Does canonical projection of commutative rings(R to R/I) always send prime ideals to prime ideals?

Question

R is a commutative ring with 1, I is an ideal of R. Consider the canonical projection f: R to R/I. Suppose p is a prime ideal of R then is f(p) always prime? I think if ab+I$\in$f(P) with ab$\in$ p, thus a or b$\in$ p, thus a+I or b+I $\in$ f(p).Is that a right proof? I think maybe there is a problem, because the prime ideals of R/I are in bijection with the prime ideals of R containing I.

Answer 1

There is indeed a problem and you seem to have the right intuition as to where it is. For example, let us look at the canonical projection $\pi\colon \mathbb{Z} \to \mathbb{Z}/2\mathbb{Z}$. Then $(3) \subset \mathbb{Z}$ is a prime ideal, but $\pi((3)) = \mathbb{Z}/2\mathbb{Z}$ is not a prime ideal. (Note that $(3)$ is a prime ideal not containing the ideal that was factored out.)

The example shows that your proof must be wrong. Indeed, the mistake is the very first step: from $ab + I \in f(p)$ it does not in general follow that $ab \in p$.