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I'm trying to prove $\mathbb C =\mathbb R(a+bi)$, where $(a+bi) \in \mathbb C$ and $b \neq0$

I'm doing like this:

$[\mathbb C:\mathbb R]=[\mathbb C: \mathbb R(a+bi)][\mathbb R(a+bi):\mathbb R]$.

If we prove $[\mathbb R(a+bi): \mathbb R]=2$ we done, since we know $[\mathbb C:\mathbb R]=2$, then $[\mathbb C:\mathbb R(a+bi)]=1$ and $\mathbb R(a+bi) =\mathbb C$.

In order to prove $[\mathbb R(a+bi): \mathbb R]=2$, I'm trying to find the minimal polynomial of (a+bi) over $\mathbb R$. I found a candidate: $p(x)=x^2 -2ax+a^2+b^2$. I know that $p(a+ bi)=0$, but how to prove this polynomial is irreducible? thanks.

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For a different approach, you could show show $i \in \mathbb{R}(a+bi)$. (Assuming that you can take $\mathbb{R}(i) = \mathbb{C}$ as prior knowledge) –  Hurkyl Nov 4 '12 at 2:02

3 Answers 3

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We have, as you write \[ [\mathbb C : \mathbb R(a+bi)] [\mathbb R(a+bi):\mathbb R] = [\mathbb C:\mathbb R] = 2 \] So $[\mathbb R(a+bi): \mathbb R] \le 2$, as $a+bi \not\in \mathbb R$, we have $[\mathbb R(a+bi): \mathbb R] > 1$, so $[\mathbb R(a+bi): \mathbb R] = 2$.

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Your polynomial $p(x)=x^2-2ax+a^2+b^2$ is irreducible over $\mathbb R$ because its discriminant is $(-2a)^2-4(a^2+b^2)=-4b^2<0.$ So your approach looks OK to me. [for $x=a+bi$ you do indeed have $p(x)=0$.]

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To show irreducibility, you need only observe that its roots aren't real.

As an alternative, you could use a double-inclusion approach. We already know that $a+bi\in\Bbb C$ and $\Bbb R\subset\Bbb C$, so since $\Bbb R(a+bi)$ is the smallest field containing $\Bbb R$ and having $a+bi$ as an element, then $\Bbb R(a+bi)\subseteq\Bbb C$. All you need to show, then, is if $x+yi\in\Bbb C$ (with $x,y\in\Bbb R$), then $x+yi\in\Bbb R(a+bi)$--that is, $x+yi=c+d(a+bi)$ for some real $c,d$.

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