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If I have a field $K$ and an extension $L$ of $K$ such that all (non-constant) polynomials in $K[X]$ have a root in $L$, is the set of algebraic elements of $L$ over $K$ (the sub-field of all the elements of $L$ which are roots of a polynomial in $K[X]$) algebraically closed ?

Do you have a counterexample ?

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I'm sorry, but are you saying that every nonconstant polynomial has a root in $L$, or are you asserting this is a consequence of $L$ being an algebraic extension of $K$? And: if $L$ is an algebraic extension, then every element of $L$ is algebraic over $K$, so "the set of algebraic elements of $L$ over $K$" would be equal to $L$, by definition of "algebraic extension". –  Arturo Magidin Mar 17 '11 at 19:10
    
Is this what you wanted to ask? "If $L$ is an extension of $K$, and $L$ is algebraically closed, and we let $E$ be the set of all elements of $L$ that are algebraic over $K$, is $E$ algebraically closed?" –  Arturo Magidin Mar 17 '11 at 19:16
    
Thanks Arturo, I thought my hypothesis on L was the definition of algebraic extension. I've edited my question. –  Marc Mar 17 '11 at 19:24
    
In English, they are called "polynomials", not "polynoms". –  Arturo Magidin Mar 18 '11 at 17:10

2 Answers 2

up vote 5 down vote accepted

I'm guessing that what you wanted to ask is the following:

Let $K$ be a field, and let $L$ be an extension of $K$ (that is, $K\subseteq L$). Suppose futher that $L$ is algebraically closed (all non-constant polynomials in $L[x]$ have a root in $L$). Is the set $$\{a\in L\mid a\text{ is algebraic over }K\}$$algebraically closed?

The answer is "yes."

Let $F = \{ a\in L\mid a\text{ is algebraic over }K\}$, and let $f(x)\in F[x]$ be a nonconstant polynomial. We may assume $f(x)$ is monic, $$f(x) = x^n + a_{n-1}x^{n-1} + \cdots + a_0,\quad a_i\in F,\quad n\gt 0.$$ Since $a_i\in F$, then $a_i$ is algebraic over $K$ for all $i$, so the field $E=K(a_0,\ldots,a_{n-1})$ is a finite extension of $K$.

Now $f(x)\in F[x]\subseteq L[x]$, so we know that $L$ contains a root $r$ of $f(x)$, since $L$ is algebraically closed. Therefore, $[E(r):E] \leq n$ is finite, so $[E(r):K] = [E(r):E][E:K]$ is also finite. Therefore, $[K(r):K] \leq [E(r):K]$ is finite, so $r$ is algebraic over $K$. Since $r$ is algebraic over $K$, then $r\in F$. Thus, $f(x)$ has a root in $F$.

This shows that every nonconstant polynomial in $F(x)$ has a root in $F$, so $F$ is algebraically closed.


Apparently not. Rather, you want:

Let $K$ be a field, and let $L$ be an extension of $K$ (that is, $K\subseteq L$). Suppose further that all non-constant polynomials in $K[x]$ have a root in $L$. Is the set $$\{a\in L\mid a\text{ is algebraic over }K\}$$algebraically closed?

This is equivalent to asking whether an algebraic extension of $K$ in which every polynomial in $K$ has a root is algebraically closed (since the subfield of $L$ of all elements that are algebraic over $K$ contains all the roots of polynomials in $K[x]$).

Again, the answer is yes.

Take $L$, $F$, and $f$ as above. Since $K(a_0,\ldots,a_{n-1})$ is algebraic over $K$, and $f(x)$ is a polynomial in $K(a_0,\ldots,a_{n-1})$, then the roots of $f(x)$ are algebraic over $K$. Thus, if $r$ is a root of $f(x)$ in some algebraic closure of $F$, then $r$ has an irreducible polynomial $g(x)\in K[x]$. Let $h(x)$ be the irreducible polynomial of $r$ in $F[x]$. In particular, $h(x)$ divides both $f(x)$ and $g(x)$ in $F[x]$. Since $g(x)$ has all its roots in $L$, and all its roots lie in $F$, then $g(x)$ has all its roots in $F$. Moreover, $h(x)|g(x)$ in $F$, so every root of $h(x)$ is a root of $g(x)$, hence every root of $h(x)$ lies in $F$. But $h(x)$ divides $f(x)$, so $f(x)$ has a root in $F$. Therefore, $F$ is algebraically closed.

Added. Hmph. Wait a second here. Every polynomial in $K[x]$ has at least one root in $L$, but we're not asserting they split. "Every polynomial in $L$ has at least one root in $L$" is equivalent to "every polynomial in $L$ splits", but this is not what we have here. We would need to show that

If $L$ is an algebraic extension of $K$ such that every every nonconstant polynomial in $K[x]$ has a at least one root in $L$, then every nonconstant polynomial in $K[x]$ splits over $L$.

On reflection, I'm not so sure this is clear.

