In the article The Classification Of Real Division Algebras (written by R. S. Palais) it is said that if D is finite dimensional division algebra over $\mathbb{R}$, then:

One way of stating the fundamental theorem of algebra is to say that if D is commutative (i.e. a field) then D is isomorphic over $\mathbb{R}$ to either $\mathbb{R}$ or the field $\mathbb{C}$ of complex numbers.

I do not understand why this is true. Can someone explain this to me?


Fundamental theorem of algebra, phrasing 1: A non-constant polynomial with coefficients in $\Bbb{C}$ is a product of linear polynomials with coefficients in $\Bbb C$.

Fundamental theorem of algebra, phrasing 2: $\Bbb C$ is algebraically closed (that is, it has no finite-degree field extensions).

Fundamental theorem of algebra, phrasing 3: If $D$ is a field extension of $\Bbb R$ that is (as a vector space) finite-dimensional, then $D = \Bbb{R}$ or $D$ is isomorphic to $\Bbb C$ over $\Bbb R$.

Phrasing 1 $\implies$ phrasing 2: Assume by contradiction that $\Bbb{C}$ has some finite-degree field extension $K$ ($K \ne \Bbb{C}$). Without loss of generality we can say that $K$ is a simple extension (otherwise we take some $\alpha \in K \setminus \Bbb C$ and replace $K$ with $\Bbb{C}[\alpha]$). So, write down $K = \Bbb{C}[\alpha]$.

Take a look at the infinite sequence $\{1, \alpha$, $\alpha^2$, $\alpha^3, ... \}$. Since $K$ has finite extension degree, it is of finite dimension as a vector space over $\Bbb{C}$ (this dimension is precisely the definition of the extension degree!). So an infinite sequence must be linearly dependent, which means (think about it...) there is some polynomial with complex coefficients which $\alpha$ is a root of. But by the assumption of phrasing 1, this polynomial is a product of linear polynomials, so there is some linear polynomial which $\alpha$ is a root of. This means $\alpha \in \Bbb C$ - contradiction.

Phrasing 2 $\implies$ phrasing 1: Let $f$ be an irreducible polynomial of degree more than $1$ with coefficients in $\Bbb C$. There is an explicit construction (which you should learn about in some algebra course) of a finite field extension of $\Bbb C$ formed by adding a root of this polynomial and closing under the field operations. This contradicts the assumption of phrasing 2.

Phrasing 2 $\implies$ phrasing 3: Let $D$ be a finite-degree field extension of $\Bbb R$. There are two possibilities:

  1. If $D$ contains a square root of $-1$, then call this $\alpha$, and $\Bbb R[\alpha]$ is isomorphic to $\Bbb C$ (therefore, algebraically closed by assumption of phrasing 2). But $D$ is then a finite-degree extension of $\Bbb R [\alpha]$, so it must be equal to $\Bbb R[\alpha]$, so isomorphic to $\Bbb C$ over $\Bbb R$.

  2. If $D$ does not contain a square root of $-1$, then add $i$ to form $D[i]$. Then $D[i]$ is a finite-degree extension of $\Bbb R[i] = \Bbb C$, so by assumption of phrasing 2, we must have $D[i] = \Bbb C$. Since $i \notin D$, $D = \Bbb R$.

Phrasing 3 $\implies$ phrasing 2: Let $D$ be a finite-degree field extension of $\Bbb C$. Then it's also a finite-degree field extension of $\Bbb R$. By the assumption of phrasing 3, $D = \Bbb R$ (impossible) or $D$ is isomorphic to $\Bbb C$. Since it already extends $\Bbb C$ the only option is $D = \Bbb C$.

  • $\begingroup$ Great explanation! $\endgroup$ – Antoine Apr 21 '13 at 20:07

Hint: Let $f(x)$ be an irreducible polynomial over $\Bbb R$, and let $\alpha$ be a root of $f(x)$. Think about the extension field $\Bbb R[\alpha]$. (The fundamental theorem of algebra states that this polynomial splits to a product of linear factors in $\Bbb C = \Bbb R[i]$.)

  • $\begingroup$ These two extensions must be isomorphic (and $g: \Bbb R[\alpha] \rightarrow \Bbb R[i]$ where $g(1)=1$ and $g(\alpha)=\alpha$ is isomorphism). Right? So D and $\Bbb R[\alpha]$ are both vector spaces over $\mathbb{R}$. There, I stop. Can't go further. $\endgroup$ – Antoine Apr 21 '13 at 13:23
  • $\begingroup$ My problem is that I am not able to make a conection between field extension and (dimension of ?) D. $\endgroup$ – Antoine Apr 21 '13 at 13:29
  • $\begingroup$ Basically, the field extension degree is defined to be the dimension as a vector space. I am writing all the details in a separate answer. $\endgroup$ – Yoni Rozenshein Apr 21 '13 at 18:43

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