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As the question says , i am confused about the notation above . At first glance it looks like i can compare it with something like $\mathbb R(i)$ . But i am still confused how the field $\mathbb F_{p^n}(y)$ looks like ? Consider a polynomial $f=x^2-y$ , it clearly belongs to $\mathbb F_{p^n}(y)[x]$ . Is $f$ irreducible in $\mathbb F_{p^n}(y)$ , if i take $p=2, n=1 $ , i know that it is irreducible buy i don't understand exactly why ?

What would be the field extension with respect to $f$ ?

I am quite comfortable with finite fields .

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$\mathbb{F}_q(y)$ (with $q = p^n$) is presumably the field of rational functions over $\mathbb{F}_q$. It consists of fractions whose numerator and denominator are polynomials over $\mathbb{F}_q$.

$f$ is irreducible over $\mathbb{F}_q(y)$, no matter what $q$ is. We can see this, because the roots are clearly $\pm \sqrt{y}$, but $y$ doesn't have any square roots in $\mathbb{F}_q(y)$. (Note that if $p=2$, this is a double root! But it doesn't behave like double roots "usually" behave, because $f$ is inseparable.)

The field extension that $f$ defines is $\mathbb{F}_q(x)$; the embedding $\mathbb{F}_q(y) \to \mathbb{F}_q(x)$ sends $y \mapsto x^2$.

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@Hurkyl : The most important thing that i was missing was "what the field of rational functions are". – Theorem Jan 13 '13 at 11:04
@Hurkyl: Well, your proof, saying that $\sqrt{y} \notin \mathbb{F}(y)$, is just a reformulation of the claim, and therefore not a complete proof. One can use either Eisenstein's criterion (see my answer), or a direct calculation with fractions of polynomials, or the following argument: $\sqrt{y}$ is integral over the factorial and therefore normal domain $\mathbb{F}[y]$. Therefore, $\sqrt{y} \in \mathbb{F}(y)$ would imply $\sqrt{y} \in \mathbb{F}[y]$. But then $1 = \mathrm{deg}(y) = \mathrm{deg}(\sqrt{y}^2) = 2 \cdot \mathrm{deg}(\sqrt{y})$, a contradiction. – Martin Brandenburg Jan 13 '13 at 11:11
@MartinBrandenburg Why do you need $\sqrt{y}\in\mathbb F[y]$? If $\sqrt{y}\in\mathbb F(y)$, then $y=f^2(y)/g^2(y)$. By looking to the degrees isn't this an obvious contradiction? – user26857 Jan 13 '13 at 11:25
@Martin: I myself would have used unique factorization of fractions over $\mathbb{F}[y]$. – Hurkyl Jan 13 '13 at 11:43
@Theorem: I thought that might have been the case! But I added the rest of the stuff just in case. – Hurkyl Jan 13 '13 at 11:44

If $K$ is any field and $n \in \mathbb{N}^+$, then $X^n - y \in K(y)[x]$ is irreducible by Eisenstein's criterion.

So this doesn't have to do anything with finite fields. On the other hand, the geometry of function fields in positive characteristic is a vast area of research in arithmetic geometry.

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