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Let $F := \mathbb{F}_p(x)$, the field of rational functions in one variable over the prime field $\mathbb{F}_p$. How can we show that $F$ is an infinite field of finite characteristic?

Thoughts so far

$F$ is clearly infinite since (for example) $1,x,x^2,x^3 \ldots$ etc. are all contained in $F$.

Suppose that $f \in F$. Then $f=\frac{p_1(x)}{p_2(x)}$ with $p_1,p_2$ having coefficients in $\mathbb{F}_p$. I'm not sure how we can show that $\text{char}F=p$ though.

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    $\begingroup$ Also, this question is extremely straightforward. Do you understand what $\mathbb{F}_p(x)$ is? Do you know the meaning of the characteristic of a field? It's okay if you don't, but you should say in your question that those are things you need to have explained. $\endgroup$ – Zev Chonoles Jan 27 '13 at 21:25
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    $\begingroup$ Perhaps I misunderstand what $F$ actually is. I've added some thoughts as you've suggested. $\endgroup$ – user59873 Jan 27 '13 at 21:36
  • $\begingroup$ $\mathbb{F}_p$ is a subfield of $\mathbb{F}_p(x)$, so... More generally, it's rather easy to prove that any field extension of a characteristic $p$ field has again characteristic $p$ (where $p$ is either a prime or $0$). $\endgroup$ – egreg May 3 '13 at 15:56
  • $\begingroup$ See also math.stackexchange.com/questions/58424/… $\endgroup$ – Martin Sleziak Feb 15 '15 at 11:51
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Hint if you take $y \in \mathbb F_p(x)$ how else can you write

$$\underbrace{y+y+\cdots+y}_{\text{p times}}?$$

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    $\begingroup$ Sorry to take the point out of the joke, but I can't help adding the answer: $y+y+\dots+y = y\cdot(1+1+\dots+1)$ $\endgroup$ – Berci Jan 27 '13 at 21:49
  • $\begingroup$ I have no idea whether I'm being completely absent-minded here (given the comments so far that seems to be the case), but I simply do not understand how the characteristic of the field $\mathbb{F}_p$ can be shown to be the same as the characteristic of the field of fractions defined above. Or indeed even see why it should be. $\endgroup$ – user59873 Jan 27 '13 at 21:53
  • $\begingroup$ I see that every element of $F$ that is not in $\mathbb{F}_p$ is transcendental. If that helps... $\endgroup$ – user59873 Jan 27 '13 at 21:57
  • $\begingroup$ What happens if you add something $p$ times? You get $p\cdot y$? What's $p$ equal to in this ring? $\endgroup$ – JSchlather Jan 27 '13 at 22:03
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Jacob's answer should be enough (and better), but you can also think of the space as $\mathbb{Z}$-module but then that $\mathbb{Z}$ also has to be modded by $p\mathbb{Z}$, right? Think about how to add polynomials on top of this and from here the characteristic being $p$ will be clear.

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