Questions about Proving R/I is not the Zero Ring

[Artusartos]

I have a question about the proof that I attached...

1) Since R/I is not the zero ring, we know that 1 \not= 0. What is the reason to say 1 + I \not= 0 + I instead of 1 \not= 0?2) Also, how do we compute something like (a+I)(b+I)? Isn't this correct (a+I)(b+I) = ab+aI+bI+I^2?

3) Finally, if we have something like R/I, how do we know if the elements in R/I are of the form a+I or aI? Or is it both (since it's a ring)?

Thank you in advance

 

[Number Nine]

Also, how do we compute something like (a+I)(b+I)? Isn't this correct (a+I)(b+I) = ab+aI+bI+I^2?

Strictly, yes; but you didn't finish. How can you simplify it further, using what you know about ideals and multiplication?

Finally, if we have something like R/I, how do we know if the elements in R/I are of the form a+I or aI? Or is it both (since it's a ring)?

R/I is the quotient of R by the normal subgroup I of (R,+) satisfying the additional constraint that RI = I. The cosets are of the form a + I.

What is the reason to say 1 + I \not= 0 + I

We know that 0 \in I by definition, so 0 + I = I. We also know that I is a proper ideal, so it can't contain 1 (what happens if an ideal contains a unit?). It follows that 1 + I \neq I.