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 .
Answer
$\mathbb{F}_q(y)$ (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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