abstract algebra rings?
let f be a homomorphism of rings, f:A1-->A2. Assume that f is surjective. Assume that A2 has no divisors of 0. Prove that ker(f) is a prime ideal in A1.
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let f be a homomorphism of rings, f:A1-->A2. Assume that f is surjective. Assume that A2 has no divisors of 0. Prove that ker(f) is a prime ideal in A1.
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Recall, ker(f) is a prime ideal if A₁ \ ker(f) is closed under multiplication.
Suppose a,b ∈ A₁ are such that f(a),f(b) ≠ 0, i.e. a,b ∈ A₁ \ ker(f).
Since A₂ has no zero divisors we have
f(ab) = f(a)f(b) ≠ 0,
meaning ab ∈ A₁ \ ker(f).
→ ker(f) is prime.