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If $K$ is algebraic closure of $F$, then as a ring, $K$ is integral over $F$. Is that true or not?

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Yes, it is true. By the definition of "algebraic closure", every $\alpha\in K$ is algebraic over $F$; that is, for any $\alpha\in K$, there is some non-zero $f(x)\in F[x]$ such that $f(\alpha)=0$. But then, letting $g(x)=\frac{1}{c}f(x)$ where $c$ is the leading coefficient of $f$, we also have $g(\alpha)=\frac{1}{c}f(\alpha)=\frac{1}{c}0=0$. Because $g$ is a monic polynomial with coefficients in $F$, we have that $\alpha$ is integral over $F$. Because every element of $K$ is integral over $F$, the extension of rings $K\supseteq F$ is integral.

Note that this did not depend on $K$ being the algebraic closure of $F$; in fact the same argument works for any algebraic extension of $F$. As Wikipedia says here,

If $A$, $B$ are fields, then the notions of "integral over" and of an "integral extension" are precisely "algebraic over" and "algebraic extensions" in field theory (since the root of any polynomial is the root of a monic polynomial).

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Very good argument . +1 . Really Nice . –  Iyengar Mar 26 '12 at 6:24
    
Thanks Chonoles –  Rajnish Mar 26 '12 at 6:37

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