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
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Comment #8870 by Manolis C. Tsakiris on
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