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If a ∈ M , we are done. So suppose that a ∈ M . Let Q be the smallest ideal containing a and M . ). Then M is properly contained in Q, so that by the definition of maximal ideals, Q = R. Hence 1 = m + ra for some m ∈ M and some r ∈ R. Then b = mb + rab ∈ M + M = M , so that b ∈ M , and we are done. 14 Every commutative ring with identity has a maximal ideal. ) Proof. We first Zornify: (1) Let S be the set of all proper ideals in the ring. (2) S is not empty as it contains the zero ideal. (3) We impose a partial order on S: if I, J ∈ S, we say I ≤ J if I ⊆ J .

Then every increasing chain of ideals I1 ⊆ I2 ⊆ · · · stabilizes, meaning that there exists an integer n such that In = In+1 = In+2 = · · ·. Proof. Let I = ∪j Ij . First verify that I is an ideal. Since R is a principal ideal domain, there exists a ∈ R such that I = (a). Then a ∈ In for some n, whence I = (a) ⊆ In ⊆ In+i ⊆ I. Z [X]. 7 Let f (X) = X 4 + X 3 + X 2 + X + 1 ∈ 2Z Z ideal. Verify that 2Z [X]/(f ) is a field with 16 elements. 8 Let R be a ring. Prove that the following are equivalent: i) Every ideal in R is finitely generated.

9 Let X 2 + X + 1 ∈ 2Z Z F = 2Z [X]/(X 2 + X + 1). Prove that F is a field with 4 elements. 10 Prove that X 3 + X + 1 has no roots in Prove that F is a field with 8 elements. Z 2Z . Let F = Z . 2Z Z 3 2Z [X]/(X + X Let + 1). 11 Prove that there is no field with 6 elements. 12 Let R and S be (non-zero) rings. Prove that R ⊕ S is not a domain. 13 The characteristic of a domain is 0 or prime. Proof. Suppose that the characteristic of a domain R is not 0. Then there exists a positive integer n such that n · 1 = 0.

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