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Below $^\ast$ denotes "nonzero elements of".

There is a problem in Jacobson's Basic Algebra 1, there is a problem to this effect: if $S$ is a subdivision ring of $\mathbb{H}$ such that $S^\ast$ is a normal subgroup of $\mathbb{H}^\ast$, then $S=\mathbb{H}$ or $S\subseteq\mathbb{R}$.

The solution that came to mind relies a bit on $\mathbb{H}$ having a norm into $\mathbb{R}$, and probably the behavior of $i,j$ and $k$.

I couldn't see a proof for general division rings, though. Does anyone know if it's true for division rings, or have a counterexample?

My gut tells me to lean toward the latter, but I have been bitterly disappointed by my gut feelings before...


Update: Two things have happened: Rankeya has given a valid answer to the written question, but I realize now I was too vague. Secondly, I looked up the correct exercise in Jacobson and found that the following exercise is precisely to show that it does hold for all division rings. Stupid gut feelings...

I'm accepting this answer and reposting the correct question.

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up vote 0 down vote accepted

Counterexample: Take the function fields $\mathbb{R}(x)$ and $\mathbb{R}(x,y)$. Since they are fields, in particular they are division rings. The former is a subfield of the latter. That $\mathbb{R}(x)^*$ is a normal subgroup of $\mathbb{R}(x,y)^*$ follows from commutativity of multiplication. But, $\mathbb{R}(x) \neq \mathbb{R}(x,y)$ and $\mathbb{R}(x)$ is not a subset of $\mathbb{R}$.

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In my answer I exploited the fact that presumably in your general statement, you still want $S \subset \mathbb{R}$. But, may be this is not the generalization you are thinking of. Certainly having the ring $\mathbb{R}$ in a general version (whatever that version is) seems specialized. –  Rankeya Nov 3 '12 at 6:19
    
Ah, sorry, you've addressed this question, but I've horribly misstated the question. –  rschwieb Nov 3 '12 at 13:17
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