Why is $X^2-3$ irreducible over $\mathbb{Q}(\sqrt[3]{2})$ ?
2026-04-09 16:33:15.1775752395
Irreducible polynomial over $\mathbb{Q}(\sqrt[3]{2})$
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Your conjecture is right: if $X^2-3$ is reducible, then $\sqrt{3}\in\mathbb{Q}[\sqrt[3]{2}]$ and you have $$ \mathbb{Q}\subseteq\mathbb{Q}[\sqrt{3}]\subseteq\mathbb{Q}[\sqrt[3]{2}] $$ Can you apply the dimension formula?