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I have to prove that $X^{4}-10X^{2}+1$ cannot be put as a product of two irreducible polynomials in $\mathbb{Z}[X]$ and show that is irreducible in $\mathbb{Q}[X]$ the exercise also requires finding every root in $\mathbb{Q}[X]$ and in $\mathbb{R}[X]$.

For the second and third part I tried to "complete squares"...

\begin{align*} X^{4}-10X^{2}+1 & =\left(X^{2}-5\right)^{2}-24=\left(X^{2}-5\right)^{2}-\left(2\sqrt{6}\right)^{2}=\\ & =\left(X^{2}-5+2\sqrt{6}\right)\left(X^{2}-5-2\sqrt{6}\right) \end{align*}

The sum of irrationals is irrational and the square root of an irrational number is irrational so the polynomials above are irreducible in $\mathbb{Q}[X]$, I don't know the proper proof for, after doing that, conclude that $X^{4}-10X^{2}+1$ is irreducible in $\mathbb{Q}[X]$.

Obviously the irrational square roots are $X=\pm\sqrt{5-2\sqrt{6}}$ and $X=\pm\sqrt{5+2\sqrt{6}}$ which are in $\mathbb{R}[X]$ and not in $\mathbb{Q}[X]$. Any ideas?

José Carlos Santos
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Jack Talion
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1 Answers1

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Let$$r_1=\sqrt{5+2\sqrt6},\ r_2=-\sqrt{5+2\sqrt6},\ r_3=\sqrt{5-2\sqrt6},\text{ and }r_4=-\sqrt{5-2\sqrt6}.$$These numbers are the roots of $p(x)$. Since this polynomial has no rational roots, if it was reducible in $\mathbb Q[x]$, one of its factors would have to be $(x-r_1)(x-r_2)$, $(x-r_1)(x-r_3)$, or $(x-r_1)(x-r_4)$. So, check that none of these polynomials belong to $\mathbb Q[x]$.

José Carlos Santos
  • 427,504