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A problem of finding a Galois group of $x^5 - 5 a^4 x + a$ appeared in a previous prelim at my school for $a \in \mathbb N$, and the only hard part seems like showing that the polynomial is irreducible, since it has 2 complex roots.

However, I do not know about any algebraic way to show that above is irreducible—other person from my school showed it by using rational root theorem and Rouche’s theorem.

So I was wondering if there is any way to show that above polynomial is irreducible in a purely algebraic way.

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  • $\begingroup$ Irreducible over what? $\mathbb{Q}$? $\endgroup$ Mar 15 at 12:48
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    $\begingroup$ Is the rational root theorem not an algebraic method? $\endgroup$
    – aschepler
    Mar 15 at 12:50
  • $\begingroup$ @Cameron Williams Yes $\endgroup$ Mar 15 at 13:11
  • $\begingroup$ @aschepler It was used to conclude there is no linear factor, but Rouche’s theorem is not an algebraic method. $\endgroup$ Mar 15 at 13:12
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    $\begingroup$ If $a$ has at least one prime factor that has an exponent of $1$, then the polynomial is irreducible by Eisenstein's criterion. $\endgroup$
    – Tob Ernack
    Mar 15 at 13:13
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Too long for a comment. For $a\in \Bbb{Z}$

  • If $a=0$ then it is reducible, if $a=\pm 1$ then it is irreducible $\bmod 3$.

  • If $a$ is not a 5th power then $f(x)=x^5+5a^4x+a$ is irreducible over $\Bbb{Q}_p$ whenever $5\nmid v_p(a)$, because $f(\gamma)=0$ gives that $v_5(\gamma)=v_5(a)/5$ so that $[\Bbb{Q}_p(\gamma):\Bbb{Q}_p]\ge 5$.

  • Otherwise $a= b^5$. Let $g(x)=a^{-1} f(xb)=x^5+5b^{16}x+1$, $h(x)=g(x-1)=x^5 - 5x^4 + 10x^3 - 10x^2 + 5xb^{16} + 5x- 5b^{16}$

    If $5\nmid b$ then it is Eisenstein at $5$.

  • It remains to check the case $a=(5c)^5$.

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    $\begingroup$ See here math.stackexchange.com/questions/193201/… to complete the proof. I assume the same method can be employed without using p-adics (since it is a qual problem); namely one can reduce to a degree $2$ and a degree $3$ factor and try to match coefficients. $\endgroup$
    – Vlad Matei
    Mar 15 at 15:22
  • $\begingroup$ Indeed I was trying to see why it didn't have a linear factor. $\endgroup$
    – reuns
    Mar 15 at 15:32
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    $\begingroup$ @reuns, your polynomial $g(x)=x^5+5b^{16}x+1$ clearly doesn't have a linear factor if $b\not=0$. (The rational root theorem says the only possible roots to check are $1$ and $-1$.) $\endgroup$ Mar 15 at 15:37

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