## Prime Ideals

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
Attached Thumbnails

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 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 $1 + I \neq I$.