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Is there an infinite field F with characteristic of the field $F$ is $p$ (p is prime) and not algebraically closed field ?

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    $\begingroup$ $\mathbb{F}_p (t)$. $\endgroup$ – Sungjin Kim Jul 14 '16 at 21:52
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    $\begingroup$ If $\Omega_p$ is the algebraic closure of $\Bbb{F}_p$, any infinite proper subfield of $\Omega_p$ satisfies this condition. For example, we can consider $$K= \{ \alpha \in \Omega_p | [\Bbb{F}_p(\alpha ): \Bbb{F}_p] \mbox{ is a power of }2 \}$$ $\endgroup$ – Crostul Jul 14 '16 at 21:55
  • $\begingroup$ FYI, the characteristic is always prime if it is non-zero (so the clarification you give is unnecessary) $\endgroup$ – MCT Jul 14 '16 at 22:00
  • $\begingroup$ The field defined by Crostul is known as the quadratic closure of $\mathbb F_p$. → en.wikipedia.org/wiki/… $\endgroup$ – PseudoNeo Jul 14 '16 at 22:39
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$\mathbb{F}_p (t)$ is an example. It is defined as

$$ \mathbb{F}_p(t) = \{\frac{f(t)}{g(t)}: f(t),g(t)\ne 0 \in\mathbb{F}_p[t] \} $$

This is the field of rational functions with coefficients in $\mathbb{F}_p$. It is obvious that $\mathbb{F}_p (t)$ is infinite. However, it has characteristic $p$. This is because it contains subfield $\mathbb{F}_p$ and $char(\mathbb{F}_p) = p$. Also, if we consider $X^2 - t$, we see that it is not algebraically closed.

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  • $\begingroup$ Why $X^2-t$ does not have roots in $\mathbb{F}_p (t)$? $\endgroup$ – MeMe Jul 14 '16 at 22:11
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    $\begingroup$ @MeMe, try $X=f/g$ and compare the parity of the degrees in $f^2=tg^2$. $\endgroup$ – lhf Jul 14 '16 at 22:32
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$\Bbb{F}_p(t)$ is a perfect example of such a field.

If you look for an algebraic extension over $\Bbb{F}_p$, I suggest you to consider the following: if $\Omega_p$ denotes the algebraic closure of $\Bbb{F}_p$, any infinite proper subfield of $\Omega_p$ satisfies this condition. For example, we can consider $$K= \{ \alpha \in \Omega_p | [\Bbb{F}_p(\alpha ): \Bbb{F}_p] \mbox{ is a power of }2 \}$$

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