Let \(p\) be a prime, let \(\mathbb F_p\) be the \(p\)-element field, and let \(K=\mathbb F_p(t)\) be the field of rational functions in \(t\) with coefficients in \(\mathbb F_p\). Consider the polynomial \(f(x)= x^p-t\in K[x]\).
Show that \(f\) does not have a root in \(K\).
Let \(E\) be the splitting field of \(f\) over \(K\). Find the factorization of \(f\) over \(E\).
Conclude that \(f\) is irreducible over \(K\).
Solution
1. The polynomial \(f(X)=X^p-t\) has no root in \(K=\mathbb F_p(t)\).
Suppose \(r\in K\) satisfies \(r^p=t\). Write \[ r=\frac{a(t)}{b(t)} \] with coprime \(a,b\in\mathbb F_p[t]\), \(b\ne0\). Then \[ a(t)^p=t\,b(t)^p. \] Let \(v_t\) denote the exponent of the irreducible polynomial \(t\) in a nonzero polynomial. Taking \(v_t\) gives \[ p\,v_t(a)=1+p\,v_t(b), \] which is impossible modulo \(p\). Hence \(f\) has no root in \(K\).
2. If \(\alpha\) is a root of \(f\) in a splitting field \(E\), then\[ f(X)=(X-\alpha)^p \]in \(E[X]\).
We have \(\alpha^p=t\). Since the characteristic is \(p\), all intermediate binomial coefficients \(\binom pj\) vanish in \(\mathbb F_p\), so \[ (X-\alpha)^p=X^p-\alpha^p=X^p-t=f(X). \] Thus \(\alpha\) is the unique root of \(f\) in an algebraic closure, with multiplicity \(p\).
3. The polynomial \(f(X)=X^p-t\) is irreducible over \(K\).
1. Therefore \(f\) is irreducible over \(K\).