# A Problem Book in Algebra by V. A. Krechmar

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By V. A. Krechmar

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Example text

2. Hence φ(a − a1 ) = 0S = φ(a) − φ(a1 ). Thus φ(a) = φ(a1 ) as required. ¯ is a ring homomorphism. 2. φ Suppose a + I, b + I are elements of R/I. Then ¯ ((a + I) + (b + I)) = φ = = = 33 ¯ ((a + b) + I) φ φ(a + b) φ(a) + φ(b) ¯ + I) + φ(b ¯ + I). φ(a So φ is additive. Also ¯ ((a + I)(b + I)) = φ = = = ¯ φ(ab + I) φ(ab) φ(a)φ(b) ¯ + I)φ(b ¯ + I). φ(a ¯ is multiplicative - φ ¯ is a ring homomorphism. So φ ¯ is injective. 3. φ ¯ Then φ(a ¯ + I) = 0S so φ(a) = 0S . This means Suppose a + I ∈ ker φ.

This means Suppose a + I ∈ ker φ. a ∈ ker φ, so a ∈ I. Then a + I = I = 0R + I, a + I is the zero element of R/I. ¯ contains only the zero element of R/I. Thus ker φ ¯ is surjective. 4. φ ¯ + I) and every Let s ∈ Imφ. Then s = φ(r) for some r ∈ R. Thus s = φ(r ¯ of some coset of I in R. element of Imφ is the image under φ ¯ : R/ ker φ −→ Imφ is a ring isomorphism, and Imφ is isomorphic to the Thus φ factor ring R/ ker φ. 4 Maximal and Prime Ideals The goal of this section is to characterize those ideals of commutative rings with identity which correspond to factor rings that are either integral domains or fields.

To see this first suppose φ is injective. Then ker φ = {0R }, otherwise if r ∈ ker φ for some r = 0 we would have φ(r) = φ(0R ), contrary to the injectivity of φ. On the other hand suppose ker φ = {0R }. Then if there exist elements r1 and r2 of R with φ(r1 ) = φ(r2 ) we must have φ(r1 − r2 ) = φ(r1 ) − φ(r2 ) = 0S . This means r1 − r2 ∈ ker φ, so r1 − r2 = 0R and φ is injective. The characterisation of injectivity in the above note can be very useful. If φ : R −→ S is an isomorphism, then S is an “exact copy” of R.