Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The following is exercise number 10 in Hungerford's Algebra, page 190:

Let $R$ be a ring with no zero divisors such that for all $r, s \in R$ there exist $a, b \in R$, not both zero, such that $ar + bs = 0$.

a) If $R = K \oplus L$ (module direct sum) then $K = 0$ or $L = 0$

b) If $R$ has an identity, then $R$ has the invariant dimension property.

I have proved part a), but I'm having difficulty with b). I realize that it suffices to show that $R^m \cong R^n$ implies $m = n$, but I can't seem to use part a) to conclude this.

share|improve this question
    
I'd try to show that the number of $a,b$ that works for a given $r,s$ isn't the same in both rings. –  xavierm02 Oct 26 '13 at 12:21
    
What do you mean, both rings? –  Pedro Oct 26 '13 at 13:37
    
Well $R^m$ and $R^n$. But it's just an idea I had looking at this problem. The only thing I know about rings is their definition. –  xavierm02 Oct 26 '13 at 15:17
    
I don't think the property keeps holding for $R^n$ as a ring, if n > 1. –  Pedro Oct 26 '13 at 17:22
    
Well from what I understand, since addition and multiplication are defined coordinate-wise, you get the existence of $a,b$ for each $r,s$. But you have more choice of $a,b$ than in $R$. –  xavierm02 Oct 26 '13 at 17:55

1 Answer 1

up vote 2 down vote accepted

I think you might have been bamboozled into thinking that you should use a) to prove b), when really you can do it independently. Let's first note that the original hypothesis could have been written as: "if $r,s$ are nonzero, then there exists nonzero $a,b$ such that $ar=bs$."

Let the ring $R$ be as above and have an identity. Without loss of generality, suppose that $m>n$. Choose bases, and express the isomorphism in terms of an $n\times m$ matrix $A$ and an $m\times n$ matrix $B$ over $R$, such that $AB=I_n$ and $BA=I_m$.

Now: the main thing about the hypothesis about $ar=bs$ that strikes me is that we can perform elementary row operations on matrices and turn entries to zero until we are in row-echelon form. We proceed to use this ability to arrive at a contradiction.

The row combining matrices will look like modified identity matrices, where one of the diagonal elements is replaced by some nonzero element of $R$, and then above or below that entry, there will be another nonzero element from $R$, and elsewhere the matrix is zero.

We claim that such an elementary matrix has kernel zero. To see why, notice we can do another row operation on the left of it to change it into a diagonal matrix with nonzero diagonal, and that surely has kernel zero, so the original row operation must also have kernel zero.

We will also need row-swapping matrices, but these are just permutation matrices with entries in $\{0,1\}$, and they all obviously have kernel zero.

So, it is possible to find a matrix $X$ (with kernel zero) such that $XB$ is in row-echelon form. Since $B$ has more rows than columns, $XB$ has rows of zeros at the bottom. But look: $(XB)A=X$. On one hand, the zeros sitting at the bottom of $XB$ should result in zero rows at the bottom of $X$ in this multiplication. On the other hand, $X$ has right kernel zero, so it can't have zero rows! This is a contradiction.

(Edit note: this solution is completely different from the first one I offered, which didn't work as anticipated.)


(More superfluous information for those who happen to be interested.)

A high level way of phrasing this problem (which is, no doubt, not Hungerford's goal at this point in his book) is that the stated condition is the left Ore condition on a domain (with identity). It turns out that this makes the domain a left uniform ring, and hence $R^n$ has left uniform dimension $n$. At this point, the original problem amounts to the task of determining that finite left uniform dimensions are well-defined.

Now, a) is actually pretty mundane: all domains with identity have that property. But left Ore domains have a much nicer property that kind of looks like that:

If $R$ is a left Ore domain, and $R^n\cong R^n\oplus N$ as left $R$ modules, then $N=\{0\}$.

This result obviously implies the invariant dimension property. Actually, this result holds for any ring $R$ with finite uniform left dimension.

share|improve this answer
    
Maybe I'm wrong, but Ore domains don't have (skew)fields of fractions and tensoring $R^m\cong R^n$ by this the problem transfers to vector spaces? –  user89712 Oct 29 '13 at 9:25
    
Dear @user : Yeah, if $Q$ is the right division ring of fractions for $R$, I think one can argue that $R\otimes_RQ_Q\cong Q_Q$, and that $_RQ$ is flat so that $Q^m_Q\cong R^m\otimes_RQ_Q\cong R^n\otimes_RQ_Q\cong Q^n_Q$. It's just a bit beyond what I wanted to say here :) –  rschwieb Oct 29 '13 at 12:52
    
Sorry. I had accepted but I noticed something odd. How can you find $r_i, s_i$ such that $r_i a_i x_i = s_i y_i$? I don't see how you can apply the ring condition since $x_i$ and $y_i$ are vectors, not elements of the ring. –  Pedro Oct 29 '13 at 22:04
    
@Pedro Urgh, yes, you're right to bring that up. I had thought that was possible by an elimination argument, but now I see that isn't an advisable route. I'll work on improving this :S –  rschwieb Oct 30 '13 at 14:18
    
@Pedro : How's it look now? –  rschwieb Oct 30 '13 at 17:19

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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