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An algebraic closure of a field K is an algebraic extension of K that is algebraically closed. It is one of many closures in mathematics. But why is it a countable set?

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  • $\begingroup$ In fact, it is generated by the roots of the polynomials in $K[x]$, i.e. the span of the roots of the polynomials is the entire algebraic closure (as a vector space over $K$). This bounds the dimension of the algebraic closure of $K$ over $K$, which in turn bounds the size of the algebraic closure as a set. $\endgroup$ – Alex G. Jun 7 '14 at 14:44
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    $\begingroup$ Irreducible polynomials of degree $n$ in $\Bbb F_p$ splits over $\Bbb F_{p^n}$. The algebraic closure is then $\bigcup_{n\geq 1} \Bbb F_{p^n} $. $\endgroup$ – Balarka Sen Jun 7 '14 at 14:53
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HINT: If $F$ is a field then there is a surjection from $F[x]\times\Bbb N$ onto the algebraic closure of $F$ (essentially count the roots of each polynomial, and map all the excessive points to $0$).

So it suffices to see that if $F$ is countable then $F[x]$ is countable.

[It should be noted that the axiom of choice is used here, and it is consistent with its failure that a countable field has an uncountable algebraic closure.]

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  • $\begingroup$ Pardon my axiomatic idiocy, but where does the axiom of choice come into play here? $\endgroup$ – user98602 Jun 7 '14 at 18:09
  • $\begingroup$ You have to choose the enumeration of the roots when you count them; and for example it turns out that $\Bbb Q$ can have two algebraic closures: the canonically defined countable one, and an uncountable one. But these models are fairly strange. $\endgroup$ – Asaf Karagila Jun 7 '14 at 18:15
  • $\begingroup$ Ah, I see. Neat! Do you have a reference for the nonstandard model of $\overline{\mathbb Q}$? $\endgroup$ – user98602 Jun 7 '14 at 18:16
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    $\begingroup$ There are a couple of mentioning of this here and on MathOverflow; keywords are Lauchli (the a should have umlauts), Hodges and "algebraic closure". $\endgroup$ – Asaf Karagila Jun 7 '14 at 18:22
  • $\begingroup$ Thank you for the reference! $\endgroup$ – user98602 Jun 7 '14 at 18:31
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It suffices to show that the prime field $\Bbb{F}_p$ has a countable algebraic closure, as $K$ shares its algebraic closure with its prime field.

We can form a nested chain of extensions $$ E_1\subset E_2\subset\cdots \subset E_i\subset E_{i+1}\subset\cdots $$ of finite fields $E_i$ for all positive integers $i$ such that $E_1=\Bbb{F}_p$ and then declare $E_i$ to be the (up to isomorphism) unique degree $i$ extension of $E_{i-1}$. By induction we see that $E_i=\Bbb{F}_{p^{i!}}$ is the unique extension of $E_1$ of degree $i!$.

Note that during this process we also gave, for all $j$, a specific way of including $E_j$ as a subfield of $E_{j+1}$. Such inclusion maps allow us to form the direct limit (it would be a union, if we had a set that contains all the fields $E_i$ as subsets) $$ E=\lim_{\to}E_i. $$ Also observe that $E_k$ contains a unique copy of all the finite fields $\Bbb{F}_{p^\ell}$ such that $\ell\mid k!$. In particular a copy of $K$ is in there.

Now

  • $E$ is a countable set, because it is a nested union of countably many finite subsets.
  • All the elements of $E$ belong to one of the fields $E_i$, and hence they are all algebraic over $\Bbb{F}_p$.
  • If $f(x)=\sum_{i=0}^n a_ix^i\in E[x]$ is a polynomial of degree $n$, then the coefficient $a_i\in E_{\ell_i}$ for some $\ell_i\in\Bbb{N}$. Thus $f(x)\in E_M$, where $M=\max\{\ell_i\mid i=0,1,\ldots,n\}$. Therefore $f(x)$ splits into linear factors in $E_{M+n}$, and thus also in $E$. Therefore $E$ is algebraically closed.
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  • $\begingroup$ I'm not sure whether you need axiom of choice, when you do it this way. Producing the direct limit in general requires countable choice (correct me if I'm wrong), but here we could map all the elements of $E$ (or all the elements of all the $E_i$ in a compatible way) to $\Bbb{N}$ and be done with it, me thinks. $\endgroup$ – Jyrki Lahtonen Jun 7 '14 at 17:46
  • $\begingroup$ Since $\Bbb F_p$ is a well-ordered field, we can construct a well-ordered algebraic closure. It might have another, non-isomorphic, algebraic closure though. I'm not 100% sure that you can uniformly enumerate the $E_i$'s, though. $\endgroup$ – Asaf Karagila Jun 7 '14 at 18:27
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The reason is that the algebraic closure is the union of all its finite subextensions of $K$. If $K$ is finite (or even countable, like the rationals), each one is finite (respectively, countable), and there are countably many. The union of countably many countable sets is countable.

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  • $\begingroup$ It's not exactly by definition... $\endgroup$ – Asaf Karagila Jun 7 '14 at 14:50
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    $\begingroup$ It's not really by definition at all. The algebraic closure is usually defined to be the smallest in (the containment sense) of all fields, $E$, containing $K$ such that every polynomial with coefficients in $K$ has roots in $E$ $\endgroup$ – Stella Biderman Jun 7 '14 at 17:49
  • $\begingroup$ One could also define it to be the largest algebraic extension, in which case, what I said is true. Nonetheless I have removed those words and, I hope, any more opportunity for quibbling. $\endgroup$ – Ryan Reich Jun 7 '14 at 18:55

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