# Noncommutative rings, finding $a$ and $b$ such that every term in the sum $(a+b)^n = a^n + a^{n-1}b + \ldots$ is distinct

This question was inspired by the binomial theorem for rings. For commutative rings, we have the identity

$$(a+b)^n = \sum_{k=0}^n {n \choose k}a^kb^{n-k}$$

which does not hold for non-commutative rings. However, we can still expand $(a+b)^n$ to get \begin{align*} (a+b)^2 &= a^2 + ab + ba + b^2 \\ (a+b)^3 &= a^3 + a^2b + aba + ab^2 + ba^2 + bab + b^2a + b^3 \\ \cdots \\ (a+b)^n &= aa\ldots a + aa \ldots ab + aa \ldots ba + aa \ldots bb + \ldots \end{align*}

This motivates the following question. For every $n$, is it possible to find a ring $R$ with elements $a$ and $b$ such that the $2^n$ terms in the expansion of $(a+b)^n$ are distinct?

There is also a stronger question: is it possible to find $a$ and $b$ such that for every $n$, this is true?

For example, taking $A = \begin{bmatrix}2 &1 \\0 &1\end{bmatrix}$ and $B = \begin{bmatrix}1 &2 \\0 & 1\end{bmatrix}$ from $M_2(\mathbb{Z})$ works for $n = 1, 2, 3$ and $4$ but fails for $n = 5$ because $ABBBA = BBAAB$.

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One way to do this is to take $R$ to be the group ring $\mathbb{Z}[F_2]$, where $F_2$ is the free group on generators $a$ and $b$. Then the fact that the summands are all different (for any $n$) is a direct consequence of the freeness of the group.

Of course, one doesn't have to explicitly embed a free group in a ring for a free group to be present. For example, $\begin{pmatrix} 1 & 2 \\ 0 & 1\end{pmatrix}$ and $\begin{pmatrix} 1 & 0 \\ 2 & 1\end{pmatrix}$ generate a free group of rank $2$, so you can take $R$ to be the ring of two-by-two integer matrices and these matrices as $a$ and $b$.

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Thanks! I'm not familiar with free groups but it seems to be the key idea here. – Mikko Korhonen Aug 23 '12 at 16:48

Take the polynomial ring over a field in two noncommuting variables: $\mathbb{F}\langle x,y\rangle$. I've put them in langle/rangle brackets to remind everyone that they do not commute with each other.

Then $x+y$ has your property.

Generally any free object on two generators is going to do something like that, because free objects don't satisfy any relations. For example, in commutative rings the relation $xy-yx=0$ allows you to "shuffle" different looking expressions until they are identical (like $aab$ and $aba$).

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Dear rschwieb: Nice answer! $+1$! $\mathbb F$ could be any nonzero commutative ring, for instance $\mathbb Z$. $\mathbb{F}\langle x,y\rangle$ is also called the tensor algebra of the free $\mathbb{F}$-module over $\{x,y\}$. – Pierre-Yves Gaillard Aug 23 '12 at 16:49
Nice, thanks. I was wondering about an example which would work for $n$, but not for $n+1$ and above. This is just a quick idea which might be really silly. Suppose we take your example, but take the exponents of $x$ and $y$ modulo $n$, ie. assume $x^n = y^n = 1$, does this give us an example which has the property up to $n$, but not for $>n$? Does it even make sense to assume $x^n = y^n = 1$ for variables? – Mikko Korhonen Aug 23 '12 at 17:01
@m.k. if I understand you correctly, then a quotient by the ideal $\langle x^n-1,y^n-1\rangle$ ought to do the trick? – Ben Millwood Aug 23 '12 at 17:26
@m.k. Modding out that ideal is certainly possible, but then you need to think of a way to check to see that no unintended relations were introduced by that process. It could be some weird relation came in that ruins the condition you want to preserve. – rschwieb Aug 23 '12 at 17:32
@m.k., a quick and easy way to make this "hold for $n$ but not for $n+1$ and above" is to mod out by all words in $x,y$ of length $n+1$. More precisely, this is the ideal $I = (x,y)^{n+1}$. Because it's a homogeneous ideal, it's easy to see that there are no "unintended consequences" to introducing the relations. And all terms in $(x+y)^m$ for $m > n$ are zero. This may help you sort out the details: en.wikipedia.org/wiki/Graded_algebra – Manny Reyes Aug 23 '12 at 19:08