Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
up vote 5 down vote accepted

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).

share|cite|improve this answer
Very good argument . +1 . Really Nice . – Iyengar Mar 26 '12 at 6:24
Thanks Chonoles – Rajnish Mar 26 '12 at 6:37

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.