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I am studying Galois theory and I have reached a section on Primitive Elements (which is short but quite interesting). We define a primitive element as:

Let $K$ be a field and $L$ an extension of $K$. If there exists an $\alpha\in L$ such that $L=K(\alpha)$, then $\alpha$ is a primitive element.

It got me thinking... is $i=\sqrt{-1}$ a primitive element? Clearly $\mathbb{R}$ is a field and $\mathbb{C}$ is an extension of $\mathbb{R}$. But $\mathbb{R}(i)=\mathbb{C}$, so does that make $i$ a primitive element? The Galois theory I've looked at only really concerns $\mathbb{Q}$ and extensions of it, so there is no mention of this.

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    $\begingroup$ Note that $i\neq \sqrt{-1}$, but $i^2=-1$. Those are two different things. The square root function only takes positive numbers as arguments (and outputs the positive root). $\endgroup$
    – user304329
    May 1, 2017 at 10:35

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Yes: $\mathbb C = \mathbb R(i)$, so $i$ is a primitive element for $\mathbb C$ over $\mathbb R$.

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  • $\begingroup$ Thank, I thought it made sense. I just wanted some confirmation. $\endgroup$
    – JSharpee
    May 1, 2017 at 10:54

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