I'm trying to understand a proof of the Frobenius Theorem on Division Algebras, but my knowledge of the relevant mathematics isn't really up to scratch. The proof I'm reading is this one.

I'm not sure what the notation $R[\alpha]$ could mean. I'm assuming that $R[\alpha]=\{a+b\alpha|a,b \in R\}$. Why should this be isomorphic to the complex numbers? Basically I want to know what a finite field extension is, and why it should be isomorphic to the complex numbers.

Thanks for any help!


In the notation there, $\bf{R}[\alpha]$ denotes the subalgebra generated by the real numbers $\bf{R}$ and the element $\alpha$.

Over fields in general, $F[\alpha]$ can look very different from the complex numbers. For example, for $\alpha=\sqrt[n]{2}$, we have ${\bf Q}[\alpha]=\{a_0+a_1\alpha+a_2\alpha^2+\dots+a_{n-1}\alpha^{n-1}\mid a,b,c\in \bf Q\}$. In the extension $\bf Q\subseteq \bf R$, the extension $\bf Q[\pi]$ is actually isomorphic to ${\bf Q}[x]$ (this is an example of an extension by a non-algebraic element.)

Since $\alpha$ is algebraic over $\bf{R}$ and not in $\bf R$, this ring $\bf{R}[\alpha]$ is a finite dimensional commutative domain over $\bf R$, so it's actually a proper field extenion. But the fundamental theorem of algebra tells us there's only one such field, and that is $\bf C$.

Basically I want to know what a finite field extension is, and why it should be isomorphic to the complex numbers.

A field extension is just a pair of two fields $E,F$ such that $E\supseteq F$. $E$ can be viewed as a vector space over $F$ in the natural way (just using multiplication in $E$.) This extension is called finite when $E$ is a finite dimensional vector space over $F$. In particular, extension by an algebraic element is a finite dimensional extension.

  • $\begingroup$ Sorry what what does "algebraic over R" mean? Thanks for the reply by the way $\endgroup$ – James Machin Jan 3 '14 at 13:38
  • $\begingroup$ @JamesMachin Given a field extension $F\subset E$, an element $\alpha\in E$ is said to be "algebraic over $F$" if $\alpha$ is the root of a polynomial with coefficients in $F$. So $\sqrt{2}$ is algebraic over $\bf Q$ since it is a root of $x^2-2$, which is a polynomial over $\bf Q$, but $\pi$ isn't algebraic over $\bf Q$. $\endgroup$ – rschwieb Jan 3 '14 at 13:53
  • $\begingroup$ Oh so basically you mean there will be a power of $\alpha$ say n when $\alpha ^n=r$ for some $r\in R$? Is that correct? Thanks again by the way. $\endgroup$ – James Machin Jan 3 '14 at 14:03
  • $\begingroup$ @JamesMachin I can see why my example suggests that, but no, that's not what I meant to convey. The right picture to carry is that there exists $n$ such that $\{1,\alpha,\alpha^2,\alpha^3,\dots,\alpha^{n-1}\}$ become linearly dependent. That means they have a nontrivial linear combination that equals zero... and that gives you your polynomial! $\endgroup$ – rschwieb Jan 3 '14 at 14:20
  • $\begingroup$ @JamesMachin If there exists no such $n$, then the set of powers of $\alpha$ looks like the set of powers of $x$, and then you can see why $F[\alpha]$ would be isomorphic to $F[x]$. $\endgroup$ – rschwieb Jan 3 '14 at 14:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.