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Let $L/K$ be a field extension such that $L$ is algebraically closed. Show that $\{a\in L\mid[K(a):K]\lt\infty\}$ defines an algebraic closure of $K$.

So this is the set of minimal polynomials $f_a\in K$ of $a\in L$ with $f(a)=0$ and with finite degree right? Can this be proven using normal or seperable extensions?

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  • $\begingroup$ This set equals $\{ a\in L\mid $a$ \text{ is algebraic over }K \}$. Hence it is an algebraic closure of $K$ in $L$. $\endgroup$ – Dietrich Burde Dec 18 '14 at 10:43
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Proposition: Suppose $L/K$ is a field extension and $L$ is algebraically closed. Let $E = \{a\in L\mid a \text{ is algebraic over K} \}$. Then $E$ is an algebraic closure of $K$ in $L$.

Proof: We have to show that $E$ is a field and that $E/K$ is algebraic. It is hence enough to show that $E$ is algebraically closed. Suppose $f(x) \in E[x]$ is a nonconstant polynomial. Then $f(x) \in L[x]$, so $f(x)$ has a root $a\in L$. Threfore $a$ is algebraic over $E$, as $f(a) = 0$. So, $E(a)/E$ is an algebraic extension, but $E/K$ is algebraic, so that $E(a)/K$ is algebraic. Thus $a$ is algebraic over $K$, so $a\in E$ by definition of $E$. So $E$ is algebraically closed.

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