Prove that every finite extension field of $\Bbb R$ is either $\Bbb R$ itself or isomorphic to $\Bbb C$.

I tried in this way.Let $E$ be a a finite extension of $\Bbb R$. Then $E$ is an algebraic extension of $\Bbb R$. Let $\alpha \in E-R$ .then $\exists p(x)\in \Bbb R[x]$ such that $p(\alpha)=0$. Surely $\deg(p(x)=2$ since otherwise it would be reducible.

Now $\Bbb R(\alpha)\cong \Bbb R[x]/<p(x)>$. Now $[\Bbb R(\alpha):\Bbb R]=2$ .Also $[\Bbb C:\Bbb R]=2\implies \Bbb R(\alpha)\cong \Bbb C$.

Since $\Bbb C$ is algebraically closed so is $\Bbb R(\alpha)$ and so it has no proper algebraic extensions.

Can I conclude from here that $E=\Bbb C$?

Please suggest steps if needed.I would be very grateful

  • $\begingroup$ The proof is almost correct. It is necessary to show that any degree $2$ extension of $\mathbb{R}$ is isomorphic to $\mathbb{C}$. $\endgroup$ – Slade Jan 11 '16 at 9:22
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    $\begingroup$ I would conclude like this: $\mathbb{C}$ is algebraic closed, thus we can embed $\mathbb{R}(\alpha)$ into the complex numbers. Now the field extensions $\mathbb{C} / \mathbb{R}$ and $\mathbb{R}(\alpha) / \mathbb{R}$ have both degree $2$, thus $\mathbb{C} / \mathbb{R}(\alpha)$ has degree $0$, i.e. they are equal. $\endgroup$ – Paul K Jan 11 '16 at 9:22
  • $\begingroup$ You need to conclude first that $\mathbb{C}$ is the algebraic closure of $\mathbb{R}$, thus every algebraic extension of $\mathbb{R}$ is contained in $\mathbb{C}$. After that, you can use the degrees of the extension like you did. $\endgroup$ – Ryan Jan 11 '16 at 9:27
  • $\begingroup$ @Slade; can't I say like this :::: since dimension of the vector space of $\Bbb C$ over $\Bbb R$ is $2 $ and dimension of the vector space of $\Bbb R(\alpha)$ over $\Bbb R$ is $2 $ and any two finite dimensional vector space over a same field are isomorphic then $\Bbb R(\alpha)\cong \Bbb C$ $\endgroup$ – Learnmore Jan 11 '16 at 10:13
  • $\begingroup$ @menag;it will be convenient if you post an answer to this because I am not getting where do you want to plug your comments in my answer $\endgroup$ – Learnmore Jan 11 '16 at 10:14

You have already correctly deduced that the polynomial $p$ is of degree $2$ if $p$ is irreducible. Therefore we have $$[\mathbb{R}(\alpha) : \mathbb{R}] = 2.$$ Moreover, since $\mathbb{C}$ is the algebraic closure of $\mathbb{R}$ there is an embedding from $\mathbb{R}(\alpha)$ to $\mathbb{C}$ and we can identify it with a subset, i.e. $\mathbb{R}(\alpha) \subseteq \mathbb{C}$. Now we have $$2 = [\mathbb{C} : \mathbb{R}] = [\mathbb{C} : \mathbb{R}(\alpha)] \cdot [\mathbb{R}(\alpha) : \mathbb{R}] = [\mathbb{C} : \mathbb{R}(\alpha)] \cdot 2.$$ We deduce $[\mathbb{C} : \mathbb{R}(\alpha)] = 1$ and thus $\mathbb{C} = \mathbb{R}(\alpha)$.

  • $\begingroup$ can you please explain why is this true? Since C is algebraic closure of R there is an embedding...$ Why is such an embedding there? $\endgroup$ – Learnmore Jan 11 '16 at 11:19
  • $\begingroup$ The polynomial $p$ has a zero $\beta \in \mathbb{C}$. Define the homomorphism $\tau : \mathbb{R}(\alpha) \to \mathbb{C}$ by $\tau(\alpha) = \beta$ (ofc, $\tau(x) = x$ for $x \in \mathbb{R}$ and the rest by homomorphy). $\endgroup$ – Paul K Jan 11 '16 at 14:16

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