfield of field $K$?

The answer is "yes" in case the extension is finite or more generally algebraic. See: Ring Inside an Algebraic Field Extension

Is this still true without that condition?

Thanks! It seems that if $K/F$ is transcendal extension, the answer is no. but if $K/F$ is algebraic, the answer is yes — ziang chen, Sep 16 '13 at 19:27
Yes, that's right. The extension $K/F$ being algebraic is exactly the condition that makes every intermediate ring extension a field. — Daniel Fischer, Sep 16 '13 at 19:30
http://math.stackexchange.com/questions/397733/algebraic-field-extension-question — ziang chen, Sep 19 '13 at 03:00

score 4 · Accepted Answer · edited Aug 17 '16 at 11:03

No:

Consider the case in which $F = \Bbb Q$, the rationals, and $K = \Bbb R$, the reals; let $\tau$ be any transcendental real number, e.g. we might take $\tau = e$ or $\tau = \pi$. Let $R$ be the ring $\Bbb Q [\tau]$, i.e. polynomial expressions in $\tau$ with rational coefficients. Then $\Bbb Q \subset R \subset \Bbb R$. $R$ is easily seen to be a subring $\Bbb R$, but it cannot be a field; it does not contain $\tau^{-1}$; if it did, we would have, for some polynomial $p(x) \in \Bbb Q [x]$,

$p(\tau) = \tau^{-1}$;

but then

$\tau p(\tau) = 1$,

so $\tau$ would satisfy the polynomial equation

$\tau p(\tau) - 1 = 0$,

which would imply that $\tau$ is algebraic over $\Bbb Q$, a contradiction. QED.

Intermediate ring of an arbitrary field extension is also a field?

1 Answers1