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We can define algebraic closure of some field $F$ in two ways, both seem logical:

(1) it is the smallest (with respect to inclusion) algebraically closed field that contains $F$.

(2) it is a field that we got from $F$ once we have extended $F$ with all zeros of all polynomials in $F[x]$.

But, these definitions are not equivalent. In fact, it is obvious that field that we got in (1) would contain a field that we would get in (2). For example why they are different, take $\mathbb{Q}$. On (1) we obviously obtain $\mathbb{C}$. I am not exactly sure what we obtain in (2), but it is a proper subset of $\mathbb{C}$: we can not get stuff like $\pi,e$ and so on...

What is then a standard definition of algebraic closure? I know there is a standard result that algebraic closure of every field exists, but according to which definition?

Dibidus
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    Why would we get $\Bbb C$ in $(1)$? We would get $\overline{\Bbb Q}$, as an algebraic closure of $\Bbb Q$. May be, this post is helpful, or this one. – Dietrich Burde May 01 '24 at 09:07
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    The most standard definition is actually none of the two. Usually algebraic closure is defined as an algebraically closed field that is an algebraic extension of $F$. It is true (but not that obvious) that this is equivalent to the two conditions you wrote. – Mark May 01 '24 at 09:18
  • Does this answer your question? Construct algebraic closure of $\mathbb{Q}$ – Anne Bauval May 01 '24 at 09:22
  • It makes no sense to talk about "smallest with respect to inclusion". Consider $\mathbb{R}$, and look at all overfields of $\mathbb{R}$ within the Hamiltonians. Which is the "smallest" algebraically closed field containing $\mathbb{R}$. Is it $\mathbb{R}[i]$, $\mathbb{R}[j]$, or $\mathbb{R}[k]$? Or $\mathbb{R}[\alpha]$, where $\alpha$ is any of the infinitely many quaternions satisfying $x^2+1=0$? – Arturo Magidin May 01 '24 at 12:40
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    @ArturoMagidin My guess is that OP probably meant a minimal algebraically closed field with respect to inclusion, i.e one that doesn't contain any other algebraically closed fields that contain $F$. If so, this is equivalent to being an algebraic closure. – Mark May 01 '24 at 13:10

1 Answers1

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The standard definition of "algebraic closure of a field $F$" is:

The field $K$ is said to be an algebraic closure of $F$ if and only if $K$ is an algebraic extension of $F$, and $K$ is algebraically closed.

Where

A field $L$ is algebraically closed if and only if every nonconstant polynomial with coefficients in $L$ has a root in $L$.

Your (1) is not stated in a correct way, as "smallest" is not something you can do in this situation. Instead, you may say "minimal", in the sense that if $F\subseteq L\subsetneq K$, then $L$ is not algebraically closed.

Your claim that using (1) and starting from $\mathbb{Q}$ you would get $\mathbb{C}$ is also incorrect, using either your incorrect phrasing or the correct one, since $\overline{\mathbb{Q}}$, the field of algebraic numbers, is algebraically closed, contains $\mathbb{Q}$, and is properly contained in $\mathbb{C}$.

Theorem. Let $F$ be a field. The following are equivalent:

  1. Every nonconstant polynomial $f(x)\in F[x]$ has a root in $F$.
  2. Every nonconstant polynomial $g(x)\in F[x]$ splits over $F$.
  3. Every irreducible polynomial in $F[x]$ has degree one.
  4. There is no algebraic extension of $F$ except $F$ itself.
  5. There is a subfield $K$ of $F$ such that $F$ is algebraic over $K$ and every polynomial in $K[x]$ splits in $F[x]$.

Proof. It is straightforward that 4$\implies$3, that 3$\implies$2, and that 2$\implies$1.

That 1$\implies$4 is as follows: let $K$ be an algebraic extension of $F$, and let $a\in K$. Then $a$ is algebraic, so if $f(x)$ is the minimal polynomial of $a$ over $F$, then $f(x)$ is irreducible. By (1), since $f(x)$ has a root $r$, it follows that $x-r\mid f(x)$, hence $f(x)=x-r$. Since $f(a)=0$, then $f(x)=x-a$, hence $a\in F$. Thus, $K\subseteq F$.

4$\implies$5 by taking $K=F$. To see that 5$\implies$4, let $L$ be an algebraic extension of $F$, and let $a\in L$. Since $a$ is algebraic over $F$ and $F$ is algebraic over $K$, then $a$ is algebraic over $K$. Let $f(x)\in K[x]$ be the minimal polynomial of $a$ over $K$. Then $f(x)$ splits in $F[x]$. By unique factorization, since $x-a$ divides $f(x)$ in $L[x]$, $x-a$ must divide of the factors of $f(x)$ in $F[x]$. But all those factors are linear, and since $f(x)$ is monic, we may assume that all the linear factors of $f(x)$ in $F[x]$ are monic. So $x-a$ is one of the factors, hence $a\in F$. Thus, $F=L$, proving 4. $\Box$

Corollary. Let $K$ be an extension of $F$. The following are equivalent:

  1. $K$ is algebraic over $F$ and $K$ is algebraically closed.
  2. $K$ is the splitting field over $F$ of the set of all nonconstant polynomials in $F[x]$.
  3. $K$ is the splitting field over $F$ of the set of all irreducible polynomials in $F[x]$.

Proof. Clearly, 2 and 3 are equivalent. If $K$ is algebraic over $F$ and algebraically closed, then every nonconstant polynomial in $F[x]$ splits in $K$, so if $L$ is the splitting field of the set of all nonconstant polynomials of $F$, then we may take $F\subseteq L\subseteq K$. Then $L$ is algebraically closed by item 5 of theorem, hence $L$ has no proper algebraic extensions. Since $K$ is algebraic over $L$, then $L=K$. This shows that 1 implies 2. Conversely, if $K$ is the splitting field of all nonconstant polynomials in $F[x]$, then $K$ is algebraic over $F$, and $K$ satisfies item 5 of the theorem, hence is algebraically closed. $\Box$

Arturo Magidin
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  • I think in 5$\implies $4 we don’t need to invoke unique factorization property of $L[x]$. We know $f(x)$ spilts in $F[x]$. Let $f(x)=(x-a_1)\cdots (x-a_n)$ where $a_i\in F$. Since $f(a)=(a-a_1)\cdots (a-a_n)=0$, by elementary property of fields $a-a_i=0$. Thus $a=a_i\in F$. – user264745 May 23 '24 at 14:58
  • Also in corollary $1\Rightarrow 2$, since $K$ is algebraic over $F$, $K=X$ where $X$ is the set of all roots of all non constant polynomial in $F[x]$. So there is no need to use item $5$ of theorem. – user264745 Jun 22 '24 at 17:11