#30, 33, 34 posted in files.

31. a. Let f: R → S be a ring homomorphism, let A and I be a subring and ideal of R, and let B and J be a subring and ideal of S. Is the image f(A) of A in S necessarily a subring? Is the image f(I) necessarily an ideal? Likewise, is the preimage f−1(B) of B in R necessarily a subring? Is the preimage f−1(J) necessarily an ideal? For each question, either prove the assertion or give a counterexample.

b. Prove the First Isomorphism Theorem for Rings: If φ: R → S is a ring homomorphism, then R/ker (φ)≅φ(R).

Since a subring is inside a bigger ring, we can prove a ring is a subring by checking closure under addition, multiplication, and zero. To prove closure under addition we know that f(a + b) = f(a) + f(b). To prove closure under multiplication we know that f(ab) = f(a)f(b).

We also know that the image of a subring is a subring.

A —> f(A) f(A + I) = ((A) + f(I) R - [φ] = [φ D]

Since f(a +b) and f(ab) are elements of A, the image f(A) of A in S is a subring because we have closure under addition, multiplication, and the fact that it is a homomorphism.

Since a subring is inside a bigger ring, we can prove a ring is a subring by checking closure under addition, multiplication, and zero. To prove closure under addition we know that f^{-1}(a + b) = f^{-1}(a) + f^{-1}(b). To prove closure under multiplication we know that f^{-1}(ab) = f^{-1}(a)f^{-1}(b). Since f^{-1}(a +b) and f^{-1}(ab) are elements of B, the image f^{-1}(B) of B in R is a subring because we have closure under addition, multiplication, and the fact that it is a homomorphism.

Given a ring homomorphism, we want to establish an isomorphism.

32. Let p be a prime. It is a fact that for each k≥1 there is a unique (up to isomorphism) finite field of cardinality pk, denoted 𝔽pk. Prove that there is no surjective ring homomorphism from ℤ onto 𝔽pk for k>1. (Hint: Use the First Isomorphism Theorem for Rings and suppose toward contradiction there were a surjective ring homomorphism φ from ℤ onto 𝔽pk for some k>1. Think about ℤ modulo the kernel of φ and what ideals of various kinds look like in ℤ.)

Prove that there is no surjective ring homomorphism from ℤ onto 𝔽pk for k>1.

Proof by contradiction:

Suppose there is a surjective ring homomorphism from ℤ onto 𝔽pk for k>1.

Using the First Isomorphism Theorem for Rings, we know that If φ: ℤ → 𝔽pk is a ring homomorphism, then ℤ/kernel (φ) is isomorphic to 𝔽pk.

For k > 1, 𝔽pk is isomorphic to ℤ / pk ℤ if and only if pk ℤ is maximal and pk is prime.

R / I is a field if and only if I is maximal.

We know that every maximal ideal C ℤ is a prime ideal and 𝔽pk is a field if and only if pk ℤ is maximal.

However, pk ℤ is not maximal because pk is not prime.

pk is not prime because this quotient is not necessarily a field.

This logical inconsistency proves that φ must not be surjective. Thus, by contradiction, we know that there is no surjective ring homomorphism from ℤ onto 𝔽pk for k > 1.