Why is $p(x)=x^2+t\in\mathbb{F}_2(t)[x]$ irreducible? Why is $p(x)=x^2+t\in\mathbb{F}_2(t)[x]$ irreducible?
The only argument I can think of is that $\sqrt t$ doesn't seem like a rational function, but I tried to prove this and got stuck when I had to square a polynomial...is this the way to prove this claim, or is there another argument ?
 A: You could try the following: If $p$ is reducible, it will neccesarily reduce to a pair of linear factors. Furthermore, as $p$ is monic, one can assume that the factors are monic as well. Thus, if $p$ were reducible, $p = (x+a)(x+b)$. By comparison of coefficients, $a + b = 0$ and $ab = t$, so $b = -a$ and thus $-a^2=t$. Now, $a = q_1/q_2$, where $q_1, q_2$ are polynomials in $t$, so $-q_1^2/q_2^2 = t$, or equivalently, $-q_1^2 = t q_2^2$. But since $\deg q_1^2$ is even and $\deg t q_2^2$ is odd, no such factorization can exist. This is, essentially, a rigourous version of your argument.
A: My algebra is rusty, so hopefully I don't make any embarassing mistake:
$\mathbb{F}_2[t]$ is an integral domain where the ideal $\langle t \rangle$ generated by $t$ is maximal, thus prime.
Then, by the Generalized Eisenstein Criterion the polynomial $x^2+t$ is irreducible over both $\mathbb{F}_2[t]$ and its field of fractions.
P.S. You actually don't even need to employ the GEC. All you need is the Gauss Lemma, which sais that since $\mathbb{F}_2[t]$ is an UFD, then $x^2+t$ is irreducible over $\mathbb{F}_2(t)$ if and only if $x^2+t$ is irreducible over $\mathbb{F}_2[t]$. The last follows from Eisenstein, but Eisenstein is overkill here :)
A: Just a thought. Say that $p(x) = x^2 + t \in \mathbb{F}_2(t)[x]$ is reducible. Then  there is a rational function $\frac{q(x)}{d(x)}$ with $q(x), d(x) \in \mathbb{F}_2(t)$ such that
$$\begin{align}
\left(\frac{q(x)}{d(x)}\right)^2 &= -t = t\quad \Rightarrow \\
q(x)^2 &= t d(x)^2.
\end{align}
$$
Now the polynomial with coefficients in the field $\mathbb{F}_2(t)$ on the left hand side has even degree (being the square of a polynomial),  but the polynomial on the right hand side has odd degree, so we have a contradiction.
