By Wilkins D.R.

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Now r ∈ I, since r = j − qn, j ∈ I and qn ∈ I. But 0 ≤ r < n, and n is by definition the smallest strictly positive integer belonging to I. We conclude therefore that r = 0, and thus j = qn. This shows that I = nZ, as required. 3 Quotient Rings and Homomorphisms Let R be a ring and let I be an ideal of R. If we regard R as an Abelian group with respect to the operation of addition, then the ideal I is a (normal) subgroup of R, and we can therefore form a corresponding quotient group R/I whose elements are the cosets of I in R.

We say that an ideal I of the ring R is finitely generated if there exists a finite subset of I which generates the ideal I. 5 Let R be a unital commutative ring, and let X be a subset of R. Then the ideal generated by X coincides with the set of all elements of R that can be expressed as a finite sum of the form r1 x1 + r2 x2 + · · · + rk xk , where x1 , x2 , . . , xk ∈ X and r1 , r2 , . . , rk ∈ R. Proof Let I be the subset of R consisting of all these finite sums. If J is any ideal of R which contains the set X then J must contain each of these finite sums, and thus I ⊂ J.

Suppose that I +g = I. Now the only factors of f are constant polynomials and constant multiples of f , since f is irreducible. But no constant multiple of f can divide g, since g ∈ I. It follows that the only common factors of f and g are constant polynomials. Thus f and g are coprime. 12 that there exist polynomials h, k ∈ K[x] such that f h + gk = 1. But then (I +k)(I +g) = I +1 in K[x]/I, since f h ∈ I. Thus I +k is the multiplicative inverse of I + g in K[x]/I. We deduce that every non-zero element of K[x]/I is invertible, and thus K[x]/I is a field, as required.

### Abstract Algebra (Course 311) by Wilkins D.R.

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