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Does the splitting field of $X^3-2$ have a real embedding?

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How could it have? – Berci Oct 8 '12 at 13:11
Note that if $x \ne y$ are two distinct roots of $X^3-2$, and $t = \frac{x}{y}$, then $t^3 = \frac{2}{2} = 1$ and $t \ne 1$. – Joel Cohen Oct 8 '12 at 13:13
up vote 6 down vote accepted

No. Briefly, because in $\mathbb C$, it has $3$ distinct roots, so the splitting field contains these, but in $\mathbb R$ there is only one root (say, because of strict monotonicity of $x\mapsto x^3$), so no embedding can be possible, as all $3$ roots should still satisfy the polynomial, hence would go into the same element, $\sqrt[3]2$.

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