If K is the kernel of a ring homomorphism, i.e. all the elements that map to 0, verify that K is an ideal. Conversely, if K is an ideal, the cosets of K in R form a factor ring, and the "coset" function implements the ring homomorphism.
To illustrate, let R be any ring with characteristic 0, and let n be any integer > 1. Let K be n times the elements of R, and verify that K is an ideal. The group homomorphism with kernel K reduces all coefficients mod n. If the original ring contained integer polynomials, for example, the quotient ring contains polynomials with coefficients ranging from 0 to n-1.
If n is composite, with prime factor p, a second homomorphism can be applied. The kernel contains all multiples of p, and the ring is reduced mod p.
A ring monomorphism is 1-1, with a kernel of 0. This is also called an embedding of one ring into another. A ring isomorphism is 1-1 and onto, essentially a relabeling of the ring elements. A ring automorphism maps a ring faithfully onto itself. All these definitions come from their counterparts in group theory.
If the characteristic of R is p, p prime, and R is commutative, the ring homomorphism f(x) = xp is valid, and is called the Frobenius homomorphism. (biography) Obviously f commutes with *. Use the binomial theorem, and the fact that all the intermediate binomial coefficients are divisible by p to show (x+y)p = xp+yp. This homomorphism is sometimes an automorphism, whence it is called the Frobenius automorphism.
The correspondence theorem applies to rings, as well as groups. Left ideals, right ideals, ideals, and subrings carry forward from R to S, and backward from S to R. The reverse map holds even if f maps R into S, rather than onto S. Restrict S to f(R) and find a smaller ideal or subring; then pull this back to its preimage in R.
If R is the integer polynomials, and f reduces coefficients mod n, the image of Z, a subring in R, is Zn, a subring in S. Also, the multiples of 7x2, an ideal in R, become multiples of 7x2 in S, an ideal in S. If n = 7 this image is 0; 0 is a perfectly good ideal. If n and 7 are coprime, the image contains all multiples of x2 in S, which has a preimage of all multiples of x2 in R. Thus the preimage of the image of an ideal can be larger than the original ideal.
Recall that the product of two ideals consists of all finite sums of xy, where x comes from the first ideal and y comes from the second. A ring homomorphism respects + and *, hence the image of the product of two ideals, or finitely many ideals, is the product of their images.
On the other hand, the product of the preimages of two ideals in S could be properly contained in the preimage of their product. Let F map Z onto Z7 by reducing the integers mod 7. The kernel of f is the multiples of 7. Multiply the kernel by itself and get the multiples of 49. This is the square of the preimage of 0 in Z7. However, the preimage of the square of 0 is all multiples of 7.