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:

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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 $\alpha \in F$.

  • 1
    $\begingroup$ Certainly, if the polynomialk is not reducible, it won't split into (several) distinct factors $\endgroup$ – Hagen von Eitzen Apr 23 '19 at 13:32
  • $\begingroup$ @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? $\endgroup$ – user193319 Apr 23 '19 at 13:35

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