Lemma 9.8.10. Let $E/F$ be a finite or more generally an algebraic extension of fields. Any subring $F \subset R \subset E$ is a field.

Proof. Let $\alpha \in R$ be nonzero. Then $1, \alpha , \alpha ^2, \ldots$ are contained in $R$. By Lemma 9.8.5 we find a nontrivial relation $a_0 + a_1 \alpha + \ldots + a_ d \alpha ^ d = 0$ with $a_ i \in F$. We may assume $a_0 \not= 0$ because if not we can divide the relation by $\alpha$ to decrease $d$. Then we see that

$a_0 = \alpha (- a_1 - \ldots - a_ d \alpha ^{d - 1})$

which proves that the inverse of $\alpha$ is the element $a_0^{-1} (- a_1 - \ldots - a_ d \alpha ^{d - 1})$ of $R$. $\square$

Comment #8870 by Manolis C. Tsakiris on

We need to say something about the coefficients $a$, e.g., that they are in $F$.

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