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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...

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  • $\begingroup$ The intuition for "what happens in a ring" during addition in multiplication is virtually no different than it is for integers. Modulo differences like the one I used in my solution... $\endgroup$
    – rschwieb
    Apr 4, 2018 at 18:52
  • $\begingroup$ Take an easy example of a ring, say $M_2(\mathbb{F}_2)$, or just $\mathbb{F}_2$. $\endgroup$ Apr 4, 2018 at 18:53
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    $\begingroup$ Write out the expansion of $(x+1)^5$ as usual. Then apply the fact that $1+1=0$ What is the value of $2x=x+x?$ What is the value of $3x$? What is the value of $nx$ for an integer $n$? Perhaps what you're missing is the convention that in any ring, $nx$ is the sum of $n$ copies of $x$ for any element $x$. $\endgroup$
    – saulspatz
    Apr 4, 2018 at 18:54
  • $\begingroup$ The ring may not be commutative, but this is not a problem here. $\endgroup$ Apr 4, 2018 at 18:55
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    $\begingroup$ @DietrichBurde good point: we should just add that Restricting to the subring generated by $1$ and $x$, you’re back in a commutative ring. $\endgroup$
    – rschwieb
    Apr 4, 2018 at 19:01

4 Answers 4

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

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  • $\begingroup$ Yup. Thank you so much! $\endgroup$
    – C. Cristi
    Apr 4, 2018 at 18:59
  • $\begingroup$ You're very welcome. When you attempt these problems, you should make sure you know what rules you're allowed to use. I used the distributive property of rings for most of this argument. We're happy to provide you with references if you'd like as well. $\endgroup$ Apr 4, 2018 at 21:30
  • $\begingroup$ I'd like some references! And some information about how to find the divisors of 0 in a ring? That would be really helpful! $\endgroup$
    – C. Cristi
    Apr 5, 2018 at 8:29
  • $\begingroup$ If you're unfamiliar with ring theory, you should probably start with: abstract.ups.edu $\endgroup$ Apr 11, 2018 at 17:23
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Using the Freshman dream theorem you have right away that $(x+1)^4=x^4+1$.

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

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  • $\begingroup$ That's a nice way to look at it but I didn't know freshman dream theorem. Nice theorem tho. $\endgroup$
    – C. Cristi
    Apr 4, 2018 at 18:59
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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 $$x^5+x^4+x+1.$$

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

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