-1

I'm currently studying field theory and working through some exercises on algebraic closures. I came across an interesting question in an advanced algebra course, and I'd like some help understanding it.

The question is:

Let $p$ be a prime number. Prove that there exists an algebraically closed field of characteristic $p$ which is not finite.

Luciano
  • 127
Lsp
  • 27
  • 3
  • 7
    In fact all algebraically closed fields are infinite. Even for F_p since every field of order p^k is algebraic over F_p, you can consider the union of all such finite fields, which gives you an infinite algebraic extension – Ubik May 23 '25 at 13:15
  • 1
    Yes, if $F$ is a finite field, then the polynomial the polynomial $1+\prod_{r\in F}(x-r)$ has no roots in $F,$ so $F$ is not algebraically closed. The hard part in constructing the algebraic closure is that "the union" is somewhat false. There are lots of choices needed to define it as a union. Specifically, you need to pick one of the inclusions of $\mathbb F_{p^n}\to\mathbb F_{p^m}$ when $n\mid m$ so that they are consistent. @Ubik Once you have the algebraic closure, it is the union of the finite subfields, but that assumes you have the closure already. – Thomas Andrews May 23 '25 at 14:30
  • @ThomasAndrews Maybe I didn't explain it well, but I didn't mean that this union is the closure, but that you can find a unbounded chain of finite field extensions, then if one takes the union one gets a infinite algebraic extension, regardless of being the closure or not, this already shows that the algebraic closure is infinite. I agree that constructing the closure is not as simple as taking the union. – Ubik May 23 '25 at 15:20
  • 2
    Yeah, the simplest way to construct is to take the direct limit of $\mathbb F_{p^{n!}},$ picking any one inclusion $\mathbb F_{p^{n!}}\to\mathbb F_{p^{(n+1)!}},$ for each $n.$ @Ubik – Thomas Andrews May 23 '25 at 15:27

1 Answers1

2

Assuming you know that every field admits an algebraic closure (it's hard to know what you know or don't know, you didn't write it down in your message), then this comes from the fact that an algebraically closed field is infinite, as Ubik said.

To demonstrate this, suppose you have a finite field K and consider $P \in K[X]$ defined by $$P = \prod_{\alpha \in K} (X-\alpha) + 1.$$ What are the roots of P? What do you deduce?

Dietrich Burde
  • 140,055
Francis H.
  • 823