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Given $R$ is a ring, and $b$ is some positive integer such that $x + x^{2b + 1} = x^{2b} + x^{10b + 1}$ for all $x \in R$, prove that $R$ is Boolean, i.e. $x =x^2$ for all $x$ in $R$.

I am not sure where to begin with the problem, and what avenue of approach I should take such that my methodology is inclusive to all $x$ in R. Would it be feasible to convert this problem into matrix form, and then prove $A = A^2$?

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You can try proving that R has characteristic 2. To do this you can try substituting x = 1, 2, 3, ... and see what happens. –  Qiaochu Yuan Dec 15 '10 at 1:42
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How would you "convert the problem" into "matrix form"? –  Arturo Magidin Dec 15 '10 at 1:44
    
@user4767: There is something amiss in your first sentence. Are you assuming that the equality holds for all $x\in R$, and are trying to deduce from that equality that $x=x^2$ for all $x$? –  Arturo Magidin Dec 15 '10 at 1:45
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TO show $2 = 0$ in $R$ evaluate the identity at $x = -1$. –  Bill Dubuque Dec 15 '10 at 4:13
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found a really great paper that could be of service: emis.de/journals/HOA/IJMMS/Volume17_4/749015.pdf –  user4767 Dec 15 '10 at 22:05
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1 Answer 1

This kind of problem usually succumbs to some sort of "clever choice" of $x$'s to plug in and manipulate. I don't have a full answer for you, but here's Bill's suggestion for showing that such a ring has characteristic $2$ (a necessary condition for being Boolean). Let $a\in R$. Evaluating the identity at $a$ and at $-a$, we have: \begin{align*} a + a^{2b+1} &= a^{2b} + a^{10b+1}\\ -a -a^{2b+1} &= a^{2b} - a^{10b+1} \end{align*} Adding both equalities we obtain $2a^{2b}=0$. Now, using the fact that $a=a^{2b}+a^{10b-1}-a^{2b+1}$, we have: $$2a = 2a^{2b} + 2a^{10b-1}-2a^{2b+1} = 0 + 2a^{2b}a^{8b-1} - 2a^{2b}a = 0,$$ so for all $a\in R$ we have $a+a=0$.

I don't have any clever manipulations to suggest for proving $a=a^2$ for all $a$, alas. I tried a few last night, and didn't get anywhere.

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Since this is only a very partial answer, and was made following Bill's hint, I've marked it "community wiki." A bit of an abuse of community wiki, I know, but still. –  Arturo Magidin Dec 15 '10 at 18:18
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