Question

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′ .

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101 Answers

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.

Letb_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)

Sincef(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.

Letb_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)

Sincef(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.

Letb \in B . Since B is a subring, -b \in B . Using the property of f :

f(-b) = -f(b)

Sincef(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'

2. Verify the closure under addition.

Let

Since

3. Verify the closure under multiplication.

Let

Since

4. Verify the presence of the additive inverse.

Let

Since

Hence,

Answer:

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