$x^3 = x$ for all $x \in R$, where $R$ is a ring. Prove that $6x = 0$ for all $x \in R$.

Question

I am very confused about this problem. Because, clearly 6x = 0 only has one solution, 0. But I can make

$x^3 = x$

$x^3 -x=0$ (Adding additive inverse of x to both sides)

$x(x^2-1)=0$ (Distributive property of the ring)

$x(x+1)(x-1) = 0$ (Distributive property of the ring again)

which gives $x=0,1,-1$ as solutions. Not also I don't really know how to prove this problem, it confuses me how this is possible? Is it because the ring could not be an integral domain?

Note that the problem states for all $x\in R$. Note also that your last conclusion does in fact assume $R$ is an integral domain. — Ted Shifrin, Dec 17 '16 at 03:11
For instance, in $\mathbb{Z}/(6)$, every $x$ is a solution to $6x=0$. — Fimpellizzeri, Dec 17 '16 at 03:17
If $1\in R$, since $2^3=2$, that is, $6=0$, which implies that $6x=0$ for all $x\in R$. — m-agag2016, Dec 17 '16 at 03:26
The information you want was already available on several solutions here, but I see it seems you have several questions about this line of thought that might not be answered there. — rschwieb, Dec 17 '16 at 12:50

score 17 · Accepted Answer · answered Dec 17 '16 at 03:27

17

You have $$(x+x)^3=x+x$$ so $$8x^3=2x$$ and thus $$8x=2x$$ and $$6x=0$$

answered Dec 17 '16 at 03:27

Rene Schipperus

40,478

score 4 · Answer 2 · answered Dec 17 '16 at 03:24

4

Lemma: If $R$ is a ring and $a\cdot x=0$ for all $x\in R$, then $a=0$.

Proof: Take $x=1$. $\square$

Corollary: Let $R$ be a ring and $a \in R$. Then $a=0$ if and only if $a\cdot x=0$ for all $x\in R$.

By the corollary, it suffices to show that $6=0$ in $R$. To that end, notice that by hypothesis $2^3=2$ in $R$.

answered Dec 17 '16 at 03:24

Fimpellizzeri

23,321

It is not given that 6 ∈ R. So it will be wrong to do so directly. – Shreedhar Bhat Dec 17 '16 at 03:36
$6x$ doesn't make sense in that context. Adding to that, $6$ means only $1+1+1+1+1+1$, which is always some element of a ring $R$. – Fimpellizzeri Dec 17 '16 at 03:38
6x is defined as x+x+x+x+x+x . – Shreedhar Bhat Dec 17 '16 at 03:40
Yes, precisely, so $6$ always belongs to $R$. – Fimpellizzeri Dec 17 '16 at 03:40
1

@ShreedharBhat $6$ is defined as $1+1+1+1+1+1$. (I'm assuming our ring is unital.) – Viktor Vaughn Dec 17 '16 at 03:40
Yes it does: add $x$ to itself six times. Also, your proof doesn't work if the ring is not assumed to have a unity. I think the simpler upvoted proof above is essentially the same idea without overextending the assumptions. – Vik78 Dec 17 '16 at 03:41
@Vik78 ? Rings have an element $1$ by definition (multiplicative identitiy). I don't know how simpler algebraic manipulations are. I think this answer shows the essence of the question (that $6=0$ in $R$). – Fimpellizzeri Dec 17 '16 at 04:01
Certainly not by everyone's definition. Ever heard of the term "unital ring?" – Vik78 Dec 17 '16 at 04:02
@Fimpellizieri In my textbook, the ring doesn't need to have multiplicative identity, 1. – user3000482 Dec 17 '16 at 04:55

score 3 · Answer 3 · answered Dec 17 '16 at 03:25

3

$(x\pm1)^3=x^3\pm3x^2+3x\pm1$ so using $x^3=x$:

$x\pm1=x\pm3x^2+3x\pm1$ so

$\pm3x^2+3x=0$

Adding for both + and -:

$6x=0$

answered Dec 17 '16 at 03:25

Momo

16,300

score 2 · Answer 4 · answered Dec 17 '16 at 03:29

2

$(x+1)^3=x^3+3x^2+3x+1=x+1=x+3x^2+3x+1$implies that $3x^2+3x=0$. $3(x^2+x)=0$.

$3((x+1)^2+x+1)=3(x^2+2x+1+x+1)=3(x^2+x)+6x+6=0$.

$(1+1)^3=1+1$, $8=2$ implies $6=0$, so $6x+6=6x=0$.

answered Dec 17 '16 at 03:29

Tsemo Aristide

89,587

$x^3 = x$ for all $x \in R$, where $R$ is a ring. Prove that $6x = 0$ for all $x \in R$.

4 Answers4