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Dummit and Foote, 13.5.5:

For any prime $p$ and nonzero $a \in \mathbb F_p$ prove that $x^p-x+a$ is irreducible and separable over $\mathbb F_p$

The question goes on to suggest two approaches to proving the irreducibility (separability follows):
1. Prove first that if $\alpha$ is a root then $\alpha + 1$ is also a root.
2. Suppose it's reducible and compute the derivatives.

Now I've solved the problem using the first hint, but only after trying for hours to find the contradiction given by the second approach (I'm stubborn). I'd really like to see if it is possible to get the irreducibility by that approach. Note that the derivative here is the algebraic definition rather than the analytic notion.

Right away we see from the second approach, that assuming $f(x)=x^p-x+a = g(x)h(x)$ and taking the derivatives on each side $D_xf(x)= g(x)D_xh(x)+(D_xg(x))h(x)=px^{p-1} -1=-1$

I've tried comparing the coefficients of each side to no avail.

user26857
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Ben
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1 Answers1

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For what it's worth, I agree with you: the derivative being $-1$ is obviously relevant for separability, but for irreducibility, not so much. But the first hint is all that's needed for irreducibility, anyway, so probably it's just some kind of typo---they intended the second hint to apply for separability.

Stephen
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