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Let $\mathbb F$ be a field and $\mathbb K $ be an extension field of $\mathbb F$ such that $\mathbb K$ is algebraically closed.

Let $\mathbb L$ be the field of all elements of $\mathbb K$ which are algebraic over $\mathbb F$. Then $\mathbb L_{|\mathbb F}$ is an algebraic extension.

My question is : Is $\mathbb L$ algebraically closed ?

I am trying to prove the existence of algebraic closure, so please don't assume that every field has an algebraic closure.

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marked as duplicate by user223391, Ben, Scientifica, José Carlos Santos, Leucippus May 8 '18 at 0:33

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Yes it is -- the key fact is that an algebraic extension of an algebraic extension is still algebraic.

Let $f(X)\in \mathbb L[X]$ be a polynomial. We need to show that $f$ has a root in $\mathbb L$.

By assumption, $f$ has a root $\alpha\in\mathbb K$, so we can consider the extensions $$\mathbb F\subset\mathbb L\subset\mathbb L(\alpha)\subset\mathbb K.$$

By construction $\mathbb L(\alpha)$ is algebraic over $\mathbb L$, which is algebraic over $\mathbb F$. Hence $\mathbb L(\alpha)$ is algebraic over $\mathbb F$, so by assumption, $\alpha\in \mathbb L$. The result follows.

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Given a poynomial $P(x) = \sum_{k=0}^n a_k x^k$ with coefficients in $\mathbb{L}$, which are algebraic over $\mathbb{F}$, we can restrict to the subfield $\mathbb{F}_1 := \mathbb{F}(a_1,\ldots,a_k)$. Conjugating all roots to $\mathbb{F}_1$ gives a finite field extenstion $\mathbb{F}_2$ of $\mathbb{F}_1$. Since $\mathbb{F}_1 / \mathbb{F}$ is a finite extension, also $\mathbb{F}_2 / \mathbb{F}$ is finite, because $$[\mathbb{F}_2:\mathbb{F}] = [\mathbb{F}_2:\mathbb{F}_1] [\mathbb{F}_1:\mathbb{F}].$$ So all roots are already algebraic over $\mathbb{F}$.

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