Let A be a ring. Rewrite ${(x+1)}^5$

Question

Let $(A,+,\times)$ be a ring for which 1 + 1 = 0 (0 and 1 are the neutral elements of it). Rewrite ${(x+1)}^5$ in terms of $x\in A$. I can't really work with rings and I don't know why... I do very well in compositions law and groups, etc.. but can't really understand what really happens in a ring...

Answer 1

A ring is an object which generalizes the integers (in that you may add and multiply objects). When $1+1=0$ has the property that $x+x=(1+1)x=0x=0.$ Using this fact, you can quickly show that $(x+1)^2=x^2+2x+1=x^2+1.$ You can then show that $$(x+1)^4=x^4+1$$ and you should use this to compute $(x+1)^4(x+1)=(x^4+1)(x+1)=x^5+x^4+x+1.$

Answer 2

Using the Freshman dream theorem you have right away that $(x+1)^4=x^4+1$.

Then $(x+1)^5=(x^4+1)(x+1)=x^5+x^4+x+1$

Answer 3

If $1+1=0$ then effectively all even numbers disappear, and all odd numbers are $1$ (since $1 + 2k = 1 + k(1+1) = 1$). You can therefore multiply step by step, throwing out multiples of $2$ as you go, as shown in the other answers. Alternatively, if you know Pascal's triangle, then you can look at the row $$1\quad 5\quad 10\quad 10\quad 5\quad 1,$$ corresponding to $x^5 + 5x^4 + 10x^3 + 10x^2 + 5x + 1$, then reduce modulo $2$ to get $$1\quad 1\quad 0\quad 0\quad 1\quad 1,$$ so the resulting polynomial is $$x^5+x^4+x+1.$$

Answer 4

Just use the binomial theorem and replace every even integer by $0$ and every odd integer by $1$: $$ (x+1)^5=x^5 + 5 x^4 + 10 x^3 + 10 x^2 + 5 x + 1 = x^5 + x^4 + x + 1 $$