Let $K=\mathbb{Q}$, and let $f(x)=x^3-2$. Consider the collection $\mathfrak{F}$ of all subfield of $\mathbb{C}$ that contain $\sqrt[3]{2}$, but do not contain the complex roots of $f(x)$. The collection is nonempty (it contains $\mathbb{Q}(\sqrt[3]{2})$). Partially order $\mathfrak{F}$ by inclusion. If we have a chain $\mathcal{C}$ of elements of $\mathfrak{F}$, $\mathcal{C}=\{F_i\}_{i\in I}$, then $\cup\mathcal{C}$ is a field, contained in $\mathbb{C}$, and does not contain the complex roots of $f(x)$. By Zorn's Lemma, $\mathfrak{F}$ contains maximal elements. Let $E$ be a maximal element of $\mathfrak{F}$.

Does every irreducible nonconstant polynomial in $\mathbb{Q}[x]$ have at least one root in $E$?

On further reflection, the result is definitely true if $K$ is perfect (in particular, in characteristic zero, which takes care of my stricken attempt at an example above). I think it is true in general (by first closing under adjunction of $p$th roots and then using the result for perfect fields to get the conclusion), but I haven't quite worked out all the details yet.

Theorem. Let $L$ be an extension of $K$, and assume that every separable nonconstant polynomial in $K[x]$ has at least one root in $L$. Then every nonconstant separable polynomial in $K[x]$ splits over $L$.

Proof. Let $f(x)$ be a separable nonconstant polynomial in $K[x]$, and let $r\in L$ be a root of $f(x)$. Let $\overline{K}$ be an algebraic closure of $K$. Then there is an embedding of the subfield $K(r)$ of $L$ into $\overline{K}$, so we may assume $K(r)\subseteq \overline{K}$. Let $F$ be the splitting field of $f(x)$ over $K$. Since $F$ is separable and of finite degree over $K$, by the Primitive Element Theorem, there exists $a\in\overline{K}$ such that $F=K(a)$. Let $g(x)$ be the irreducible polynomial of $a$ over $K$. Then we know that $g(x)$ has a root $\alpha\in L$, and that the subfield $K(\alpha)$ of $L$ embeds into $\overline{K}$. Since $K(a)$ is normal, we must have that the image of $K(\alpha)$ in $\overline{K}$ is $K(a)$; that is, $K(\alpha)\cong K(a)$. But $f(x)$ splits in $K(a)$, hence $f(x)$ splits in $K(\alpha)\subseteq L$ Thus, $f(x)$ splits in $L$, as claimed. QED

Corollary. Let $K$ be a field, and let $L$ be an extension of $K$ such that every nonconstant polynomial in $K[x]$ has at least one root in $L$. If $K$ is perfect, then the subfield $E=\{a\in L\mid a\text{ is algebraic over }K\}$ is algebraically closed.

Proof. If $f(x)$ is any nonconstant polynomial in $K[x]$, then $f(x)$ is separable (since $K$ is perfect). Since $L$ satisfies the hypothesis of the theorem, then $f(x)$ splits over $L$, hence over $E$. Thus, $E$ is a splitting field of all nonconstant polynomials in $K[x]$, and algebraic over $K$, hence $E$ is algebraically closed. QED


Final addition: And Keith Conrad has been kind enough to point us in the comments to his notes on the subject, which shows that, indeed, first adjoining all the $p^k$th roots to all elements of $K$ will work to give the result in general. So the result does hold: if every polynomial in $K[x]$ has at least one root in $L$, then $L$ contains an algebraic closure for $K$.

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Thank you very much for your very clear (and quick) answer. So if I replace my question with : Let K be a field, and let L be an extension of K. Suppose further that every polynomials in K[x] splits over L. Is the set of algebraic elements in K algebraically closed ? The answer would be yes. Am I right ? –  Marc Mar 17 '11 at 20:47
    
@Marc: Yes, the answer then would be an unqualified "Yes." I'll think about the weaker claim during my run today. –  Arturo Magidin Mar 17 '11 at 20:54
    
The last corollary does not require that K be perfect. See Theorem 2 at math.uconn.edu/~kconrad/blurbs/galoistheory/algclosure.pdf. –  KCd Mar 18 '11 at 1:01
    
Thank you again guys for your comments and answer. I thought the answer would be more immediate. My goal was to use model theory to build the algebraic closure of a field. It is not very difficult to build L from K with the compactness theorem and I wanted to use that to prove the existence of an algebraic closure of K. So it is possible, but it is not easier than Artin's proof even when you know model theory. –  Marc Mar 18 '11 at 15:09
    
@KCd: Aha; yes, pretty much what I thought, but couldn't quite carry out in practice. I'm also interested to note that my thought process went through precisely the process described in those notes. Thanks! –  Arturo Magidin Mar 18 '11 at 15:15

This is a standard result that follows by factorization into separable and inseparable extensions. See the post below from Ask an Algebraist. Alas, the thread has since been deleted (due to vandals), but the references below to Isaacs will help you to locate the standard conceptual proof (to be contrasted with the non-conceptual elementary form given below by a student - cf. my remarks following it). See Isaacs: Algebra: a graduate course. Ch. 19: Separability and Inseparability and see esp. the remark preceding Theorem 19.18.


