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I was looking at a question from Artin from Algebra which says that an irreducible polynomial g in F[x] where F is subfield of $\mathbb{C}$. So as per the proofs I have seen so far says as - $$g(x)=(x-\alpha)^2*p(x)$$ and $$g'(x) = (x-\alpha)*q(x)$$ where neither $(x-\alpha) \in F[x]\; nor\; p(x)$. Now finally to conclude they say that gcd(f, f') = 1 but how can f' will belong to F[x] this is not necessary because if we consider $\mathbb{Z}/2\mathbb{Z}$ here let $g(x) = x^5+1$ then $g'(x) = 5x^4+1$. and clearly $5\in \mathbb{Z}/2\mathbb{Z}$ until I write 5 as 1 but then g'(x) would not be derivative of g(x).

Thanks in advance.

user3138843
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1 Answers1

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As $[5]=[1]$ (equivalence classes) in $\Bbb F_2$, $g'(x)=x^4$ in $\Bbb F_2[x]$.

See Formal derivatives over finite fields., Introduction to Finite Fields and Appendix B Finite Fields.

Quote from the last link:

The derivative of the polynomial $X^q-X$ is the constant polynomial -1.
Martín-Blas Pérez Pinilla
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  • In $\Bbb F_2[x],$ $g'(x)=x^4\ne x^4+1.$ Anyway all this seems completely off topic since the OP's question had a second flaw: $\Bbb F_2\not\subset\Bbb C.$ – Anne Bauval Jan 24 '23 at 22:26