Say I have a Ring with set $G$ and binary operations $+$ and $\times$. If $G$ has order 2 under addition (meaning $A+A=0,\forall A\in G$, where $0$ is the additive identity), how can I reproduce the binomial theorem for this ring? What I mean is, if for regular numbers we have: $$(x+y)^n = \sum_{i=0}^n={n \choose i}x^{n-i}y^i$$ Can we have a similar expression for $(A+B)^n,\forall A,B\in G$ that can cancel out the $A+A$ terms?

Much appreciated.

  • 2
    $\begingroup$ What you mean is characteristic, not "order". A hint, see what happens for $(a+b)^2$, and try to extrapolate from there. $\endgroup$ – fkraiem May 23 '14 at 18:53
  • $\begingroup$ I've heard be referred to as Order before, there's even a wikipedia article on it. I have actually tried extrapolating, and concluded that $(A+B)^2=A^2+B^2$, but I can't seem to find a pattern as the power goes up. I assume it might have something to do with when the binomial coefficient is even or odd, but then again I could be wrong. $\endgroup$ – Disousa May 23 '14 at 18:58
  • $\begingroup$ You could express $n$ as a power of $2$ times an odd number, but that's probably not going to give you something very pretty. However, $(a+b)^2 = a^2+b^2$ (and its analogues in any prime characteristic) is very important. $\endgroup$ – fkraiem May 23 '14 at 19:12
  • $\begingroup$ Given that $(A+B)^n=A^n+B^n$, it's quite easy to realize that $(A+B)^{(2^n)}=A ^{(2^n)} +B ^{(2^n)}$, but that's all I got really... $\endgroup$ – Disousa May 23 '14 at 19:25
  • $\begingroup$ If your ring is commutative (i.e. $AB=BA$), then you only need to reduce the binomial coefficients modulo two following the link in jdc's answer. What leads to the same end result is that if $n=\sum_ia_i2^i$ with $a_i=\{0,1\}$, then write $S$ for the set of indices such that $a_i=1$. And you have $$(A+B)^n=\prod_{i\in S}(A^{2^i}+B^{2^i}).$$ $\endgroup$ – Jyrki Lahtonen May 23 '14 at 20:58

${n \choose i}$ is going to be 1 in your ring if it is odd, and otherwise 0, so you just need to know when ${n \choose i}$ is even. For that, see http://en.wikipedia.org/wiki/Lucas%27_theorem.


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