# Show that if $F \subset D \subset E$ then D is a field [duplicate]

Exercise: Let $E$ be a finite extension field of $F$, and $D$ an integral domain such that $F \subset D \subset E$. Show that $D$ is a field.

Attempt: All of the field axioms for D are inherited by the inclusion relationship and ring axioms except for the existence of multiplicative inverse. $\alpha \in D \cap F \implies \alpha^{-1} \in F \implies \alpha^{-1} \in D$, so consider $\beta \in D \backslash F$.

Since $E/F$ is finite it's also algebraic, but then $F[\beta] = F(\beta)$ and so $\beta^{-1} \in F(\beta) \subset D$.

I'm not really sure this proof is correct, but in case it is there are two things that bug me:

• If we just ask $D$ to be a ring then from $D \subset E$ follows $D$ is an integral domain.
• We only used the fact that $E/F$ is an algebraic extension in the proof.

## marked as duplicate by Bill Dubuque, Dietrich Burde, Alex M., rschwieb abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Nov 6 '16 at 16:54

• $D\setminus E$ is empty, perhaps you mean $D\setminus F$. – Adam Hughes Nov 6 '16 at 16:34
• Yes, I tend to mix them – cronos2 Nov 6 '16 at 16:36
• Since $E$ has no zero divisors, also $D\subset E$ has no zero divisors. For the multiplicative inverse, compare also with this question. – Dietrich Burde Nov 6 '16 at 16:37
• @Bill Dubuque thanks, didn't find that one – cronos2 Nov 6 '16 at 16:52

It's already an integral domain, all you're missing is inverses. Let $p_\beta(x) = a_0+a_1x+\ldots + a_nx^n$ be the minimal polynomial for $\beta$ over $F$. Then note
$$-a_0^{-1}(a_1+a_2\beta+a_3\beta^2+\ldots + a_n\beta^{n-1})\cdot\beta=1$$
So $\beta^{-1}\in D$, showing inverses. Here we use that all such $\beta\in E$ which is finite, so that there is a minimal polynomial for it over $F$.
• Yes, that's the kind of argument I was thinking of. But, again, we're only using $E/F$ is algebraic, not necessarily finite, am I right? – cronos2 Nov 6 '16 at 16:52
• @cronos2 it doesn't matter if $E/F$ is globally finite, true, each element just needs to be contained in some finite extension. But there are plenty of examples of theorems not 100% as sharp as possible, usually this is because the form you use them in is not the most general one. – Adam Hughes Nov 6 '16 at 17:00