$K=\mathbb Z_{p}(t)$ is the field of all rational polynomials over $\mathbb Z_{p}$ .

The polynomial $$f(x)=x^{p} -t $$ has to be irreducible over $K[x]$.

So the polynomial is in $\mathbb Z_{p} (t)[x]$

Now $f'(x)=0$ so I guess this polynomial will not be separable in its splitting field, shall have repeated roots.

Can I arrive at a contradiction from here $?$



Much more generally, if $k$ is a completely arbitrary field of any characteristic and $n\geq 1$ an arbitrary integer, then the polynomial $P=P(x,t)=x^n-t$ is irreducible over $k(t)$.

Indeed $P$ is irreducible in $k[x,t]$ (since it is of degree one in $t$), thus also equivalently in $(k[t])[x]$ and finally also in $k(t)[x]$.
The last step is the crucial one: if A is a UFD (here $A=k[t]$) with fraction field $K=\operatorname {Frac}(A)$, then a non constant polynomial irreducible in $A[x]$ is automatically irreducible in $K[x]$.


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