What is the form of the binomial theorem in a general ring? I mean what's the expression for $(a+b)^n$ where $n$ is a positive integer?

Question

What is the form of the binomial theorem in a general ring? I mean what's the expression for $(a+b)^n$ where $n$ is a positive integer

Answer 1

The usual formula is $$\sum_{k=0}^n{\binom{n}{k}a^kb^{n-k}}$$ In a general commutative ring with identity it is the same, except you interpret $\binom{n}{k}$ as the element $1+1+1+\cdots+1$, where there are $\binom{n}{k}$ terms.

Answer 2

It's the usual formula $\sum \left(\begin{matrix} n\\ k\end{matrix}\right) a^k b^{n-k}$, if the ring is commutative. Terms of the form $\left(\begin{matrix} n\\ k\end{matrix}\right)$ are interpreted in a general ring $A$ via the unique homomorphism $\mathbb{Z}\to A$.

In a noncommutative ring the formula is harder to write down concisely: for $n=2$ we have $a^2+ab+ba+b^2$, for $n=3, a^3+a^2b+aba+ba^2+b^2a+bab+ab^2+b^3$, and indeed in general we now have $2^n$ terms instead of just $n+1$. So the "formula" is just that $(a+b)^n$ is the sum of all words in $a$ and $b$ of length $n$, which isn't much help to anybody-though you do need to worry about such expressions, for instance, in matrix calculus. In this case the fact that the $b$s can't all be moved to one side explains the difference of the Frechet derivative from the one you learn in ordinary calculus.