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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.

Thanks for your help.

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marked as duplicate by Bill Dubuque, Dietrich Burde, Alex M., rschwieb abstract-algebra Nov 6 '16 at 16:54

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

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  • $\begingroup$ 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? $\endgroup$ – cronos2 Nov 6 '16 at 16:52
  • $\begingroup$ @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. $\endgroup$ – Adam Hughes Nov 6 '16 at 17:00

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