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I have solved the problem as follows-
As per definition, $\overline{F}$ is the subset of $K$ consisting of all the elements that are algebraic over $F$. We have to show $\overline{F}$ is algebraically closed i.e. there is no proper algebraic extension of $\overline{F}$ i.e. equivalently every polynomial in $\overline{F}[x]$ has a root in $\overline{F}$.
So, choose $p(x)\in\overline{F}[x]$
If $p(x)$ is irreducible over $\overline{F}$, then let $u$ be a root of $p(x)$. Now extend $\overline{F}$ to a field $\overline{F}(u)$ and since $[\overline{F}(u):\overline{F}]$ is finite, $\overline{F}(u)$ is algebraic over $\overline{F}$. So we get, $F\subseteq\overline{F}\subseteq\overline{F}(u)$, where $\overline{F}(u)$ is algebraic over $\overline{F}$ and $\overline{F}$ is algebraic over $F\implies$ $\overline{F}(u)$ is algebraic over $F\implies u$ is algebraic over $F\implies u\in\overline{F}$(it follows from the definition of $\overline{F}$)$\implies p(x)\in\overline{F}[x]$ has a root in $\overline{F}$.
Now, if $p(x)$ is reducible over $\overline{F}$, then we will just work with an irreducible factor of $p(x)$ in $\overline{F}[x]$.
Hence, it is proved that $\overline{F}$ is algebraically closed.
Is this okay? But, I have not used the algebraic closed property of $K$.
Can anybody clear my doubts? Thanks for help in advance.

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You never used the fact that $K$ is agebraically closed. Let $p\in \bar F[X]$, since $\bar F\subset K$, $p\in K[X]$, thus $p$ is a product of polynomials of degree $1$ of $K[X]$ since $K$ is algebraicaly closed. By definition of the agebraic closure, the roots of $p$ contained in $K$ are contained in $\bar F$, we deduce that the degree of $p$ is $1$ and $\bar F$ is algebraically closed.

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  • $\begingroup$ I can't understand, what you want to show with $p(x)$, please clarify. $\endgroup$ – Biswarup Saha Sep 6 '18 at 14:37

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