Exercise 1028Number theory › IrrationalityL1Added 3 Nov 2026

Rational coefficients and irrational roots

#rational coefficients#irrational roots#conjugates

Statement

Let $P\in\mathbb{Q}[X]$ . Show that if $\sqrt2$ is a root of $P$ , then $-\sqrt2$ is also a root.

Hint

Use the conjugate roots theorem for real polynomials.

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Conjugation. The map $\sigma:\mathbb{Q}(\sqrt2)\to\mathbb{Q}(\sqrt2)$ , $a+b\sqrt2\mapsto a-b\sqrt2$ , is a field automorphism.

Apply to $P$ . $P(\sqrt2)=0\Rightarrow\sigma(P(\sqrt2))=P(\sigma(\sqrt2))$ , i.e. $P(-\sqrt2)=0$ .

Conclusion. The conjugate $-\sqrt2$ is also a root.

Consequence. $X^2-2$ divides $P$ if $\sqrt2$ is a root of $P$ .