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I've been solving problems from my Galois Theory course, and at some point I sated that, given $p$ prime, $$\frac{X^p-1}{X-1} = X^{p-1}+X^{p-2}+\cdots+X+1$$ is irreducible in $\mathbb{Q}[X]$, because I felt I saw that in my course classes, but I'm not sure how can I prove or if it's even true.

Is it really irreducible in $\mathbb{Q}[X]$ for any $p$ prime? If that's the case, how can I prove it? Any help will be appreciated, thanks in advance.

Alejandro Bergasa Alonso
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  • This is not an answer (just simple observation) for $p≤5$, the polynomial is irreducible over $\mathbb Q$. This mean the special case is irreducible. So, if for any $p$ the polynomial is reducible over $\mathbb Q$, this should have been true for $p≤5$ as well. But this is a contradiction..It's just the first thing I think about.. – lone student Mar 31 '21 at 06:49

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Applying the linear substitution $X=Y+1$ to $f(X)=\frac{X^p-1}{X-1}$ we obtain $$f(X)=Y^{p-1}+{{p}\choose{1}}Y^{p-2}+\dots +{{p}\choose{p-2}}Y+{{p}\choose{p-1}}.$$ By Eisenstein's criterion, the latter is irreducible in $\mathbb{Q}[X]$.

Mathematician 42
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