Let f : A −→ A′ be a ring homomorphism. If B is a subring of A, then f(B) = {f(x) : x ∈ B} is a subring of A′ .

1. Check if f(B) is non-empty.
B is a subring of A, so it contains at least the zero element. Since f is a homomorphism of rings:
f(0) = 0, hence f(B) is non-empty.

2. Verify the closure under addition.
Let b_1, b_2 \in B. Because B is a subring, b_1 + b_2 \in B. Using the property of f:
f(b_1 + b_2) = f(b_1) + f(b_2)
Since f(b_1) \in f(B) and f(b_2) \in f(B), we have:
f(b_1) + f(b_2) \in f(B)

3. Verify the closure under multiplication.
Let b_1, b_2 \in B. Since B is a subring, b_1 \cdot b_2 \in B. Using the property of f:
f(b_1 \cdot b_2) = f(b_1) \cdot f(b_2)
Since f(b_1) \in f(B) and f(b_2) \in f(B), we have:
f(b_1) \cdot f(b_2) \in f(B)

4. Verify the presence of the additive inverse.
Let b \in B. Since B is a subring, -b \in B. Using the property of f:
f(-b) = -f(b)
Since f(b) \in f(B), we have -f(b) \in f(B)

Hence, f(B) is a subring of A'

Answer:
f(B) \text{ is a subring of } A'