Solutions to Problems

2. Since A[b] is a subring of B, it is an integral domain. Thus if bz = 0 and b = 0, then z = 0. 3. Any linear transformation on a finite-dimensional vector space is injective iff it is surjective. Thus if b ∈ B and b = 0, there is an element c ∈ A[b] ⊆ B such that bc = 1. Therefore B is a field. 4. Since P is the preimage of Q under the inclusion map of A into B, P is a prime ideal. The map a + P → a + Q is a well-defined injection of A/P into B/Q, since P = Q ∩ A. Thus A/P can be viewed as a subring of B/Q. 5. If b + Q ∈ B/Q, then b satisfies an equation of the form