# How can I see that $\Bbb F_{p^2}(t)$ is a separable extension of $\Bbb F_p(t)$?

I am given 2 examples to see that a imperfect field can have both separable and inseparable extensions.

I am told that $\Bbb F_{p^2}(t)$ is a separable extension of $\Bbb F_p(t)$, where $\Bbb F_p(t)$ is imperfect.

But how can I see that fact that $\Bbb F_{p^2}(t)$ is a seperable extension? May I please ask for a proof?

Hint: How do you get the field $\mathbb{F}_{p^2}$ from the field $\mathbb{F}_p$? The addition of the transcendent variable $t$ does not change things much.
• Is it the fact that $[\Bbb F_{p^2}:\Bbb F_p]=2$ so if we write $\Bbb F_{p^2}=\Bbb F_p(\alpha)$, the minimal polynomial of $\alpha$ is of degree 2. By the quadratic formula, once we have one root $\alpha\in\Bbb F_p(\alpha)$, the other root of the polynomial is also in $\Bbb F_p(\alpha)$. So the minimal polynomial of the $\alpha$ we adjoint to $\Bbb F_p$ splitts in $\Bbb F_{p^2}$? – PropositionX May 17 '17 at 1:55
• I thought for while and I am worring about how to proceed. I think to conclude that it has no multiple roots, I need to conclude that the minimal polynomial is not of form $(x-\alpha)^2$ in its splitting field $\Bbb F_p(\alpha)$.May I please ask is that correct? What to do to conclude it? – PropositionX May 17 '17 at 2:12
• @PropositionX Yes, you are correct. For fields of characteristic $p>2$, then you are essentially adjoining the square root of an element, and in that case the minimal polynomial is $f(x)=(x-\alpha)(x+\alpha)$. In characteristic $2$ it is a little bit more subtle, but it turns out that $f(x)=x^2+x+1$ is the minimal polynomial which works and is separable. – TomGrubb May 17 '17 at 2:19