I claimed that $\mathbb{Q}[X]/(X^3-2)$ is a field because $(X^3-2)$ is a maximal ideal of $\mathbb{Q}[X]$.
Show that $Y^2+Y+1$ is irreducible over the field $\mathbb{Q}[X]/(X^3-2)$
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ring-theory
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2That's fine, if you can prove $(x^3-2)$ is a maximal ideal. But what does that have to do with the question about $y^2+y+1$? – 2012-11-17
1 Answers
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Solutions of $Y^2+Y+1=0$ are $\dfrac{-1\pm \sqrt{-3}}{2}$. Call one of them $\alpha$. Certainly $\alpha$ is not rational. If $Y^2+Y+1$ splits over our field, $\mathbb{Q}(\alpha)$ is isomorphic to a subfield of $\mathbb{Q}[X]/(X^3-2)$. This is impossible, since $\mathbb{Q}(\alpha)$ has degree $2$ over the rationals, while $\mathbb{Q}[X]/(X^3-2)$ has degree $3$, and $2$ does not divide $3$. The fact that $\mathbb{Q}[X]/(X^3-2)$ has degree $3$ over the rationals follows from the fact that $X^3-2$ is irreducible over the rationals. To show this, show that $\sqrt[3]{2}$ is not rational.