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I have a question on the proof of the following proposition in Dummit and Foote (3rd edition, Proposition 31, p.544).


This section is what I entirely quote from the book.

Proposition. Let $K$ be an algebraically closed field and let $F$ be a subfield of $K$. Then the collection of elements $\overline{F}$ of $K$ that are algebraic over $F$ is an algebraic closure of $F$. An algebraic closure is unique up to isomorphism.

Proof. By definition $\overline{F}$ is an algebraic extension of $F$. Every polynomial $f(x) \in F[x]$ splits completely over $K$ into linear factors $x - \alpha$ (the same is true for every polynomial even in $K[x]$). But each $\alpha$ is a root of $f(x)$, so is algebraic over $F$, hence is an element of $\overline{F}$. It follows that all the linear factors $x - \alpha$ have coefficients in $\overline{F}$, i.e., $f(x)$ splits completely in $\overline{F}[x]$ and $\overline{F}$ is an algebraic closure of $F$.

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It seems like the authors showed that every polynomial in $F[x]$ splits over $\overline{F}$, but in order to show that $\overline{F}$ is algebraic closure of $F$, shouldn't we show that $\overline{F}$ is algebraically closed? (That is, shouldn't we take $f(x)$ from $\overline{F}[x]$ instead of $F[x]$?) I don't know what I am missing, but I thought the proof should have been proceeded as follows:

Fix any $f(x) \in \overline{F}[x]$, which splits over $L$. Let $\alpha \in L$ be any root of $f$. Then $\alpha$ is algebraic over $\overline{F}$ and since $\overline{F}$ is algebraic extension of $F$, it follows that $\alpha$ is algebraic over $F$ so that $\alpha \in \overline{F}$.

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Perhaps it depends what definition of "alegbraic closure of $F$" you're using. You could be using "smallest algebraically closed field containing $F$", in which case you're correct. Or you could be using "smallest field containing all roots of polynomials in $F$", in which case the proposition's proof is fine (and your comment would be part of the proof of a lemma that algebraic closures are always algebraically closed). –  Greg Martin Jun 18 '13 at 17:07
    
@GregMartin That's an answer. Can you copy that as an answer? –  user123412 Jun 18 '13 at 17:58
    
@GilYoungCheong, notice that in prop. 29 D&F already proved that an algebraic closure is algebraically closed... –  DonAntonio Jun 18 '13 at 18:52
    
@DonAntonio Thank you. I guess this happens when I do not stick to one book for reading (and I usually can't). I should have looked at the section more carefully! –  user123412 Jun 19 '13 at 1:23

2 Answers 2

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Perhaps it depends what definition of "alegbraic closure of $F$" you're using. You could be using "smallest algebraically closed field containing $F$", in which case you're correct. Or you could be using "smallest field containing all roots of polynomials in $F$", in which case the proposition's proof is fine (and your comment would be part of the proof of a lemma that algebraic closures are always algebraically closed).

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An algebraic closure of a field $F$ is an extension $K$ of $F$ which is (a) algebraic over $K$ and (b) algebraically closed, meaning that every non-constant polynomial over $K$ has a root in $K$. That is the definition. One also sometimes talks about the algebraic closure of $F$ in some fixed extension of $F$, but without reference to some larger extension, an algebraic closure of $F$ is an algebraically closed extension of $F$ which is algebraic over $F$.

Now, regarding your question, yes, to prove that an extension $K$ of $F$ is an algebraic closure, you must verify that $K$ is algebraic over $F$ and that every non-constant polynomial in $K[X]$ has a root in $F$. However, it turns out that if $K/F$ is algebraic and every non-constant polynomial in $F[X]$ has a root in $K$, then in fact every non-constant polynomial in $K[X]$ has a root in $K$, and the argument is essentially your last sentence. I'll give the argument with more detail.

Suppose $f\in K[X]$ is an irreducible monic polynomial, and form the extension $E=K[X]/(f(X))$ (this is a field because $f$ is irreducible over $K$) which contains the root $\alpha=X+(f(X))$ of $f$. Now, if we write $f(X)=\sum_{i=0}^n a_i X^i$, then each $a_i$ is algebraic over $F$, by assumption, so the extension $F(a_1,\ldots,a_n)\subseteq K$ is finite over $F$ (because it can be written as a tower of extensions where we adjoin a single algebraic element at each step). Note that $f$ has coefficients in $F(a_1,\ldots,a_n)$, so the extension obtained by adjoining $\alpha$ to $F(a_1,\ldots,a_n)$ is finite over $F(a_1,\ldots,a_n)$, and hence also over $F$. So $\alpha$ is algebraic over $F$. This means that there is a polynomial $g\in F[X]$ such that $g(\alpha)=0$. Since $f\in K[X]$ is the minimal polynomial for $\alpha$ over $K$, and $g\in F[X]\subseteq K[X]$ has $\alpha$ as a root, we must have $f\mid g$ in $K[X]$. But by assumption, $g$ splits over $K$, so $f$ splits over $K$ as well, and by irreducibility, we must have $\deg(f)=1$. It follows then (by unique factorization) that every non-constant polynomial over $K[X]$ splits over $K$, as desired.

I could have avoided messing around with the coefficients of $f$ by simply invoking the fact that for fields $F\subseteq K\subseteq E$, $E/K$ and $K/F$ algebraic implies $E/F$ algebraic, but the bit with the coefficients I gave is essentially the proof of that anyway, so I included it (the point being that adjoining a single algebraic element always yields an extension of finite degree, and any finite degree extension is algebraic essentially by inspection).

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