From: Bill Dubuque; Date: July 6, 2010
In reply to: "Thanks & Summary", posted by student on July 6, 2010:
student wrote:

Thanks for your help, it's all clear now. The proof you gave provides good intuition! For my own understanding I have written it up in total and tried to make the paralles between the separable and the inseparable case clear. I post the proof below, maybe someone reading this will benefit from it.

BTW you said that this theorem occurs in many textbooks. Actually I could only locate it in Isaacs book (pp. 303-304), however with a different proof (as you suggested). Are there any specific books about algebra in general or field theory in particular you would recommend?

OK here comes the proof (hopefully without typos), which is basically a summary of the previous posts in this thread:

THEOREM
Let $L$ and $K$ be fields s.t. $L$ is an algebraic extension of $K$. Assume every irreducible polynomial in $K[X]$ has at least one zero in $L$. Then $L$ is an algebraic closure of $K$.

REMARK
If in the situation of the theorem $\operatorname{char}(L) = p$, then for every integer $k \geq 1$ every element $c \in K$ has a unique $p^k$-th root $d$ in $L$ (i.e. $d^{p^k} = c$).

PROOF (REMARK)
Existence: The polynomial $X^{p^k} - c$ is in $K[X]$ and hence has a root $d$ in $L$. Uniqueness: Assume $d^{p^k} = c = e^{p^k}$. Then $(d - e)^{p^k} = 0$ [freshman's dream] and hence $d = e$.

PROOF (THEOREM)
Suppose $f$ is an irreducible polynomial with coefficients in $K$. To prove the theorem we have to show that $f$ splits over $L$. The proof is divided into two cases: $f$ separable and $f$ inseparable.

*) 1st case: $f$ separable.
Let $S$ denote a splitting field for $f$ over $K$ lying inside a splitting field for $f$ over $L$. Being the splitting field for a separable polynomial over $K$, $S$ has a primitive element: $S = K(a)$. Let $g$ be the minimal polynomial of $a$ over $K$. Then $g$ has a root, say $b$, in $L$. Since $g$ is irreducible and both $a$ and $b$ are roots of $g$, $K(a)$ is isomorphic to $K(b)$. [In fact $K(a) = K(b)$: Since $S$ (being the splitting field for a polynomial in $K[X]$) is normal over $K$, and since $g$ is irreducible and has a root in $S$ (namely $a$), $g$ must split over $S$, hence $b$ is in $K(a)$. Since $K(a)$ and $K(b)$ both have the same degree (namely $\deg(g)$) over $K$, $K(a) = K(b)$.] Thus $L \supseteq K(b) = K(a) = S$ contains all roots of $f$.

*) 2nd case: $f$ inseparable.
Then $\operatorname{char}(K) = p > 0$, and $f = g(X^{p^n})$ for a separable irreducible polynomial $g\in K[X]$ and some $n > 0$. By the 1st case $g$ splits in $L$, and we may let $a_1,...,a_m$ be the roots of $g$ in $L$. Then $S = K(a_1,...,a_m)$ (being the splitting field for a separable polynomial over $K$) has a primitive element: $S = K(a)$. Let $h$ be the minimal polynomial of $a$ over $K$. The polynomial $h(X^{p^n})$ is in $K[X]$ and hence has a root in $L$, say $b$. Then $b^{p^n}$ is a root of $h$ in $L$. Since $h$ is irreducible and both $a$ and $b^{p^n}$ are roots of $h$, $K(a)$ is isomorphic to $K(b^{p^n})$. In fact $K(a) = K(b^{p^n})$ [same argument as in the 1st case].

Now let $r$ denote a root of $f$ lying inside a splitting field for $f$ over $L$. Then $r^{p^n}$ is a root of $g$, hence $r^{p^n}$ is in $S = K(a) = K(b^{p^n})$. Thus $r^{p^n} = F(b^{p^n})$ for some polynomial $F \in K[X]$. By the remark, every coefficient of $F$ has a $p^n$-th root in L, so $F(b^{p^n}) = G(b)^{p^n}$ for some $G \in L[X]$ [freshman's dream]. Thus $(r - G(b))^{p^n} = 0$, hence $r = G(b)$ is in $L(b) = L$.

Q.E.D.

This is just an "elementary" form of the standard proof (e.g. cf. Isaacs ref. above). But, alas, by eliminating the innate structural elements (factorization into separable and inseparable extensions) in favor of "simpler" calculations with elements, the innate structure of the proof is greatly obfuscated. The elementary proof has been constructed by directly inlining the lemmas leading up to Isaacs proof vs. invoking them by name. This is not good pedagogically. Decomposing algebraic extensions into their separable and inseparable parts is an essential tool required to study general algebraic extensions. Such key ideas should not be obscured such as above - esp. when, as here, the structural form is just as short and simple as is the elementary form. Please read the section of Isaacs leading up to the proof to better understand the innate structure. You've a much better chance of remembering the structural proof years later because it has been abstracted into simple steps using key structures/lemmas that will often be reused. But few if any students would probably remember the non-conceptual elementary form as presented above.

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