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Show that the polynomial:

$$g(X)=1+X+X^2+\ldots+X^{p-1}$$

is irreducible over $\mathbb{Q}$, where $p$ is prime.

I am not sure how to approach this.

Chris Brooks
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cf12418
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  • You could substitute $x=y+1$ and apply Eisenstein's criterion... (I wouldn't have come up with this myself neither, so don't worry. There should be a more straightforward way.) – Bart Michels May 19 '14 at 17:40
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    Barto's suggestion is typical and has been done many, many times on MSE. – Git Gud May 19 '14 at 17:40
  • The Eisenstein trick is the simplest and quickest way to prove this (possibly except for the cases $p=2$ or $3$) – Andrea Mori May 19 '14 at 17:46
  • can you show the next step on this trick? im not sure wether i should be expnding out the brackets or not, thanks – cf12418 May 19 '14 at 17:52

1 Answers1

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Substitute $X=u+1$. $g(u)=g(X+1)=\dfrac{(X+1)^p-1}{X}=X^{p-1}+pX^{p-2}+\cdots +p$. Then use Eisenstein's criterion. The constant term $p$ is not divisible by $p^2$ though the other coefficients are divisible by $p$.

Pedro
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  • please could you explain how you get from the fraction to the final line? thanks – cf12418 May 19 '14 at 18:10
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    you develop $(X+1)^p$ using binomial formula: the constant term is obviously $1$ and it is a classical result that $C_p^k$ is divisible by $p$ for $0<k<p$. –  May 19 '14 at 18:16