$x^3-\sqrt{2}$ is Irreducible Over the Field $\Q(\sqrt{2})$

Field theory problems and solution in abstract algebra

Problem 82

Show that the polynomial $x^3-\sqrt{2}$ is irreducible over the field $\Q(\sqrt{2})$.

 
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Hint.

Consider the field extensions $\Q(\sqrt{2})$ and $\Q(\sqrt[6]{2})$.

Proof.

Let $\sqrt[6]{2}$ denote the positive real $6$-th root of of $2$.
Then since $x^6-2$ is irreducible over $\Q$ by Eisenstein’s criterion, hence it is the minimal polynomial for $\sqrt[6]{2}$ over $\Q$. Therefore, we have $[\Q(\sqrt[6]{2}): \Q]=6$.
Similarly, we have $[\Q(\sqrt{2}):\Q]=2$.

Since $\Q(\sqrt[6]{2}) \ni (\sqrt[6]{2})^3=\sqrt{2}$ and $[\Q(\sqrt{2}):\Q]=2$, we must have
\[[\Q(\sqrt[6]{2}): \Q(\sqrt{2})]=3.\]

Note that $\sqrt[6]{2}$ is a root of the polynomial $x^3-\sqrt{2} \in \Q(\sqrt{2})[x]$.
Hence it must be the minimal polynomial for $\sqrt[6]{2}$ over $\Q(\sqrt{2})$.
In particular, the polynomial is irreducible over $\Q(\sqrt{2})$.


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