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How do you prove that algebraically closed fields of characteristic $p$ exist?

I have also read:

For a finite field of prime power order $q$, the algebraic closure is a countably infinite field that contains a copy of the field of order $q^n$ for each positive integer $n$ (and is in fact the union of these copies).

Why would the algebraic closure have to be characteristic $p$?

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    $\begingroup$ If adding $1$ added to itself a prime number of times yields zero in the finite field $F$, then it also does so in any field $K$ containing the finite field $F$! $\endgroup$ – anon Jun 24 '15 at 2:36
  • $\begingroup$ Oh, I answered the title, not the body. Those are two entirely different things. Can you please find a title that is closer to the question? $\endgroup$ – Thomas Andrews Jun 24 '15 at 2:48
  • $\begingroup$ If you are interested: Keith Conrad carried out Artin's classical construction of an algebraic closure (of any field) and made it available online. $\endgroup$ – Stefan Mesken Jun 24 '15 at 2:49
  • $\begingroup$ The characteristic of a subfield is the same as the characteristic of the field of which it is a subfield. $\endgroup$ – Michael Hardy Jun 24 '15 at 2:52
  • $\begingroup$ @ThomasAndrews, I meant for the questions to be different. $\endgroup$ – user240033 Jun 24 '15 at 3:01
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This only answers the main question.

Just take a sequence of inclusions:

$$\mathbb F_p\to\mathbb F_{p^2}\to\mathbb F_{p^6}\to\cdots\to \mathbb F_{p^{n!}}\to\cdots$$

Then the direct limit (essentially the union) is algebraically closed and a field.

That's because any polynomial in elements of this field has coefficients contained in one $\mathbb F_{p^{k!}}$. Thus, it splits in $\mathbb F_{p^{d\cdot k!}}$ for some $d$, and then note that $d\cdot k!\mid(dk)!$. So the polynomial splits in $\mathbb F_{p^{(dk)!}}$.

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Let $F$ be the field of prime power order. Let $p$ be that prime. The field $F$ is a subfield of it algebraic closure, which let us call $G$. your question seems to be: why must $G$ have characteristic $p$? We could as why $F$ must have characteristic $p$. And then, supposing $F$ does have characteristic $p$, why must $G$ have characteristic $p$?

Let's answer the second question first: Since $F\subseteq G$, the "$1$" in $F$ is the same as the "$1$" in $G$. It satisfies $\underbrace{1+\cdots+1}_p= 0$. Since this is the "$1$" in $G$, the field $G$ has characteristic $p$.

Now the first question: If $F$ has $p^n$ elements, why must $F$ have characteristic $p$? If you know the characteristic is some prime number $r$, then you have $\underbrace{1+\cdots+1}_r=0$, and $\{0, 1, 1+1, 1+1+1, \ldots, \underbrace{1+\cdots+1}_{r-1}\,\}$ is a subgroup of the additive group. The order of the subgroup divides the order of the whole group, which is $p^n$. If $p$ and $r$ are prime and $r$ divides $p^n$, then $r=p$.

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