# Reducibility Unnecessary Hypothesis?

Problem: Let $$F$$ be a field with $$\text{char} F = p$$ for some prime $$p$$. Show that if $$X^p - X - a$$ is reducible in $$F[X]$$, then it splits into distinct factors in $$F[X]$$

Solution in back of the book:

This solution confuses me slightly. Why is $$f(X)$$ splitting as two monic polynomials? Why is $$-a_1$$ the sum of $$r$$ roots? If $$f$$ shouldn't a priori be factored into two monic polynomials, shouldn't it be that $$-a_1$$ times sum unit is the sum of a certain number of roots? Also shouldn't it be the sum of $$p$$ roots? $$\alpha, \alpha +1,..., \alpha + p-1$$ are $$p$$ distinct roots. Why only $$r < p$$ roots?

Finally, why is reducibility necessary? If $$\alpha$$ is a root of $$f$$ in some extension $$E$$, then every $$\alpha + i$$ lies in $$E$$ too, so $$f(X) = (X- \alpha)(X-(\alpha +1))...(X-(\alpha + p - 1))$$ in $$E[X]$$. But we also know $$f(X) = 1_F \cdot X^p - 1_F \cdot X - a$$, so $$1_F = \sum_{i=0}^{p-1} (\alpha + i1_E) = (p-1)1_E \cdot \alpha + \frac{(p-1)p}{2} 1_E$$. But $$1_F = 1_E$$, because the multiplicative identity of a subfield is the same as the field extension, so the equation reduces to $$1_F = (p-1)1_F \cdot \alpha$$ which implies $$\alpha \in F$$.

Did I make a mistake?

EDIT I was being a knucklehead. Clearly the polynomial has to be reducible, otherwise the statement is incoherent. However, I am still wondering about my proof that $$\alpha \in F$$.

• Certainly, if the polynomialk is not reducible, it won't split into (several) distinct factors – Hagen von Eitzen Apr 23 '19 at 13:32
• @HagenvonEitzen Oh, yeah, you're right. So we have to tack on that hypothesis for the statement to even make sense. So, is what I did wrong? – user193319 Apr 23 '19 at 13:35