Algebra Homework 7

Certainly 0R ∈ A. Let a, b ∈ A. Then a ∈ An and b ∈ Am for some n,m ∈ N. Without loss of generality, assume n ≤ m. This means An ⊆ Am. So we have a, b ∈ Am. Since Am is a subring, it follows that −a ∈ Am, a + b ∈ Am, and ab ∈ Am. So also −a ∈ A, a + b ∈ A and ab ∈ A. This means A is closed under additive inverses, addition, and multiplication, so A is a subring of R. Let r ∈ R. Since An is an ideal, ra ∈ An. This implies ra ∈ A and so A is an ideal by definition. Every Ai is contained in I by the definition of I. Hence, A, which is the union of such Ai, is also contained in I. So A is an ideal contained in I which means A ∈ I. But A is an upper bound of the chain, and the chain was arbitrary, so every chain in I has an upper bound. By Zorn’s Lemma, I must have a maximal element, say M . We must show that M is a prime ideal in R. By 3.2.16, since R has identity, it will suffice to show that M is maximal in R. Let B be an ideal in R such that M ( B ⊆ R. Since M is maximal in I, B must not be contained in I so there exists 0R 6= b ∈ B such that b is not a zero-divisor. I did not have enough time to show that B = R. If I had, then I would know that M ⊂ I is maximal in R and thus also prime.