# Thread: Proving Rings (sorry for the Tex error)

1. ## Proving Rings (sorry for the Tex error)

$\displaystyle Prove that $$\left<\frac{R}{I}\,,+\,,\cdot\right>$$ is a ring.$

2. Originally Posted by DanielThrice
$\displaystyle Prove that $$\left<\frac{R}{I}\,,+\,,\cdot\right>$$ is a ring.$

Perhaps you mean to prove that if $\displaystyle R$ is a conmutative ring, $\displaystyle I\subset R$ ideal of $\displaystyle R$, then $\displaystyle R/I$ is a commutative ring with the standard operations $\displaystyle (x+I)+(y+I)=(x+y)+I$ and $\displaystyle (x+I)(y+I)=xy+I$ . If so, start proving that the sum is well defined that is, if $\displaystyle x+I=x'+I$ and $\displaystyle y+I=y'+I$ then, $\displaystyle (x+I)+(y+I)=(x'+I)+(y'+I)$ .

Fernando Revilla

3. Sorry, (I was half asleep last night) here is what we already know about the problem:

Let R = < R, +, *> (* means multiplication) be a ring, and let I < R be an additive subgroup of < R, + >.
Consider the set of cosets

R/I = {a + I: a is an element of R}

equipped with its own operations + and * defined by

(a + I) + (b + I) = (a + b) + I
(a + I) (b + I) = ab + I

Assume that the operations + and are well-defined <---> the
additive subgroup I satises the following conditions:

ab is an element of I for all a in R and b in I

ba is an element of I for all a in R and b in I