Let $\mathbb{Z}_2[t]$ be a polynomial ring over $\mathbb{Z}_2 $ and $F = \mathbb{Z}_2(t)$ its field of fractions.
Show that the polynomial $x^2 − t \in F[x]$ is irreducible over $F$, but that it has a
multiple root in some extension field of $F$. In particular, conclude that $F$ is not a
perfect field.
In this answer shows why $x^2-t$ is irreducible over $F[x]$ using a generalization of the Eisenstein criterion. I can understand how this works, but I wonder if in this case irreducibility can be determined in some more elementary way. For example, conclude that this polynomial cannot factor as a product of lower degree polynomials (?)
For the second part of the exercise, if $a$ is a root of $x^2-t$ in a splitting field $K$ of $F$. Note that $a^2=t\in F$. So $x^2-t=x^2-a^2=(x-a)^2\in F[x]\subset K[x]$. Thus, $a$ is a multiple root of $x^2-t$ in $K$, and hence $F$ is not a perfect field. In this part, Am I right?
The definition of perfect field in my book is
A field $F$ is said to be perfect if no irreducible $f (x) \in F[x]$ has
multiple roots in any splitting field of $f (x)$ over $F$
And the definition of multiple root is
Let $F$ be a field, $f (x) \in F[x]$ and $a \in F$. We say that $a$ is a
multiple root of $f (x)$ if $(x − a)^2\mid f (x)$
Best Answer
For your first question, there is indeed an elementary way: if $X^2-T$ is not irreducible, it factors as the product of two monic linear (in $X$) polynomials, and hence it has a root in $\mathbf F_2[T]$. However, this is impossible for degree reasons.
For your second question, your argument is correct.
Note that the extension in which the polynomial $X^2-T$ splits into linear factors is (isomorphic to) the field $$\mathbf F_2(T)[X]/(X^2-T),$$ in which the root is the congruence class $\xi$ of $X\bmod X^2-T$.