A (algebraic) number field is defined to be a finite-degree extension field of $\mathbb{Q}$.

Let $f$ be a monic, irreducible polynomial in $\mathbb{Z}[X]$ and $\alpha$ be an complex-but-not-rational root of $f$. It is known that $\mathbb{Q}[\alpha]$ is a number field with extension degree of $\deg f$.

My question is that, for any number field $K$, if it's true that $K$ is isomorphic to some $\mathbb{Q}[\alpha]$ as defined above.

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    $\begingroup$ Yes. This is a consequence of the theorem stating that a finitely generated algebraic extension of $\Bbb{Q}$ is simple. It holds for any perfect field in place of $\Bbb{Q}$, but does not hold for fields like $\Bbb{F}_p(x)$. $\endgroup$ – Jyrki Lahtonen Dec 4 '16 at 7:17
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    $\begingroup$ See the Primitive Element Theorem $\endgroup$ – Hayden Dec 4 '16 at 7:18

For the question exactly as stated, the answer is no: $\mathbb{Q}$ is a finite-degree extension field of $\mathbb{Q}$, and thus an algebraic number field, but it cannot be written as a field extension $\mathbb{Q}[\alpha]$ where $\alpha$ is irrational.

The answer becomes yes if you remove the clause "complex-but-not-rational". As noted in the comments, this is the content of the primitive element theorem when given the fact that $\mathbb{Q}$ is perfect, a consequence of being characteristic zero.

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    $\begingroup$ (I am assuming that the question is about proving $K/\mathbb{Q}$ is a simple extension, rather than the simpler question about normalizing the representation to be defined by a monic integer polynomial) $\endgroup$ – Hurkyl Dec 4 '16 at 8:16

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