Ring with three idempotents satisfying $\alpha+ \beta+ \gamma=0$. Prove that $\alpha=\beta= \gamma$.

Question

Problem

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Let be $(R,+,\cdot)$ a ring.

Moreover, $(R,+,\cdot)$ has the property that if $$2022x=0$$ then $$x=0.$$

Let be $\alpha$, $\beta$, $\gamma$ three idempotent elements such that $$\alpha+ \beta+ \gamma=0.$$

Prove that $\alpha=\beta= \gamma$.

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My attempt:

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We know that $$x^2=x, \forall x\in \{ \alpha, \beta, \gamma \}$$

I then proved via noetherian induction that: $$x^n=x, \forall x\in \{ \alpha, \beta, \gamma \}, \forall n \text{ positive integer}$$

After that I observed that $$\alpha ^n+ \beta^n+ \gamma^n=\alpha+ \beta+ \gamma=0$$

I got stuck there. I didn't use the property that if $$2022x=0$$ then $$x=0$$

We have also got that $$\sum_{sym} \alpha \beta=0$$

I have not the faintest idea how to use it. Any help would be appreciated. How's it going?

I have the article only in a pdf file. I could not upload the pdf file, so I attached the first page of the article.

Can it be shown that $R$ is an Algebra by using if $2022x=0$ then $x=0$?

What is an algebra? May $R$ be an algebra? May we consider a homomorphism from the noncommutative polynomial ring $C[x_1, x_2, x_3]$ into $R$ defined by $\phi (x_1) = \alpha$ and the same for $x_2, x_3$?

Answer 1

This answer builds up on previous work by @shangq_tou and @winic92046.

First one observes that, for all $n\in \mathbb{N}$, we have that

$$\alpha^n + \beta^n + \gamma^n = 0$$

This can be used to show that all elementary symmetric polynomials:

$$a_1 =\alpha + \beta + \gamma,$$

$$a_2 = \alpha\beta+ \alpha \gamma + \beta\gamma,$$

$$a_3 =\alpha\beta\gamma$$

are equal to zero. The first one is zero by hypothesis. In order to prove that the other two are zero, we use the Newton's identities:

$$0 = \alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta+ \gamma)^2 - 2(\alpha\beta+ \alpha\gamma + \beta \gamma) = - 2(\alpha\beta+ \alpha\gamma + \beta \gamma) $$

This implies that

$$2(\alpha\beta+ \alpha\gamma + \beta \gamma) = 0$$

Multiplying this equation by $1011$, we get that

$$2022(\alpha\beta+ \alpha\gamma + \beta \gamma) = 0$$

Which implies that $\alpha\beta+ \alpha\gamma + \beta \gamma = 0$.

Another one of Newton's identities says that:

$$0 = \alpha^3 + \beta^3 + \gamma^3 = (\alpha+ \beta+ \gamma)^3 - 3(\alpha+\beta+ \gamma)(\alpha\beta + \beta\gamma + \alpha\gamma) + 3\alpha\beta\gamma$$

And thus,

$$3\alpha\beta\gamma = 0$$

Multiplying this by $2022:3 = 674$ gives:

$$2022\alpha\beta\gamma = 0$$

And it follows that $\alpha\beta\gamma = 0$.

Now, consider the following polynomial:

$$p(x) = (x-\alpha)(x-\beta)(x-\gamma) = x^3 - a_1x^2 +a_2x -a_3 = x^3$$

(We have already shown that $a_1, a_2, a_3$ are all equal to zero).

Since obviously $\alpha, \beta, \gamma$ are roots of this polynomial, we get that

$$\alpha^3 = \beta^3 = \gamma^3 = 0$$

And, because all of them are idempotent, this implies that $\alpha = \beta = \gamma = 0$.

The hypothesis about $2022$ is useful to show that, for all $r\in R$, $2r = 0$ or $3r = 0$ implies $r=0$ (because $2$ and $3$ divide $2022$).

Answer 2

Here is a proof that works for non-commutative rings. As in Compacto's proof, you don't really need $2002r=0 \Rightarrow r=0$, you only need $2r=0 \Rightarrow r=0$ and $3r=0 \Rightarrow r=0$.

I'll write $a,b,c$ for $\alpha,\beta,\gamma$. From $c=-(a+b)$ we deduce

$$c^2=(a+b)(a+b)=a^2+ab+ba+b^2=a+b+ab+ba\tag{1}$$,

and hence $-(a+b)=a+b+ab+ba$, or

$$ 2(a+b)+ab+ba=0 \tag{2} $$

Mutiplying by $a$ on the left in (2), we deduce $2(a^2+ab)+a^2b+aba=0$, or

$$ 2a+3ab+aba=0 \tag{3} $$

Mutiplying by $a$ on the right in (2), we deduce $2(a^2+ba)+aba+ba^2=0$, or

$$ 2a+3ba+aba=0 \tag{4} $$

Comparing (3) with (4), we see that $3(ba-ab)=0$, so $2022(ba-ab)=0$ and hence

$$ ba=ab \tag{5} $$

Injecting this last identity into (2), we see that $2(a+b+ab)=0$, so $2022(a+b+ab)=0$ and hence

$$ ab=-(a+b) \tag{6} $$

Mutiplying by $a$ on the left in (6), we deduce $a^2b=-(a^2+ab)$, or $ab=-(a+ab)$, or $2ab+a=0$. Combining this with (6), we obtain $a-2a-2b=0$, or

$$ a=-2b \tag {7} $$

It follows that $a^2=4b^2$, or $a=4b$. From (7) we deduce $4b=a=-2b$, so $6b=0$, so $2022b=0$ and hence $b=0$. This finishes the proof.