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In Hungerford's Algebra text, it is stated that a field $K$ is algebraically closed iff there exists a subfield $F$ such that $K$ is algebraic over $F$ and all polynomials in $F[x]$ split in $K[x]$. This seems to be a much weaker condition that $K$ being algebraically closed. I don't see why it is true (it is not proven in the text)

Is there a particular reason why this is true?

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2 Answers 2

up vote 4 down vote accepted

If $\alpha$ is algebraic over $K$, then $\alpha$ is algebraic over $F$. The minimal polynomial splits over $K$. Thus, $\alpha \in K$.

So this is quite immediate. However, there is even a weaker condition: Every non-constant polynomial in $F[x]$ has some root in $K$. Then it also follows that $K$ is algebraically closed; but this is a nontrivial theorem (Isaacs, Roots of polynomials in algebraic extensions of fields; or: Gilmer, A Note on the Algebraic Closure of a Field).

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Let $\bar{K}$ be the algebraic closure of $K$ and fix $\alpha\in \bar{K}$. Then $\alpha$ is algebraic over $F$ because algebraic extensions are transitive. In particular this implies that the minimum polynomial of $\alpha$ over $F$ splits in $K[x]$ so that $\alpha \in K$. So we have $K=\bar{K}$ and $K$ is algebraically closed.

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