Prove that for every $x,y\in R$ exist $z\in R$ such that $\langle x,y\rangle=\langle z\rangle$

Question

Let $R$ a commutative ring (with $1$) such that $x^2=x$ for all $x\in R$.

Any idea with this? the exercise had an item a) that said to prove that $2x=0$ for all $x\in R$, but I can't see how to use it.

@RobArthan the ideal generated by $x$ and $y$ – Martín Vacas Vignolo May 08 '18 at 21:42 — Martín Vacas Vignolo, May 08 '18 at 21:42

score 3 · Answer 1 · answered May 08 '18 at 21:41

3

Specifically, you can show that $\langle x,y\rangle = \langle x + y + xy\rangle$.

(It's enough to show that $x$ and $y$ are multiples of $x + y + xy$.)

answered May 08 '18 at 21:41

Misha Lavrov

score 0 · Answer 2 · answered May 09 '18 at 01:23

0

From $-x = (-x)^2 = x^2 = x$ we got $2x = 0$. Now we can use this to show that

$x(x+y+xy) = x^2 + xy + x^2y = x + 2xy = x$ and similarly

$y(x+y+xy) = y$.

answered May 09 '18 at 01:23

Badam Baplan

2 Answers2