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Is the quaternion ring an EAS Division ring?

An EAS Division ring is a ring $D$ such that each endomorphism of $D$ is surjective.

I know that $\mathbb{R}$ and $\mathbb{Q}$ are EAS Division rings.

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Which quaternions? Integer quaternions? Rational quaternions? Real quaternions? –  Arturo Magidin Sep 1 '11 at 17:04
    
@Arturo: Agree with you in that a little bit clarification is needed. But integer quaternions (neither Hurwitz or Lipschitz order) don't form a division ring, so that possibility is surely ruled out. –  Jyrki Lahtonen Sep 1 '11 at 18:55
    
integer or rational or real quaternions –  kaveh Sep 1 '11 at 20:00
    
I added the division-algebras tag. Who knows, we might attract an expert? :-) –  Jyrki Lahtonen Sep 1 '11 at 20:30

1 Answer 1

up vote 10 down vote accepted

The kernel of any endomorphism is an ideal. A division ring $D$ has only a single ideal $\{0\}$, so any endomorphism of a division ring is injective. Any ring endomorphism of $D$ is also an algebra homomorphism over the prime field $F$ of the center $K$. This is because $f(1)=1$, and the standard exercise then shows that $f(a)=a$ for all the elements of $F$. Consequently $f(ad)=f(a)f(d)=af(d)$ for all $a\in F$ and $d\in D$. As a homomorphism of rings $f$ respects the sums, so it is $F$-linear.

If your division ring has a finite dimension over $F$, then an injective linear mapping $f$ is necessarily also surjective. This follows from the rank-nullity theorem of linear algebra: if $\dim_FD=m$, and $\dim_F \ker f=0$, then by rank-nullity, $\dim_F f(D)= m-\dim_F \ker f=m$, and hence $f$ is surjective.

The general result proven by the above argument is:

Theorem. If $D$ is a division ring, $F$ is the prime field of its center $K=Z(D)$, and $\dim_F D<\infty$, then $D$ has the EAS-property.

This settles your question in the affirmative in the case of the rational quaternions $$\mathbf{H}_\mathbf{Q}=\{a+bi+cj+dk\mid a,b,c,d\in\mathbf{Q}\},$$ because in that case $K=F=\mathbf{Q}$ and $\dim_{\mathbf{Q}}\mathbf{H}_\mathbf{Q}=4<\infty.$

The division ring of Hamiltonian (real) quaternions $$\mathbf{H}=\{a+bi+cj+dk\mid a,b,c,d\in\mathbf{R}\},$$ also has the EAS-property. But that fact does not follow from the above theorem, because this time $K=\mathbf{R}$, so $F=\mathbf{Q}$ and $\dim_{\mathbf{Q}}\mathbf{H}=\infty.$

My argument depends on the property of $\mathbf{R}$ that it has no non-trivial endomorphisms (a quick sketch of this is at the end). It also uses the following two properties of the quaternions (either known, or an exercise for you):

  1. If $q$ is any quaternion, then the element of the form $a+bq, a,b\in\mathbf{R}$ commute with $q$.
  2. If $q\notin\mathbf{R}$, then all the quaternions commuting with $q$ are of this form, i.e. $C_{\mathbf{H}}(q)=\mathbf{R}[q]$.

Let $f:\mathbf{H}\rightarrow\mathbf{H}$ be a homomorphism of rings. We already know that $f$ must be injective. We also know that $f$ is linear over $\mathbf{Q}$. If we manage to show that $f$ must be linear over $\mathbf{R}$, then the argument leading to the above theorem works, and we have our result. This is equivalent to showing that $f(a)=a$ for all $a\in\mathbf{R}$. We need a couple of intermediate steps to get there.

I first claim that the vector space (over $\mathbf{R}$) spanned by $f(\mathbf{H})$ has dimension at least three. Assume contrariwise that $f(\mathbf{H})\subseteq M$, where $\dim_{\mathbf{R}}M=2$. We know that $1=f(1)\in M$, so we can assume that $M$ has a basis $\{1,q\}$. But by fact #1 this means that all the elements of $M$ commute with each other. This is a contradiction. For if, e.g. $f(i)$ and $f(j)$ commute with each other, then $$ f(2k)=f(ij-ji)=f(i)f(j)-f(j)f(i)=0, $$ but we knew that $\ker f=\{0\}$.

Next I claim that $f(\mathbf{R})\subseteq \mathbf{R}$. Let $x\in\mathbf{R}$. From the preceding step we can infer that there exist two quaternions $u,v$ such that $\{1=f(1),f(u),f(v)\}$ is linearly independent over $\mathbf{R}$. Because $[x,u]=[x,v]=0$, we see that $[f(x),f(u)]=[f(x),f(u)]=0$. Now fact 2 tells that $$f(x)\in (\mathbf{R}\oplus\mathbf{R}f(u))\cap(\mathbf{R}\oplus\mathbf{R}f(v))=\mathbf{R}$$ and the claim is proven.

We next claim (a well-known textbook result) that a ring endomorphism $f$ of $\mathbf{R}$ is necessarily the identity mapping. As always $f$ maps any rational number to itself. Because $f$ maps any square to a square, it maps all the positive real numbers to positive real numbers. Therefore $f$ must be an increasing function. Because it's restriction to the dense subset $\mathbf{Q}$ is the identity mapping, it must be continuous and thus the identity mapping.

Now we have (finally) shown that $f$ is linear over $\mathbf{R}$. Therefore the image $f(\mathbf{R})$ is a 4-dimensional subspace of $\mathbf{H}$ (because $f$ was injective). The claim follows.

I might be willing to wager (not the family fortune, but may be a beer or a coffee?) that a version of this argument will show that if the division ring $D$ is finite dimensional over its center $K$, and the field $K$ has the EAS property, then so does $D$. Has anybody seen such a result?

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I think that finite dimention quaternions over \mathbb{R} is nessesary in an injective linear mapping is necessarily also surjective –  kaveh Sep 1 '11 at 20:05
    
@kaveh and Jyrki: excuse me, I can't quite follow the comment thread. If some of the comments are no longer applicable, would you mind cleaning them up a bit (by clicking on the little 'x' mark next to the comment to delete them). Thanks. –  Willie Wong Sep 2 '11 at 22:23
    
I have some question about Theorem : Why necessary used claim that the vector space (over R ) spanned by f(H) has dimension at least three and claim that f(R)⊆/mathbb{R} , ?Why We need a couple of intermediate steps to get f is \mathbb{R} linear –  kaveh Sep 3 '11 at 16:45
    
@kaveh: I needed $dim\ge3$ to show that $f(R)\subseteq R$. If you see an easier way of proving this bit, go ahead and do it! In general it is possible for an endomorphism of a division ring not to be linear over the center. As an example consider the quaternion algebra with $a,b,c,d\in \mathbf{R}(t)$, where $t$ is transcendental. Then the $\mathbf{R}$-linear ring endomorphism given by $i\mapsto i$, $j\mapsto j$, $k\mapsto k$, $t\mapsto t^2$ is not $\mathbf{R}(t)$-linear, and is not surjective either, so that quaternion algebra does not have the EAS-property. –  Jyrki Lahtonen Sep 3 '11 at 17:41
    
i understand thanks , please introduce refrence of the theorem –  kaveh Sep 4 '11 at 11:39

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